美赛2013A题论文(终极布朗尼盘)

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2013年美赛MCM题目A评委点评中文翻译

介绍今年的焦点问题是如何实现质量和数量的平衡。

在质量方面，尽可能使热量均匀地分布。

目标是降低或避免矩形烤盘四个边角发生热量聚集的情况。

所以解决热量均匀分布这方面的问题，使用圆形烤盘是最佳的选择。

在数量方面，应该使烤盘充分的占据烤箱的空间。

所以我们的目的是使用尽可能多的烤盘来充分占据烤箱的空间，此时矩形烤盘是最佳选择。

对于这方面的问题的解决，就要考虑烤盘在烤箱水平截面上所占的比率。

在这个评论中，我们首先描述判断步骤，然后再讨论队伍对于三个问题的求解。

下一个话题就是论文的灵敏度和假设，紧随其后讨论确定一个给定方法的优势和劣势。

最后，我们简短的讨论一下参考和引用之间的区别。

过程第一轮的判别被称为“分流轮”。

这些初始轮的主要思想是确定论文应被给予更详细的考虑。

每篇论文应该至少阅读两次。

在阅读一篇论文的时候，评审的主要问题是论文是否包含所有必要的成分,使它成为一个候选人最详细的阅读。

在这些初始轮中，评审的时间是有限制的，所以我们要尽量让每一篇论文得到一个好的评判。

如果一篇论文解决了所有的问题，就会让评审觉得你的模型建立是合理的。

然后评审可能会认为你的论文是值得注意的。

有些论文在初轮评审中可能会得到不太理想的评论。

特别值得注意的是，一篇好的摘要应该要对问题进行简要概述，另外，论文的概述和方法，队员之间应该互相讨论，并且具体的结果应该在某种程度上被阐述或者表达出来。

在早期的几轮中，一些小细节能够有突出的表现，包括目录，它更便于评委看论文，同时在看论文的时候可能会有更高的期待。

问题求解也很重要。

最后，方法和结果要清晰简明的表达是至关重要的。

另外，在每个部分的开始，应该对那个部分进行一个概述。

在竞赛中，建模的过程是很重要的，同时也包括结论的表达。

如果结果没有确切和充分的表达，那么再好的模型和再大努力也是没有用的。

最后的回合最后一轮阅读的第一轮开始于评委会会议。

在这个会议中，评委将进行讨论，他们会分享他们各自认为的问题的关键方面。

2013美国大学生数学建模竞赛论文

summaryOur solution paper mainly deals with the following problems:·How to measure the distribution of heat across the outer edge of pans in differentshapes and maximize even distribution of heat for the pan·How to design the shape of pans in order to make the best of space in an oven·How to optimize a combination of the former two conditions.When building the mathematic models, we make some assumptions to get themto be more reasonable. One of the major assumptions is that heat is evenly distributedwithin the oven. We also introduce some new variables to help describe the problem.To solve all of the problems, we design three models. Based on the equation ofheat conduction, we simulate the distribution of heat across the outer edge with thehelp of some mathematical softwares. In addition, taking the same area of all the pansinto consideration, we analyze the rate of space utilization ratio instead of thinkingabout maximal number of pans contained in the oven. What’s more, we optimize acombination of conditions (1) and (2) to find out the best shape and build a function toshow the relation between the weightiness of both conditions and the width to lengthratio, and to illustrate how the results vary with different values of W/L and p.To test our models, we compare the results obtained by stimulation and our models, tofind that our models fit the truth well. Yet, there are still small errors. For instance, inModel One, the error is within 1.2% .In our models, we introduce the rate of satisfaction to show how even thedistribution of heat across the outer edge of a pan is clearly. And with the help ofmathematical softwares such as Matlab, we add many pictures into our models,making them more intuitively clear. But our models are not perfect and there are someshortcomings such as lacking specific analysis of the distribution of heat across theouter edge of a pan of irregular shapes. In spite of these, our models can mainlypredict the actual conditions, within reasonable range of error.For office use onlyT1 ________________T2 ________________T3 ________________T4 ________________ Team Control Number18674 Problem Chosen AFor office use only F1 ________________ F2 ________________ F3 ________________ F4 ________________2013 Mathematical Contest in Modeling (MCM) Summary Sheet(Attach a copy of this page to your solution paper.)Type a summary of your results on this page. Do not includethe name of your school, advisor, or team members on this page.The Ultimate Brownie PanAbstractWe introduce three models in the paper in order to find out the best shape for the Brownie Pan, which is beneficial to both heat conduction and space utility.The major assumption is that heat is evenly distributed within the oven. On the basis of this, we introduce three models to solve the problem.The first model deals with heat distribution. After simulative experiments and data processing, we achieve the connection between the outer shape of pans and heat distribution.The second model is mainly on the maximal number of pans contained in an oven. During the course, we use utility rate of space to describe the number. Finally, we find out the functional relation.Having combined both of the conditions, we find an equation relation. Through mathematical operation, we attain the final conclusion.IntroductionHeat usage has always been one of the most challenging issues in modern world. Not only does it has physic significance, but also it can influence each bit of our daily life. Likewise,space utilization, beyond any doubt, also contains its own strategic importance. We build three mathematic models based on underlying theory of thermal conduction and tip thermal effects.The first model describes the process and consequence of heat conduction, thus representing the temperature distribution. Given the condition that regular polygons gets overcooked at the corners, we introduced the concept of tip thermal effects into our prediction scheme. Besides, simulation technique is applied to both models for error correction to predict the final heat distribution.Assumption• Heat is distributed evenly in the oven.Obviously, an oven has its normal operating temperature, which is gradually reached actually. We neglect the distinction of temperature in the oven and the heating process, only to focus on the heat distribution of pans on the basis of their construction.Furthermore, this assumption guarantees the equivalency of the two racks.• Thermal conductivity is temperature-invariant.Thermal conductivity is a physical quantity, symbolizing the capacity of materials. Always, the thermal conductivity of metal material usually varies with different temperatures, in spite of tiny change in value. Simply, we suppose the value to be a constant.• Heat flux of boundaries keeps steady.Heat flux is among the important indexes of heat dispersion. In this transference, we give it a constant value.• Heat conduction dom inates the variation of temperature, while the effects ofheat radiation and heat convection can be neglected.Actually, the course of heat conduction, heat radiation and heat convectiondecide the variation of temperature collectively. Due to the tiny influence of other twofactors, we pay closer attention to heat conduction.• The area of ovens is a constant.I ntroduction of mathematic modelsModel 1: Heat conduction• Introduction of physical quantities:q: heat fluxλ: Thermal conductivityρ: densityc: specific heat capacityt: temperature τ: timeV q : inner heat sourceW q : thermal fluxn: the number of edges of the original polygonsM t : maximum temperaturem t : minimum temperatureΔt: change quantity of temperatureL: side length of regular polygon• Analysis:Firstly, we start with The Fourier Law:2(/)q gradt W m λ=- . (1) According to The Fourier Law, along the direction of heat conduction, positionsof a larger cross-sectional area are lower in temperature. Therefore, corners of panshave higher temperatures.Secondly, let’s analyze the course of heat conduction quantitatively.To achieve this, we need to figure out exact temperatures of each point across theouter edge of a pan and the variation law.Based on the two-dimension differential equation of heat conduction:()()V t t t c q x x y yρλλτ∂∂∂∂∂=++∂∂∂∂∂. (2) Under the assumption that heat distribution is time-independent, we get0t τ∂=∂. (3)And then the heat conduction equation (with no inner heat source)comes to:20t ∇=. (4)under the Neumann boundary condition: |W s q t n λ∂-=∂. (5)Then we get the heat conduction status of regular polygons and circles as follows:Fig 1In consideration of the actual circumstances that temperature is higher at cornersthan on edges, we simulate the temperature distribution in an oven and get resultsabove. Apparently, there is always higher temperature at corners than on edges.Comparatively speaking, temperature is quite more evenly distributed around circles.This can prove the validity of our model rudimentarily.From the figure above, we can get extreme values along edges, which we callM t and m t . Here, we introduce a new physical quantity k , describing the unevennessof heat distribution. For all the figures are the same in area, we suppose the area to be1. Obviously, we have22sin 2sin L n n n ππ= (6) Then we figure out the following results.n t M t m t ∆ L ksquare 4 214.6 203.3 11.3 1.0000 11.30pentagon 5 202.1 195.7 6.4 0.7624 8.395hexagon 6 195.7 191.3 4.4 0.6204 7.092heptagon 7 193.1 190.1 3.0 0.5246 5.719octagon 8 191.1 188.9 2.2 0.4551 4.834nonagon 9 188.9 187.1 1.8 0.4022 4.475decagon 10 189.0 187.4 1.6 0.3605 4.438Table 1It ’s obvious that there is negative correlation between the value of k and thenumber of edges of the original polygons. Therefore, we can use k to describe theunevenness of temperature distribution along the outer edge of a pan. That is to say, thesmaller k is, the more homogeneous the temperature distribution is.• Usability testing:We use regular hendecagon to test the availability of the model.Based on the existing figures, we get a fitting function to analyze the trend of thevalue of k. Again, we introduce a parameter to measure the value of k.Simply, we assume203v k =, (7) so that100v ≤. (8)n k v square 4 11.30 75.33pentagon 5 8.39 55.96hexagon 6 7.09 47.28heptagon 7 5.72 38.12octagon 8 4.83 32.23nonagon9 4.47 29.84 decagon 10 4.44 29.59Table 2Then, we get the functional image with two independent variables v and n.Fig 2According to the functional image above, we get the fitting function0.4631289.024.46n v e -=+.(9) When it comes to hendecagons, n=11. Then, v=26.85.As shown in the figure below, the heat conduction is within our easy access.Fig 3So, we can figure out the following result.vnActually,2026.523tvL∆==.n ∆t L k vhendecagons 11 187.1 185.8 1.3 0.3268 3.978 26.52Table 3Easily , the relative error is 1.24％.So, our model is quite well.• ConclusionHeat distribution varies with the shape of pans. To put it succinctly, heat is more evenly distributed along more edges of a single pan. That is to say, pans with more number of peripheries or more smooth peripheries are beneficial to even distribution of heat. And the difference in temperature contributes to overcooking. Through calculation, the value of k decreases with the increase of edges. With the help of the value of k, we can have a precise prediction of heat contribution.Model 2: The maximum number• Introduction of physical quantities:n: the number of edges of the original polygonsα: utility rate of space• Analysis:Due to the fact that the area of ovens and pans are constant, we can use the area occupied by pans to describe the number of pans. Further, the utility rate of space can be used to describe the number. In the following analysis, we will make use of the utility rate of space to pick out the best shape of pans. We begin with the best permutation devise of regular polygon. Having calculated each utility rate of space, we get the variation tendency.• Model Design:W e begin with the scheme which makes the best of space. Based on this knowledge, we get the following inlay scheme.Fig 4Fig 5According to the schemes, we get each utility rate of space which is showed below.n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 shape square pentagon hexagon heptagon octagon nonagon decagon hendecagon utility rate(％)100.00 85.41 100.00 84.22 82.84 80.11 84.25 86.21Table 4Using the ratio above, we get the variation tendency.Fig 6 nutility rate of space• I nstructions:·The interior angle degrees of triangles, squares, and regular hexagon can be divided by 360, so that they all can completely fill a plane. Here, we exclude them in the graph of function.·When n is no more than 9, there is obvious negative correlation between utility rate of space and the value of n. Otherwise, there is positive correlation.·The extremum value of utility rate of space is 90.69％,which is the value for circles.• Usability testing:We pick regular dodecagon for usability testing. Below is the inlay scheme.Fig 7The space utility for dodecagon is 89.88％, which is around the predicted value. So, we’ve got a rather ideal model.• Conclusion:n≥), the When the number of edges of the original polygons is more than 9(9 space utility is gradually increasing. Circles have the extreme value of the space utility. In other words, circles waste the least area. Besides, the rate of increase is in decrease. The situation of regular polygon with many sides tends to be that of circles. In a word, circles have the highest space utility.Model 3: Rounded rectangle• Introduction of physical quantities:A: the area of the rounded rectanglel: the length of the rounded rectangleα: space utilityβ: the width to length ratio• Analysis:Based on the combination of consideration on the highest space utility of quadrangle and the even heat distribution of circles, we invent a model using rounded rectangle device for pans. It can both optimize the cooking effect and minimize the waste of space.However, rounded rectangles are exactly not the same. Firstly, we give our rounded rectangle the same width to length ratio (W/L) as that of the oven, so that least area will be wasted. Secondly, the corner radius can not be neglected as well. It’ll give the distribution of heat across the outer edge a vital influence. In order to get the best pan in shape, we must balance how much the two of the conditions weigh in the scheme.• Model Design:To begin with, we investigate regular rounded rectangle.The area224r ar a A π++= (10) S imilarly , we suppose the value of A to be 1. Then we have a function between a and r :21(4)2a r r π=+--(11) Then, the space utility is()212a r α=+ (12) And, we obtain()2114rαπ=+- (13)N ext, we investigate the relation between k and r, referring to the method in the first model. Such are the simulative result.Fig 8Specific experimental results arer a ∆t L k 0.05 0.90 209.2 199.9 9.3 0.98 9.49 0.10 0.80 203.8 196.4 7.4 0.96 7.70 0.15 0.71 199.6 193.4 6.2 0.95 6.56 0.20 0.62 195.8 190.5 5.3 0.93 5.69 0.25 0.53 193.2 189.1 4.1 0.92 4.46Table 5According to the table above, we get the relation between k and r.Fig 9So, we get the function relation3.66511.190.1013r k e -=+. (14) After this, we continue with the connection between the width to length ratioW Lβ=and heat distribution. We get the following results.krFig 10From the condition of heat distribution, we get the relation between k and βFig 11And the function relation is4.248 2.463k β=+ (15)Now we have to combine the two patterns together:3.6654.248 2.463(11.190.1013)4.248 2.463r k e β-+=++ (16)Finally, we need to take the weightiness (p) into account,(,,)()(,)(1)f r p r p k r p βαβ=⋅+⋅- (17)To standard the assessment level, we take squares as criterion.()(,)(1)(,,)111.30r p k r p f r p αββ⋅⋅-=+ (18) Then, we get the final function3.6652(,,)(1)(0.37590.2180)(1.6670.0151)1(4)r p f r p p e rββπ-=+-⋅+⋅++- (19) So we get()()3.6652224(p 1)(2.259β 1.310)14r p f e r r ππ--∂=-+-+∂⎡⎤+-⎣⎦ (20) Let 0f r∂=∂,we can get the function (,)r p β. Easily,0r p∂<∂ and 0r β∂>∂ (21) So we can come to the conclusion that the value of r decreases with the increase of p. Similarly, the value of r increases with the increase of β.• Conclusion:Model 3 combines all of our former analysis, and gives the final result. According to the weightiness of either of the two conditions, we can confirm the final best shape for a pan.• References:[1] Xingming Qi. Matlab 7.0. Beijing: Posts & Telecom Press, 2009: 27-32[2] Jiancheng Chen, Xinsheng Pang. Statistical data analysis theory and method. Beijing: China's Forestry Press, 2006: 34-67[3] Zhengshen Fan. Mathematical modeling technology. Beijing: China Water Conservancy Press, 2003: 44-54Own It NowYahoo! Ladies and gentlemen, please just have a look at what a pan we have created-the Ultimate Brownie Pan.Can you imagine that just by means of this small invention, you can get away of annoying overcookedchocolate Brownie Cake? Pardon me, I don’t want to surprise you, but I must tell you , our potential customers, that we’ve made it! Believing that it’s nothing more than a common pan, some people may think that it’s not so difficult to create such a pan. To be honest, it’s not just a simple pan as usual, and it takes a lot of work. Now let me show you how great it is. Here we go!Believing that it’s nothing more than a common pan, some people may think that it’s not so difficult to create such a pan. To be honest, it’s not just a simple pan as usual, and it takes a lot of work. Now let me show you how great it is. Here we go!Maybe nobody will deny this: when baked in arectangular pan, cakes get easily overcooked at thecorners (and to a lesser extent at the edges).But neverwill this happen in a round pan. However, round pansare not the best in respects of saving finite space in anoven. How to solve this problem? This is the key pointthat our work focuses on.Up to now, as you know, there have been two factors determining the quality of apan -- the distribution of heat across the outer edge of and thespace occupied in an oven. Unfortunately, they cannot beachieved at the same time. Time calls for a perfect pan, andthen our Ultimate Brownie Pan comes into existence. TheUltimate Brownie Pan has an outstandingadvantage--optimizing a combination of the two conditions. As you can see, it’s so cute. And when you really begin to use it, you’ll find yourself really enjoy being with it. By using this kind of pan, you can use four pans in the meanwhile. That is to say you can bake more cakes at one time.So you can see that our Ultimate Brownie Pan will certainly be able to solve the two big problems disturbing so many people. And so it will! Feel good? So what are you waiting for? Own it now!。

2013年美赛真题题目

现在需要他们的解决方案文件太solutions@为Word或PDF附件的电子邮件提交电子副本（汇总表和解决方案）队（由学生或者指导教师）。

COMAP的提交截止日期为2013年2月4日美国东部时间下午8:00，必须在收到您的电子邮件。

主题行COMAP是你的控制示例：COMAP 11111点击这里下载PDF格式的完整的竞赛说明。

点击这里下载Microsoft Word中的格式汇总表的副本。

*请务必变更控制之前选择打印出来的页面的数量和问题。

团队可以自由选择之间MCM问题MCM问题A，B或ICM问题C.COMAP镜像站点：更多：/undergraduate/contests/mcm/MCM：数学建模竞赛ICM：交叉学科建模竞赛2013年赛题MCM问题问题A：终极布朗尼潘当在一个矩形的锅热烘烤时的4个角落中浓缩，并在拐角处（以及在较小程度上在边缘处）：产品会过头。

在一个圆形盘的热量被均匀地分布在整个外缘和在边缘处的产品不过头。

然而，因为大多数烤炉使用圆形平底锅的形状是矩形的是效率不高的相对于使用在烘箱中的空间。

开发一个模型来显示横跨平底锅平底锅不同形状 - 矩形之间的圆形和其他形状的外边缘的热量分布。

假设1。

的宽度与长度之比的W / L的形状是矩形的烘箱。

2。

- 每个盘必须具有的区域A的3。

最初，两个机架在烤箱，间隔均匀。

建立一个模型，可用于选择最佳的泛类型（形状）在下列情况下：1。

适合在烤箱的锅，可以最大限度地提高数（N）2。

最大限度地均匀分布热量（H），泛3。

优化的组合的条件（1）和（2）式中的权重p和（为1 - p）被分配的结果来说明如何随不同的值的W / L和p。

在除了MCM格式解决方案中，准备一到两页的广告片的新布朗尼美食杂志突出自己的设计和结果。

问题B：水，水，无处不在新鲜的白开水是在世界大部分地区的发展限制约束。

建立一个数学模型，为确定有效的，可行的和具有成本效益的水资源战略于2013年，以满足预计的用水需求，从下面的列表]中选择一个国家，到2025年，确定最佳的水战略。

2013美国数学建模A题优秀论文

终极布朗尼烤烤盘一、摘要根据题意，我们把把要解决的分成三个问题；第一个就是建立一个模型来表示整个烤盘的外边缘热量的分布。

第二个就是优化组合题目中条件1和条件2，使得权重p和(1- p)能够描述随着W/L和p值的改变，最佳的烤烤盘形状和热量分布情况是如何改变的第三个问题就是为布朗尼美食家杂志准备一到两页的宣传广告，需要突出设计和结果。

对于第一个问题，我们结合傅里叶定律构建了二维热传导模型；然后通过模型中的S来限定范围得到六种不同形状烤盘对应的热传导偏微分方程。

然后对模型赋值和第二类边界条件（Neumann边界条件）下，应用comsol得出六种烤盘稳定热量分布图像和烤盘外边缘热量分布图像。

通过输出的图像，我们得出结论：矩形四角处温度较高，圆形外边缘热量分布比较均匀；随着烤盘边数的增加，烤盘外边缘热量分布愈加均匀，但在角处温度仍然会高一些对于问题二对于问题三关键词：二、问题重述当用一个长方形的平底烤盘（盘）烘烤时，热量被集中在4个角，在角落处，食物可能被烤焦了，而边缘处烤的不够熟。

在一个圆形的平底烤盘（盘）热量被均匀地分布在整个外边缘，在边缘处食物不会被烤焦。

但是，大多数的烤箱的形状是矩形的，采用了圆形的烤盘（盘）相对于烤箱的使用空间而言效率不高。

为所有形状的烤盘（盘）----包括从矩形到圆形以及中间的形状，建立一个模型来表示整个烤盘（盘）的外边缘热量的分布。

假设：1. 形状是矩形的烤箱宽长比为W/L；2. 每个烤烤盘（盘）的面积为A；3. 每个烤箱最初只有两个均匀放置的烤架。

根据以下条件，建立一个能使用的最佳类型或形状的烤烤盘（盘）：1.放入烤箱里的烤烤盘（盘）数量的最大值为(N)；2.烤烤盘（盘）的平均分布热量最大值为(H)；3.优化组合条件1和条件2，使得权重p和(1- p)能够描述随着W/L和p值的改变，最佳的烤烤盘形状和热量分布情况是如何改变的。

除了完成规定的解决方案，为布朗尼美食家杂志准备一到两页的宣传广告，需要突出你的设计和结果。

美赛6种题型及通关详解

所谓6种题型，提示了部分题目的内容，但如果作为选题依据，作用非常有限。

如果是为了更好的选题，搞清楚MCM与ICM的区别，可能更有帮助。

选哪道题不是特别重要，重要的是应该“尽快”选题。

竞赛时间是固定的，选题的时间越长，做题的时间越少。

选题多花1小时，意味着建模和写论文的时间就少了1小时。

能获什么奖主要看实力，其次看运气。

准备越充分，胜算越大。

如果不想碰运气的话，早点动手准备吧。

六种题型怎么理解首先，MCM/ICM（2016年起）每年共有6道题，不是6种题，MCM是ABC三题，ICM是DEF三题。

对6道题目类型的描述，不是严格的划分，角度和依据都不相同。

continuous和discrete是指模型的类型，data insights是指问题数据的特征，operations research/network science和environmental science是指问题涉及到的学科，而environmental science和policy又是指问题本身的背景。

这不是按照同一标准对题目进行划分，之间有重叠。

最显然的，如果认为continuous和discrete是互补的，那么其他4道题目应该可以分别归入其中某一类。

其次，这些一两个词的描述过于笼统、宽泛，无法体现题目的具体特征，特别是A、B、F 题的描述，提供的信息非常少，说了几乎等于没说。

continuous、discrete把所有的模型全包括了。

policy范围也太广，人类主宰世界，方方面面都可能涉及政策问题。

而且F题也是2016年新增加的，只有2016年一年的题目（难民问题），暂时还看不出来什么规律。

而C题和D题的特征相对具体一些。

比如，针对2016年起MCM新增加的C题，COMAP （Consortium for Mathematics and Its Applications）专门发布了一份文档（中文简介）说明其特征。

概括起来，MCM的C题与数据有关，虽然称不上大数据，但压缩包也在100MB 以上，与MCM/ICM其他题目相比，数据量算是大的（实际上以往MCM/ICM的题目很少给数据），这就要求选这一题的参赛队要熟悉数据处理的基本方法，包括预处理、后处理等，并掌握相应的编程技能或是相关软件的使用方法。

2013年美国大学生数学建模大赛A题一等奖

最终的布朗尼蛋糕盘Team #23686 February 5, 2013摘要Summary/Abstract为了解决布朗尼蛋糕最佳烤盘形状的选择问题，本文首先建立了烤盘热量分布模型，解决了烤盘形态转变过程中所有烤盘形状热量分布的问题。

又建立了数量最优模型，解决了烤箱所能容纳最大烤盘数的问题。

然后建立了热量分布最优模型，解决了烤盘平均热量分布最大问题。

最后，我们建立了数量与热量最优模型，解决了选择最佳烤盘形状的问题。

模型一：为了解决烤盘形态转变过程中所有烤盘形状热量分布的问题，我们假设烤盘的任意一条边为半无限大平板，结合第三边界条件下非稳态导热公式，建立了不同形状烤盘的热量分布模型，模拟出不同形状烤盘热量分布图。

最后得到结论：在烤盘由多边形趋于圆的过程中，烤焦的程度会越来越小。

模型二：为了解决烤箱所能容纳最大烤盘数的问题，本文建立了随烤箱长宽比变化下的数量最优模型。

求解得到烤盘数目N 随着烤箱长宽比和烤盘边数n 变化的函数如下：AL W L W cont cont cont N 4n2nsin 1222⎪⎭⎫ ⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+⋅--=π模型三：本文定义平均热量分布H 为未超过某一温度时的非烤焦区域占烤盘边缘总区域的百分比。

为了解决烤盘平均热量分布最大问题，本文建立了热量分布最优模型，求解得到平均热量分布随着烤箱长宽比和形状变化的函数如下：n sin n cos -n 2nsin 22ntan1H ππδπδπ⎪⎪⎪⎪⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛⋅-=A结论是：当烤箱长宽比为定值时，正方形烤盘在烤箱中被容纳的最多，圆形烤盘的平均热量分布最大。

当烤盘边数为定值时，在长宽比为1:1的烤箱中被容纳的烤盘数量最多，平均热量分布H 最大。

模型四：通过对函数⎪⎭⎫ ⎝⎛n ,L W N 和函数⎪⎭⎫⎝⎛n ,L W H 作无量纲化处理，结合各自的权重p 和()p -1，本文建立了数量和热量混合最优模型，得到烤盘边数n 随p值和LW的函数。

2013年美赛a题h奖论文_大学论文

SummaryWe build two basic models for the two problems respectively: one is to show the distribution of heat across the outer edge of the pan for different shapes, rectangular, circular and the transition shape; another is to select the best shape for the pan under the condition of the optimization of combinations of maximal number of pans in the oven and the maximal even heat distribution of the heat for the pan.We first use finite-difference method to analyze the heat conduct and radiation problem and derive the heat distribution of the rectangular and the circular. In terms of our isothermal curve of the rectangular pan, we analyze the heat distribution of rounded rectangle thoroughly, using finite-element method. We then use nonlinear integer programming method to solve the maximal number of pans in the oven. In the even heat distribution, we define a function to show the degree of the even heat distribution. We use polynomial fitting with multiple variables to solve the objective function For the last problem, combining the results above, we analyze how results vary with the different values of width to length ratio W/L and the weight factor p. At last, we validate that our method is correct and robust by comparing and analyzing its sensitivity and strengths /weaknesses.Based on the work above, we ultimately put forward that the rounded rectangular shape is perfect considering optimal number of the pans and even heat distribution. And an advertisement is presented for the Brownie Gourmet Magazine.Contents1 Introduction (3)1.1Brownie pan (3)1.2Background (3)1.3Problem Description (3)2. Model for heat distribution (3)2.1 Problem analysis (3)2.2 Assumptions (4)2.3 Definitions (4)2.4 The model (4)3 Results of heat distribution (7)3.1 Basic results (7)3.2 Analysis (9)3.3 Analysis of the transition shape—rounded rectangular (9)4 Model to select the best shape (11)4.1 Assumptions (11)4.2 Definitions (11)4.3 The model (12)5 Comparision and Degree of fitting (19)6 Sensitivity (20)7 Strengths/weaknesses (21)8 Conclusions (21)9 Advertisement for new Brownie Magazine (23)10 References (24)1 Introduction1.1Brownie panThe Brownie Pan is used to make Brownies which are a kind of popular cakes in America. It usually has many lattices in it and is made of metal or other materials to conduct heat well. It is trivially 9×9 inch or 9×13 inch in size. One example of the concrete shape of Brownie pan is shown in Figure 1Figure 1 the shape of Brownie Pan (source: Google Image)1.2BackgroundBrownies are delicious but the Brownie Pan has a fetal drawback. When baking in a rectangular pan, the food can easily get overcooked in the 4 corners, which is very annoying for the greedy gourmets. In a round pan, the heat is evenly distributed over the entire outer edge but is not efficient with respect to using in the space in an oven, which most cake bakers would not like to see. So our goal is to address this problem.1.3Problem DescriptionFirstly, we are asked to develop a model to show the distribution of heat across the outer edge of a pan for different shapes, from rectangular to circular including the transition shapes; then we will build another model to select the best shape of the pan following the condition of the optimization of combinations of maximal number of pans in the oven and maximal even distribution of heat for the pan.2. Model for heat distribution2.1 Problem analysisHere we use a finite difference model to illustrate the distribution of heat, and it has been extensively used in modeling for its characteristic ability to handle irregular geometries and boundary conditions, spatial and temporal properties variations1. In literature 1, samples with a rectangular geometric form are difficult to heat uniformly,particularly at the corners and edges. They think microwave radiation in the oven can be crudely thought of as impinging on the sample from all, which we generally acknowledge. But they emphasize the rotation.Generally, when baking in the oven, the cakes absorb heat by three ways: thermal radiation of the pipes in the oven, heat conduction of the pan, and air convection in the oven. Considering that the influence of convection is small, we assume it negligible. So we only take thermal radiation and conduction into account. The heat is transferred from the outside to the inside while water in the cake is on the contrary. The temperature outside increase more rapidly than that inside. And the contact area between the pan and the outside cake is larger than that between the pan and the inside cakes, which illustrate why cakes in the corner get overcooked easily.2.2 Assumptions● We take the pan and cakes as black body, so the absorption of heat in eacharea unit and time unit is the same, which drastically simplifies ourcalculation.● We assume the air convection negligible, considering its complexity and thesmall influence on the temperature increase .● We neglect the evaporation of water inside the cake, which may impede theincrease of temperature of cakes.● We ignore the thickness of cakes and the pan, so the model we build istwo-dimensional.2.3 DefinitionsΦ: heat flows into the nodeQ: the heat taken in by cakes or pans from the heat pipesc E ∆: energy increase of each cake unitp E ∆: energy increase of the pan unit,i m n t : temperature at moment i and point (m,n)C 1: the specific heat capacity of the cakeC 2: the specific heat capacity of the panipan t : temperature of the pan at moment iT 1: temperature in the oven, which we assume is a constant2.4 The modelHere we use finite-difference method to derive the relationship of temperatures at time i-1 and time i at different place and the relationship of temperatures between the pan and the cake.First we divide a cake into small units, which can be expressed by a metric. In the following section, we will discuss the cake unit in different places of the pan.Step 1；temperatures of cakes interior(m,n+1)x△Figure 2 heat flow According to energy conversation principle, we can get 0up down left right c Q E Φ+Φ+Φ+Φ+-∆=(2.4.1) Considering Fourier Law and △x=△y, we get1,,1,,1,,1,,,1,,1,,1,,1,()()()()i i m n m n i i left m n m n i i m n m n i i right m n m n i i m n m n i i up m n m n i i m n m ni i down m n m n t t y t t xt t y t t x t t x t t y t t x t t y λλλλλλλλ--++++---Φ=-∆=-∆-Φ=∆=-∆-Φ=∆=-∆-Φ=∆=-∆ (2.4.2)According to Stefan-Boltzman Law,441,[()]i m n Q Ac T t σ=- (2.4.3)Where A is the area contacting, c is the heat conductance.σis the Stefan-Boltzmann constant, and equals 5.73×108 Jm -2s -1k -4.21,,()i i c m n m n E cm t t -∆=-(2.4.4) Substituting (2.4.2)-(2.4.4) into (2.4.1), we get441,1,,1,1,1,1,,4[()]()0i i i i i i m n m n m n m n m n m n i i m n m n Ac t t t t t T t cm t t σλλ-++--+++-+---=This equation demonstrates the relationship of temperature at moment i and moment i-1 as well as the relationship of temperature at (m,n) and its surrounding points.Step 2: temperature of the cake outer and the pan● For the 4 cornerscakeFigure 3 the relative position of the cake and the pan in the first cornerBecause the contacting area is two times, we get4411,1,122[()]2()i i i i m n pan pan A c T t t c M t t σλ---=-● For every edgecakeFigure 4 the relative position of the cake and the pan at the edgeSimilarly, we derive4411,,2[()]()i i i i m n m n pan pan Ac T t t c M t t σλ---=-Now that we have derived the express of temperatures of cakes both temporally and spatially, we can use iteration to get the curve of temperature with the variables, time and location.3 Results of heat distribution3.1 Basic resultsRectangularPreliminarily, we focus on one corner only. After running the programme, we obtain the following figure.Figure 5 heat distribution at one cornerFigure 5 demonstrates the temperature at the corner is higher than its surrounding points, that’s why food at corners get overcooked easily.Then we iterate globally, and get Figure 6.Figure 6 heat distribution in the rectangular panFigure 6 can intuitively illustrates the temperature at corners is the highest, and temperature on the edge is less higher than that at corners, but is much higher than that at interior points, which successfully explains the problem “products get overcooked at the corners but to a lesser extent on the edge”.After drawing the heat distribution in two dimensions, we sample some points from the inside to the outside in a rectangular and obtain the relationship between temperature and iteration times, which is shown in Figure 7Figure 7From Figure 7, the temperatures go up with time going and then keep nearly parallel to the x-axis. On the other hand, temperature at the center ascends the slowest, then edge and corner, which means given cooking time, food at the center of the pan is cooked just well while food at the corner of the pan has already get overcooked, but a lesser extent to the edge.RoundWe use our model to analyze the heat distribution in a round, just adapting the rectangular units into small annuluses, by running our programme, we get the following figure.Figure 8 the heat distribution in the circle panFigure 8 shows heat distribution in circle area is even, the products at the edge are cooked to the same extent approximately.3.2 AnalysisFinally, we draw the isothermal curve of the pan.●RectangularFigure 9 the isothermal curve of the rectangular panFigure 9 demenstrates the isothermal lines are almost concentric circles in the center of the pan and become rounded rectangles outer, which provides theory support for following analysis..●CircularFigure 10 the isothermal curve of circularThe isothermal curves of the circular are series of concentric circles, demonstrating that the heat is even distributed.3.3 Analysis of the transition shape—rounded rectangularFrom the above analysis, we find that the isothermal curve are nearly rounded rectangulars in the rectangular pan, so we perspective the transition shape between rectangular and circular is rounded rectangular, considering the efficiency of using space of the oven and the even heat distribution. In the following section, we will analyze the heat distribution in rounded rectangular pan using finite element approach.During the cooking process, the temperature goes up gradually. But at a certain moment, the temperature can be assumed a constant. So the boundary condition yields Dirichlet boundary condition. And the differential equation is:22220T T x y∂∂+=∂∂ Where T is the temperature, and x, y is the abscissa and the ordinate.And the boundary condition is T=constant.After running the programme, we get the heat distribution in a rounded rectangular pan, the results is in the following.Figure 11 the heat distribution in a rounded rectangular pan From 11, we can see the temperature of the edge and the corner is almost the same, so the food won ’t get overcooked at corners. We can assume the heat in a rounded rectangular is distributed uniformly. We then draw the isothermal curve of the rounded rectangular pan.Figure 12 the isothermal curve of the rounded rectangular pan To show the heat distribution more intuitively, we also draw the vertical view of the heat distribution in a rounded rectangular pan.Figure 13 the vertical view of the heat distribution on a rounded rectangular platform4 Model to select the best shape4.1 AssumptionsBesides the assumptions given, we also make several other necessary assumptions.●The area of the even equals the area of the pan with small lattices in it.●There is no space between lattices or small pans on the pan.4.2 DefinitionsS: the area of the ovenk : the width of the external rectangular of the rounded rectangularh :the length of the external rectangular of the rounded rectangulara1: the ratio of the width and length of the external rectangular of the rounded rectangular, equals k/ha2: the ratio of the width and length of the external rectangular of the rounded rectangular, equals W/Ln: the amounts of the rounded rectangular in each rowm: the amounts of the rounded rectangular in each columnr: the radius of the rounded rectangularIn order to illustrate more clearly, we draw the following sketch.Figure 144.3 The modelProblem ⅠWe use nonlinear integer programming to solve the problem.Figure 15 the configuration of the rounded rectangular and the pan:max objective function N n m =⋅222:,004subject to n k m h k h A r r A r r Sππ⋅≤⋅≤==≥≥≤-+≤Where n, m are integers.Both a1 and a2 are variables, we can study the relationship of a1 and N at a given a2. Here we set a2=0.8.Considering the area of the oven S and the area of the small pan A are unknown, we collect some data online, which is shown in the following table.Table 1(source: / )Then we calculate the average of S and A , respectively 169.27 inch 2 and 16.25 inch 2. After running the programme, we get the following results.Figure 16 the relationship of the maximal N and the radius of the rounded rectangular r From the above figure, we know as r increases, the optimal number of pans decreases. For the data collected can not represent the whole features, the relationship is not obvious .Then we change the area of the oven and the area of each pan, we draw another figure.Figure 17This figure intuitively shows the relationship of r and N.Appearently, the ratio of the width to length of the rounded rectangular has a big influence on the optimization of N. In the following section, we will study this aspect.Figure 18 the relationship of N and a1 Figure 18 illustrates only when the ratio of the width to the length of the rounded rectangular a1 equals the ratio of the width and length of the oven a2, can N be optimized.Finally, we take both a1 and a2 as variables and study the relationship of N and a1, a2. The result is as follows.Figure 19 the relationship of N and a1, a2 Problem ⅡTo solve this problem, we first define a function u(r) to show the degree of the even heat distribution for different shapes. ()s u r AWhere s is the area surrounded by the closed isothermal curve most external of the pan.Now we think the rationality of the function. The temperature of the same isothermal curve is equivalent. We assume the temperature of the closed isothermal curve most external of the pan is t 0 , for the unclosed isothermal, the temperature is higher than t 0, and the temperature inside is lower than t 0. So we count the number of the pixel points inside the closed isothermal curve most external of the pan num 1 and the number of the whole pixel points num 2. Consequently, u(r)=num 1/num 2. To illustrate more clearly, we draw the following figure.The following table shows the relationship of u and r. And we set a=1Figure 21 the scatter diagram of u and rFirst, we consider u and r is linear, and by data fitting we deriveu r r=+⨯()0.85110.021Then, we consider u and r is second -order relationship, and we derive anothercurve.Figure 22 another relationship of u and r2 =+⨯-⨯u r r r()0.85110.02880.001According to the points we count, we can get the following figure, demonstrating the relationship u and r, a.Figure 23 the relationship of u and r, We aFrom the analysis above, we find with r increasing, u increases, which means when the radius of the rounded rectangular r increases, the degree of the even heat distribution. Given one extreme circumstance, when r gets its maximal value, the pan becomes a circular, and the degree of the even heat distribution is also the most.Finally, we can derive that the degree of the even heat distribution increases from the rectangular, rounded rectangular with smaller r, the rounded rectangular with bigger r and the circular.Figure 24 the comparision of degree of even heat distribution for different shapes Problem ⅢIn this section, we add a weight factor p to analyze the results with the varying values of W/L and p.:max (1)(,)objective function pN p u r a +-222:,004subject to n k m h k h A r r A r r Sππ⋅≤⋅≤==≥≥≤-+≤01p ≤≤In this problem, there are three variables , a, r and p .what we need to do is to select the best shape for the pan, namely, to select a and r. Firstly, we set a and get the relationship of objective function, y and r, p.Figure 25 the relationship of y and r, p From the figure, we can see with p decreasing and r increasing, which means the degree of the even heat distribution is larger, y increases.Then we set p, and get the relationship of the y and r, a.Figure 26 the relationship of y and a, rThis figure shows that the value a has little influence on the objective function.5 Comparision and Degree of fittingComparisionWhen solving the problem to select the best shape of the pan, we only take the rounded rectangular into account and ignore other shapes, Here, we concentrate on one of the polygon—the regular hexagon as an example to demonstrate our model is correct and retional.We use finite element method to derive the heat distribution, just like analyzing the rounded rectangular.Figure 27 the heat distribution of the regular hexagon panFigure 28 the heat distribution of the regular hexagon pan in two dimensions From the above two pictures, we can see the temperature of the corners is much higher than that in other sections of the pan. So the food at corners gets overcooked more easily. In fact, the longer the distance from the center to the corner is , the higher the temperature becomes.● Degree of fittingIn the second problem of selecting the best shapes of the pan, we get two equations of u and r by data fitting. Now, we will analyze the degree of fitting, which is expressed by the residual errors.To calculate the residual errors, we use the following equations.1ˆ()ˆˆT T Y X X X X Y YX βεββ-=+== Finally, we can get the residual errors by2ˆ()L i i iS y y=-∑ Where i is the number of the data sampled, here i is 11.S 1=5.021 and S 2=0.018. Obviously, the second equation is more accuracy.6 Sensitivity● From figure 16 and 17, we know that the larger the ratio of the area of theoven and the area of the pan is, the better our model fits.● We differentiate the equation derived by figure 22,and find that with rincreasing, the value of the differential goes down, which means u will become steady as r increases.● We analyze figure 23, and find the higher the value of r is, the moreobvious the influence a has on u.●We draw another figure as follows in comparision with figure 25, and getwhen the area of the oven S is very small, the relationship of y and r, p andbe seen more apparently.Figure 29 the relationship of y and r, p7 Strengths/weaknessesStrengths:●We use different methods, infinite difference and infinite element, to buildthe model, and the conclusions are consistent with each other.●We compare the heat distribution of the rounded rectangular and the regularhexagon and then calculate the residual errors, validating our model iscorrect.●The results generated by our model agree with empirical results.●Our model is straight, common and easy to understand.Weaknesses:●We didn’t give an analytic solution for the optimal number of the pans in theoven, N.●The model doesn’t take into account detailed things, like the air convectionin the oven.8 ConclusionsWe propose several models to solve the problem of the heat distribution and the optimization of the pan’s shape combining the maximal number of pans in the ovenand the maximal even heat distribution. After detailed analysis, we can get the following conclusions:●Rectangular can best fit in the oven considering the best efficiency of usingspace in an oven.●To get the maximal number of the pans in an oven, we should set the ratio ofthe width to the length of the oven equals that of the small pans.●Generally, the heat distribution of the circular is the most even. But when itcomes to the combination of the efficiency of the using space and the evenheat distribution, the rounded rectangular fits well. And with the radiusincreasing, the degree of even heat distribution increases, resulting in lesserefficiency of using space.9 Advertisement for new Brownie MagazineLove Brownies? Of course, follow us to see our new-designed ultimate Brownie Pan!Almost every Brownie gourmet may encounter the same problems when baking Brownies, cakes or other gourmets. And the most annoying thing may lie in the uneven cooked gourmets. For the heat is distributed uneven in the present pans, after baked, the cakes often can’t be get out of the pan easily or the edge is always difficult to cut because it is too filmsy. What’s worse, the overcooked food taste bad and become unhealthy containing bad things. But now, things are different. All of these trouble problems will disappear for we have ultimate pans. After careful calculation and analysis, we designed a new Brownie Pan—the rounded rectangular shape pan. We study the heat distribution thoroughly of different shapes of pans and find that, the rounded rectangular shape is almost perfect in terms of even heat distribution. The cakes at the corner of the pan will never be overcooked as long as you set the temperature appropriately. And you can get out of your edge-crisp and chewy-inside cake whenever you want. So, is it wonderful?Another troublesome thing is that, the traditional pan usually can’t get clean easily for its straight angle., which bring about many complaints from customers at American Amazon online shop. But for our rounded rectangular pan, you won’t worry about this trifle! We guarantee our ultimate Brownie pan is simple and time-saving to clean. And we recommend aluminum as the material of the pan, for it’s light and portable.In our model, we optimize the number of pans in each oven and derive the relationship of the number N and the radius r , the ratio of the width to the length of the pan . So given the ratio of the width to the length, we can get a certain r, and the number is also determined. Or given the radius r, we can also design the pan. This wonderful because different people have different demands for the number of the pans in each oven. Considering a family party or doing baby food, we need different number, of course.I believe the following merits may be attractive for most manufactures. We guarantee the rounded rectangular shape pan can save many materials. And the simple style confirms to the values of beauty. The environment-friendly, low –carbon style pan bring a new try for customers.Bring our ultimate Brownie pan to your home, you will find more surprises!10 References1Shixiong Liu, Mika Fukuoka, Noboru Sakai, A finite element model for simulating temperature distributions in rotating food during microwave heating, Journal of food engineering, V olume 115, issue 1, March 2013 Page 49-62 2Heat transfer theory /unitoperations/httrtheory.htm。

2013年数学建模美赛A题二等奖作品

The perfect pan for ovenThe heat transfer in the oven includes heat conduction, heat radiation and heatconvection. We use two-dimensional Fourier heat conduction equation ∂u∂t −α(ð2u∂x2+ð2u∂y2)=f(x, y, t) to make a research on distribution of heat for the pan. Heat source heats the pan by heat radiation. The pan interacts with air in the oven in the way of natural convection, so the pan realizes heat dissipation.We calculate heat radiation based on radiation ability of heat source and heating tube area. We use heat dissipation function to show the pan's different parts' loss of heat caused by natural convection. Both of them consist in heat source function f.The area of the pan is fixed at 0.085m2in this paper. When comparing temperatures at the edges of rectangular pans with different length to width ratios ξ, we can get that the smaller ξ is, the lower the temperature of the edges is. But as long as it is still a rectangle, the amplitude of its drop won't be very big. When we make the pans with fixed area vary from square to round square to round, we find that the bigger the fillet radius is, the lower the temperature of its corners is and the extent of temperature's reducing is large.We fix the bottom area of the oven and area of the pan. Through study, we find that round square's capacity for uniform distribution of heat is far higher than other shape's (except round). The larger the fillet radius of the round square is, the larger the pan’s waste of space is. But heat distribution is more uniform. We work out the optimal solution of pan’s size under different weights p through optimizing the relationship between two conditions. Then we get several oven's width to length ratios of W/L by arranging the pans with the optimal size.I. IntroductionThe temperature of each point in the pan is different. For a rectangular pan, the corners have the highest temperature, so the food is easily overcooked. While the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges in the round pan.To illustrate the model further, the following information is worth mentioning1.1 Floor space of the panThe floor space of each pan is not the square itself necessarily. In this paper, there are 3 kinds of pans with different shapes, as rectangular pans, round pans and round rectangle pans.For rectangular pan, the floor space is the square itself, and the pans can connect closely without space.For round pans, the diagrammatic sketch of the floor space is as follows:shade stands for the round pan; square stands for the floor spaceFigure 1Round pans have the largest floor space for a certain area. The space between each pan is larger than other two kinds of pans. The coefficient of utilization for the round pans is the lowest.For round rectangle pans, the diagrammatic sketch of the floor space is as follows:shade stands for the round rectangle pan; square stands for the floor spaceFigure 2If the area of the round rectangle is the same as the other two, its floor space is between them. The coefficient of utilization of oven decreases with the radius of expansion.1.2 Introduction of ovenThe oven is usually a cube, no matter it is used in home or for business. A width to length ratio for the oven is not a certain number. There are always two racks in the oven, evenly spaced. There are one or more pans on each rack. To preserve heat for the oven, food is heated by radiation. Heating tube can be made of quartz or metal. The temperature of the tube can reach 800℃ high when the material is quartz. The heating tube is often in the top and bottom of the oven. Heating mode can be heating from top or heating from bottom, and maybe both[1].1.3 Two dimensional equation of conductionTo research the heat distribution of pan, we draw into two dimensional equation[2]of conduction:∂u ∂t −α(ð2u∂x2+ð2u∂y2)=f(x, y, t)In this equation:u- temperature of the pant- time from starting to heatx- the abscissay-ordinateα- thermal diffusivityf- heat source functionThe heat equation is a parabolic partial differential equation which describes the distribution of heat (or variation in temperature) in a given region over time. The heat equation is of fundamental importance in diverse scientific fields. In mathematics, it is the prototypical parabolic partial differential equation. In probability theory, the heat equation is connected with the study of Brownian motion via the Fokker–Planck equation. The diffusion equation, a more general version of the heat equation, arises in connection with the study of chemical diffusion and other related processes.II. The Description of the Problem2.1 The original problemWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient withrespect to using the space in an oven.Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes -rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.2.2 Problem analysisWe analyze this problem from 3 aspects, showing as follows:2.2.1 Why the edge of the pan has the highest temperature?The form of heat transfer includes heat radiation, heat conduction and heat convection. Energy of heat radiation comes from heat resource. The further from heat resource, the less energy it gets. Heat conduction happens in the interior of the pan, and heat transfers from part of high temperature to the part of low with the temperature contrast as its motivation. With the two forms of heat transfer above, we find the result is that pan center has the high temperature and the boundary has the low. D epending on that, we can’t explain why the product gets overcooked at the corners while at the edges not. We think that there is natural convection between pan and gas, because the temperature of pan is much higher than that of gas. The convection is connected with the contact area. The point in pan center has a larger contact range with air, so the energy loss from convection is more. While the point in the corners of a rectangular pan has a narrow contact area, the energy loss is less than that in the inner part. Because that above, the energy in the pan center is more than that in corner, and the corners have higher temperature.2.2.2 Analysis of heat distribution in pans in different shapesThe shape of pan includes rectangle, round rectangle and round. When these pans' area is fixed, the rectangles with different shapes can be shown with different length to width ratios. Firstly, we study temperature (maximum temperature) in the corners of rectangles with different length to width ratios. Then we study how temperature in the corners changes when the pans vary from square to round square to round. After that, we select some rectangular panand make it change from rectangle to round rectangle to study changes of temperature in the corners. To calculate distribution of heat for the pan, we would use three main equations. The first one is the Fourier equation, namely heat conduction equation, the second one is the radiation transfer equation of heat source and the third one is the equation of heat dissipation through convection. In the radiation transfer equation of heat source, we take heat source as a point. We get its radiating capacity through its absolute temperature, blackening and Stefan-Boltzmann law. We combine radiating capacity with surface area of quartz heating tube to get quantity of heat emitted by heat source per second, then we can get heat flux at each point of the pan. In the equation of heat loss through convection, Heat dissipating capacity is proportional to area of heat dissipation, its proportional coefficient can be found from the related material. Through the establishment of above three main equations, we can use pdetool in matlab to draw the figure about distribution of heat for the pan.2.2.3 How to determine the shape of the pan?To make heat distribution of the pan uniform, we must make it approach round. But under the circumstances of the pan's fixed area, the closer the pan approaches round, the larger its floor space is. In other words, the closer the pan approaches round, the lower the utilization rate of the oven is.More uniform distribution of heat for the pan is, the lower temperature in the corners of the pan is. Assuming the bottom area of the oven is fixed, the number of most pans which the oven can bear is equal to the quotient of the bottom area of the oven divided by floor space per pan. So in a certain weight P, we can get the best type of pan (shape) by optimizing the relationship between temperature in the corners and the number of most pans which the oven can bear. Then we get the oven's width to length ratio of W/L by arranging the pans with the optimal size.2.3 Practical problem parameterizationu: temperature of each point in the oven;t: heating time;x: the abscissa values;y: the ordinate value;α:thermal diffusivity;ε: degree of blackness of heat resource;E: radiating capacity of heat resource;T: absolute temperature of heat resource;T max: Highest temperature of pans’ edge;ξ: the length to width ratio for a rectangular pan;R: radius for a round pan;L: length for the oven;W: width for the oven;q: heat flux;k: coefficient of the convective heat transfer;Q: heat transfer rate;P: minimum distance from pan to the heat resource;Other definitions will be given in the specific models below2.4 Assumption of all models1. We assume the heat resource as a mass point, and it has the same radiation energy in all directions.2. The absorbtivity of pan on the radiation energy is 100%.3. The rate of heat dissipation is proportional to area of heat dissipation.4. The area of heat dissipation changes in a linear fashion from the centre of the pan to its border.5. Each pan is just a two-dimensional surface and we do not care about its thickness.6. Room temperature is 25 degrees Celsius.7.The area of the pan is a certain number.III. ModelsConsider the pan center as origin, establishing a coordinate system for pan as follows:Figure 33.1 Basic ModelIn order to explain main model better, the process of building following branch models needs to be explained specially, the explanation is as follows:3.1.1 Heat radiation modelFigure 4The proportional of energy received by B accounting for energy from A is4π(x2+y2+P2) The absolute temperature of heat resource A (the heating tube made of quartz) T= 773K when it works. The degree of blackness for quartz ε=0.94.The area of quartz heating tube is 0.0088m2.Depending on Stefan-Boltzmann law[3]E=σεT4 (σ=5.67*10-8)we can get E=18827w/m2.The radiation energy per second is 0.0088E.At last, we can get the heat flux of any point of pan. =0.0088E4π(x2+y2+P2)Synthesizing the formulas above, we can get:=13.24π(x2+y2+P2)(1) 3.1.2 Heat convection model3.1.2.1 Round panheat dissipation area of round panFigure 4When the pan is round, the coordinate of any point in the pan is (x,y). When the point is in the centre of a circle, its area of heat dissipation is dxdy. When the point is in theboundaries of round, its area of heat dissipation is 12dxdy. According to equation of heatPdissipation through convection dQ=kdxdy and assumption that the area of heat dissipation changes in a linear fashion along the radius[4], we can get the pan's equation of heat dissipation:q =k[1-0.5(x2+y2)0.5/R] (2) 3.1.2.2 Rectangular panHeat dissipation area of rectangular pan(length: M width: N)Figure 5When the pan is rectangular, the coordinate of any point in the pan is (x,y). When the point is in the centre of a rectangle, its area of heat dissipation is largest, namely dxdy. Whenthe point is in the center of the rectangular edges, its area of heat dissipation is 1dxdy. When2the point is in rectangular vertices, its area of heat dissipation is minimum, namely According to equation of heat dissipation dQ=kdxdy and assumption that the area of heat dissipation changes in a linear fashion from the centre of a rectangle to the center of the rectangular edges.The area of any other point in the pan can be regarded as the result of superposing two corresponding points' area in two lines.Heat dissipating capacity of any point is:=1−y/N(3)1−x/M3.2 Pan heat distribution Model3.2.1 Heat distribution of rectangular pansAssume that the rectangle's length is M and its width is N,the material of the pan is iron.For rectangular pans, we change its length to width ratio, establishing a model to get thetemperature of the corners (namely the highest temperature of the pan).We assume the area of pan is a certain number 0.085m2,The distance from the pan above to the top of the oven P=0.23m,By checking the data, we can know that the coefficient of the convective heat transfer k is approximately 25 if the temperature contrast between pan and oven is 100~200℃.then get a several kinds rectangular pans following:In the two dimensional equation of conduction,we get the thermal diffusivity of iron is 0.000013m2/s through checking data.Heat source function equals received thermal radiation minus loss of heat caused by heat dissipation through ly equation (1) minus equation (3).In this way, we get a more complicated partial differential equation. For example, through analyzing the NO.1 pan, we can get the following partial differential equation.∂u ∂t −α(ð2u∂x2+ð2u∂y2)=13.24π(x2+y2+0.529)−1−y/0.291551−x/0.29155It is hardly to get the analytic solutions of the partial differential equation. We utilize the method of finite element partition to analyze its numerical solution, and show it in the form of figures directly.Pdetool in matlab can solve the numerical solution to differential equation in the regular form quickly and show distribution of heat by the three-dimensional image[5]. We enter partial differential equation of two-dimensional heat conduction into it, then we get the figures about distribution of heat in different pans.When we use pdetool, we take Neumann condition as boundary condition, and we suppose that the boundary is insulated, In fact, it is not insulated, and the heat dissipation will show in the heat source function.We heat the pan for 8 minutes no matter what kind of shape the pan is. By Pdetool, the heat distribution of each pan shows as follows:Figure 6(heat distribution for NO.1 pan) Figure 7(heat distribution for NO.2 pan) Figure 8(heat distribution for NO.3 pan)This pan is just a square pan, with the lowest length to width ratio. From the figure, we can know that corners have the highest temperature which is 297.7℃.The corners have the highest temperature for the pan, which is 296℃The corners have the highest temperature for the pan, which is 294.8℃The temperature ofcorners for the pan isslightly lower than thehighest temperature, andthe highest temperature is293.7 ℃.Figure 9(heat distribution for NO.4 pan)To get a more accurate relationship between the length to width ratio(ξ) and highest temperature(T max), we make several more figures of heat distribution based on the different length to width ratio. At last, we can get its highest temperature. The specific result is as follows:ξ is the argument and T max is the dependent variable. The points in the chart are scaled out in the coordinate system by mathematical software Origin. Connecting the points by smooth curve, we can get the figure following:Figure 10(the relationship between ξ and T max )From the figure we can find that, with the length to width ratio increases, the temperature of corners will sharply fall at first, and it is namely that the heat distributes evenly . When the length to width ratio increases further, the temperature of corners drops obscurely . When the ratio reaches about 2.125, the length to width ratio of rectangular pans has little effect on the heat distribution. In contrast, the ratio is too big, it is difficult for practical application.In addition, we can also find that as long as the shape of the pan is a rectangle. The highest temperature in the corners of the pan won't change a lot whether its length to width ratio changes, From the figure 10, we can find that maximum range is about 6 degrees Celsius. So in general, it is very difficult to change high temperature in the corners of the rectangular pan.3.2.2 Heat distribution of square pans to round pans 3.2.2.1 Size definition of round squareWe need some sizes to define round square, our definition isas follows:T maxThe size of round square isdecided by l and r (l stands forthe length of the straight flange;r stands for the radius of thefillet).Figure 11We assume that the area of the pan is 0.085m2. The r of round square has its range, which is 【0，0.164488】，When the r reaches the two extremums, round square becomes square and circle.3.2.2.2 ModelWhen we make this kind of pans’ heat distribution figures, we take the heat source function as (1) and (3). When the round square becomes circle, the heat source function is (1) and (2).We get a several round squares with different r and l, and take a circle as an example. The specific examples show in the table below:The pan is still made of iron. Here we also heat the pan for 8 minutes.By Pdetool, we can obtain their heat distribution figures as follows:Figure 11(heat distribution for NO.1 pan) Figure 12(heat distribution for NO.2 pan) Figure 13(heat distribution for NO.3pan)The temperature of the corners about the pan is relatively low, and the highest temperature is 264℃. On the contrary, the heat distributes quite evenly.The highest temperature of the pan is 271.1℃The highest temperature of the pan is 274℃The highest temperatureof the pan is 279.8℃Figure 14(heat distribution for NO.4pan)The highest temperatureof the pan is 283.7℃Figure 15(heat distribution for NO.5pan)To get the different relationship between l/r and T max, we take l/r as argument, and T max as dependent variable. We change l and r of round square, and make several heat distribution figures. In this way, we can get the highest temperature T max, and the results show in the table below:Here, we alsouse the mathematical drawing software origin. We use l/r and T max to express coordinates. We use smooth curve to connect points, then we can get the trend line.Figure 16(the relationship between l/r and T max )From the figures we can know that, the value of l/r is smaller, the temperature of the corners about pan is higher. It is namely that the pan is more closely to circle, and heat distribution is more evenly. With the value of l/r increases, the temperature of corners rise very quickly at first, then the amplitude is getting smaller. When the value of l/r is infinitely great, the highest temperature of the pan go to a certain number.Analyzing in a theoretical way , the shape of pan goes to square when the value of l/r is infinitely great. At the same time, the temperature of the round square’s corners approach to that of square’s. From the figures, we can know that this function has a upper boundary ,T maxwhose value is close to the corner temperature of square. Through this, we can verify the correctness of our models.3.2.3 Heat distribution of round rectangle (except round square)From the model about heat distribution of rectangular pan, we can learn that drop of temperature in the corners of rectangular pans will be very little when its length to width ratio is bigger than 2.125. So we select the rectangle with length to width ratio of 2.125. We let the pan vary from the rectangle to round rectangle. So we can study changes of temperature in the corners of the pan.3.2.3.1 Size definition of round rectangleThe specification of the roundsquare is decided by l and r. Weset its width the same as therectangular pan before, namely0.2m.(l stands for the length ofstraight long side, and r stands forthe radius of the fillet.)Figure 17We assume that the area of the pan is 0.085m2, For round rectangle pans, with r decreases constantly, the pan finally approaches the rectangular pan before .If the r increases constantly, its shape will become that of playground. The range of r is 【0,0.1】3.2.3.2 ModelWhen we draw the figure about heat distribution of this kind of pan, heat source function equals equation (1) minus equation (2).We can get some different round rectangles by changing the value of r and l. Their detailed specifications are shown in the following table:The pan is still made of iron. Here we also heat the pan for 8 minutes.By Pdetool, we can obtain their heat distribution figures as follows:The corner temperaturewhich is 291.7℃andslightly lower than thehighest temperature ofthe pan.Figure 18(heat distribution for NO.1pan)The corner temperaturewhich is 292.54℃andslightly lower than thehighest temperature ofthe pan.Figure 19(heat distribution for NO.2pan)The corner temperaturewhich is 293.5℃andslightly lower than thehighest temperature ofthe pan.Figure 20(heat distribution for NO.3pan)To get the different relationship between l/r and T max, we take l/r as argument, and T max as dependent variable. We change l and r of round square, and make several heat distribution figures. In this way, we can get the highest temperature T max, and the results show in the table below:Here, we also use the mathematical drawing software origin. We use l/r and T max to express coordinates. We use smooth curve to connect points, then we can get the trend line.Figure 21(T max)From the figures we can get that, with l/r increases, the highest temperature of pan goes down. That is to say, the bigger radius of the fillet is, the more evenly heat distributes. In addition, with l/r increases, T max rises quickly at first, then the extent is smaller. When l/r is infinitely great, the round rectangle pans become rectangle pans, and the temperature approaches to that of the rectangle pan before.Secondly, from the figure, we can learn that change of the largest temperature is very little and its largest temperature is all very high when the pan varies from rectangle to round rectangle. compared with round square pan, round rectangle pan has much worse capacity of distributing heat.3.3 Best type of pan selection ModelFrom the model of heat distribution, we can know that the extent of heat distribution for round square pan is more than the extent of other pans with other kinds of shape( except T maxcircle). It is difficult to accept it for people because the food is easily overcook, no matter what the number of pan is. Depending on that, people will choose the round square pan.In the following models, we mainly discuss the advantages and disadvantages of round square pans with different specifications.3.3.1 Local parametersFor study's convenience, we take commercial oven of bottom area 1.21m2as an example. The distance between heating tube on the top and the nearest rack is P=0.23m.The number of pans which the oven can contain is n;The fillet radius of round square is r;The floor space of pan is S;The weight of the number of pans in the oven is P;The area of pan is still 0.085m2, the material is still iron.3.3.2 The relationship between T max and rThrough the models before, we know the relationship between l/r and T max. We transform it as the function of r and T max, and shows in the form of table below:We take r as the argument, and T max is the dependent variable. Making the dots in the coordinate system by the mathematical software Origin[6].The figure is as follows:Figurebetween r and T max )To our surprise, we can find that there is nearlylinear relation between r and T max fromthe figure. We may fit the relation with a linear function, so we can get the function of r and T max .FigureThe function we are getting is: T max =-174.5r+291.5 (4)3.3.3 The relationship between n and rWe have introduced that the floor space is not the area of the pan itself in theT maxT maxIntroduction. So we can get the formula below of the area of round square pan and r:S=r2(4−π)+0.085We have already known the floor space of the oven. So we can know the maximum number of pans that the oven can hold , in which condition the shape of pan is sure. Above all, we can get the function of n and r.n= 1.21r2(4−π)+0.085(5) 3.3.3 The optimum solutionThe weight of the number of pans which the oven can hold is P, while the weight of heat distribution is (1-P). The dimensions of the two are different. The effects are also different with the unit change of r. Depending on the message above, in order to induce the weight P. we need to eliminate their dimensions[7].Through observing the figure 23, we can know that the range of T max is 27℃with the domain of r.Then we change the value of r in its domain of definition, then we can get n's approximate range: 3.05mThen we eliminate their dimension, so they are transformed into value which could be compared.They are T max/27 and n/3.5 respectively.We hope that we can get a smaller T max and a larger n. We let T max/27 multiply by -1, then add them (T max/27 and n/3.5) together, the final result is K. K has no practical significance, and we just want to know its relative value.K=-(1-P)T max/27+P n/3.5After simplification, we can obtain:K=-(1-P)(-6.463r+10.796)+0.345Pr2(4−π)+0.085(6)How to get the pan we want with the idealized shape and its corresponding width to length ratio of the oven by using this formula?We explain it by an exampleIf some one’s ideal weight P is 0.6, the function(6) of K becomes:K=-0.4(-6.463r+10.796)+0.207r2(4−π)+0.085We can get the figure of K within the domain of r, by using the matlab. The figure is presented as follow:Figure 24(the relationship between K and r)We plugged the value of r into the equation (5), then we can get n=13.76Because the number of the pan should be an integer, we round up n, namely n=13.The largest number of pans which the oven can contain is prime number, the oven has only a width to length ratio of 1/13.So at last, we get the following conclusion:When P=0.6, n=13, W/L=1/13, r=0.058m is the best solution.To make the coefficient of oven reaches the top (namely without space), we take the radius of round square into equation (5). We should notice that n must be an integer. We adopt the method of exhaustion, and the result shows in the table below:This table will be used in the following advertizing.3.3.4 Model verificationWe can see the equation (6) , when the P tends to 0, which means the largest number of pans the oven can contain make no sense, the equation becomes: K=- (-6.463r+10.796) ; the optimum solution is r=0.164488m. which means maximize even distribution of heat for the pan is most important.On the contrary, when the P tends to 1, which means the maximize even distribution of heat for the pan make no sense, the equation becomes: K =0.345r 2(4−π)+0.085; the optimum solution is r=0m. which means the largest number of pans the oven can contain is most The value of r for thecorresponding peak valuein the figure is 0.058m。

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For office use onlyT1________________ T2________________ T3________________ T4________________Team Control Number18263Problem ChosenAFor office use onlyF1________________F2________________F3________________F4________________2013Mathematical Contest in Modeling(MCM/ICM)Summary Sheet(Attach a copy of this page to your solution paper.)Type a summary of your results on this page.Do not includethe name of your school,advisor,or team members on this page.There are plenty of people who are fond of Brownies.Industrial manufacturing of them requires to bake the biggest amount of Brownies in the same batch and also to avoid overcooking of the batter next to the corners of the pan,in order to avoid losses. Overcooking can affect the volumes of batter that are in direct contact with more than one hot surface of the pan,for instance the bottom of it and one of its sides.The corners of the pan are the hottest regions,as three surfaces deliver heat simultaneously(two sides plus the bottom).That’s why overcooking is frequent in the corners.Round pans distribute heat in a more even way,due to the fact that there are never three surfaces delivering heat to a volume of batter(two at most),.The problem with them is that they cause a big loss of oven space as compared to square ones,even to hexagonal ones.As round pans minimize overcooking and rectangular pans minimize oven-space loss,our model posits a compromise between these two shapes,namely modified rectangular pans having a 90°centred-angle arc instead of being standard rectangles.In order to clarify which factor will be advantaged by variations of this type of shape(either space saving or heat evenness),we put weights p and(1-p)on the sum of the line lengths and on the sum of the arc lengths.As we consider that the area of the pan is fixed and its width-to-length ratio W/L is given,we propose3equations expressing the effect of having longer or shorter arcs on the pans performances(the longer the arcs,the evener the heat distribution;the shorter the arcs,the better the space in the oven is filled).We use matlab to solve these equations.Once we give p and W/L two values between0to1, we can get a shape of pan.By changing the weights p by small steps while the ratio W/L is fixed,we can plot the figure in a three-dimentional space layer by layer,so that we can see clearly how the figure changes with weights changing for a given W/L ratio.From these graphs,we can conclude that the space the pan takes increases with increasing weights on the arc factor(we provide some charts to make clear).In general,if the producer wants to increase the quality of the brownies by avoiding overcooking of volumes placed in critical locations,he needs to decrease p.If he wants to increase the production by filling the oven as much as possible,he needs to increase p. Large p values amount to short arcs,small p values mean long arcs.Part1.OverviewObtaining perfectly baked brownies depends much on the shape of the pan used to cook them,as cakes tend to get overcooked at the corners in angular pans.After having visited a few bakeries,we found out that most pans are rectangular with rounded corners.We noticed there were stains showing overcooking in the four corners and,to a less extent,on the sides,which shows that the heat absorbed by the batter had been higher for the volumes placed at the corners and against the faces of the pan.We therefore engaged in finding a pan shape able to prevent overcooking,keeping in mind that bakeries need to fit as many pans as possible into their ovens,in order to take the best advantage of each batch of brownies.After discussions and some study,we discovered that shapes maximizing the evenness of heat distribution waste oven space.We thus moved to the goal of both preventing overcooking and avoiding waste of oven space in order to define the best suited pan for productive and flawless cooking of brownies.As will be seen,our model tends to make the best compromise between these two constraints.Part2.AssumptionsAssumption1(Space optimization)Given a rectangle of fixed width-to-length ratio(W/L)and unknown area,to maximize the number of figures fitting into the rectangle requires the shape of these figures to satisfy the following conditions:1)As n figures are put together in the rectangle,the area left between figures must be the least possible one,2)No arcs are necessary,as replacing arcs by angles consumes the same area.Assumption2(Heat distribution optimization)The heat each volume unit gets from the pan is proportional to the area unit of the pan it is in contact with.Area units are considered to be the bottom and each one of the faces of the pan.As a volume unit placed in a corner receives heat from3area units of the pan,its heat value equals3units(bottom and2sides).By contrast,a volume unit placed against one of the faces of the pan receives heat from2area units of it(bottom and side),so its heat valueequals2units.For a volume unit placed in the middle of the pan,the heat received comes from a single area unit of the pan(the bottom),thus its heat value equals1unit.[Figure2-1]Assumption3(Model oven and relevant pan properties)1.The heating devices in our model oven being evenly distributed,the pan absorbs the same amount of heat value from all directions.2.The pan being evenly heated,every point of the pan shares the same temperature.3.Only the pan provides the heat,as we ignore the heat in the air surrounding it.4.Only the shape of the pan will be taken into consideration.Part3.Goals of the modelAs the design of the pan is meant to fit the requirements of energy-saving while perfect cooking of brownies(no burned or uncooked pastry),the following modelling goals have been set:(1)Maximize number of pans that can fit in the oven(N),(2)Maximize the evenness of heat distribution(H)for a pan.In order to achieve an optimal balance between goals(1)and(2),putting weights on them [p and(1-p)],will allow to decide which one should be given more attention.Part4.Model4.1Discussion of objective1:how to maximize the filling of the oven According to assumption1,we have first identified rectangles and regular hexagons as the best shapes for figures:whereas both of them meet the requirement of minimizing the space left between figures,rectangles not only satisfy this constraint but also minimize the lost space within the oven when the figures are in place,due to the fact that the W/L ratio of rectangles is more consistent with the W/L ratio of a rectangular oven than the equivalent ratio of hexagons.From this perspective,our final choice is thus a rectangle.4.2Discussion of objective2:how to maximize the evenness of heat distributionAs to the constraint to maximize the evenness of the heat distribution,we interpret it as the fact that the difference between the quantity of heat every volume unit of the batter receives is the lowest possible.As this difference increases with the number of faces of the pan, volume-type ensuring the most even heat distribution should be a cylinder.From a two-dimensional perspective,the ideal shape is thus a circle.circle..From this perspective,our final choice is thus a circle4.3Putting perspectives togetherIn order to combine both shapes,we replaced the four angles of the rectangle with four arcs with a central angle of90degrees,as shown in figure4-1below.If the central angle of the arc were greater than 90degrees (figure 4-2),the lines and the arc could not be put together in a way that satisfies assumption 1.Conversely,if the central angle of the arc was less than 90degrees (figure 4-3)space would be wasted as the figure would get bigger because the area A isfixed.4.4Building the modelIn order to satisfy 4.1and 4.2,rectangle and circle can be combined and enhanced with weights p and (1-p),so as to obtain the shape model illustrated by figure 4-1.The model is expressed via the following equations:(Each pan must have an area of A)()22............................................................................r r a b ab A π+++=①(Weights p and (1-p)are assigned to the perimeter of lines and arcs so as to allow decisions about the most critical factor)(This shape takes as much space as the same sized rectangle.According to Assumption 1,we have equation ③)We assume that p and W/L are independent variables,a,b and r being dependent ones.We solve the equations by expressing a,b and r with p and W/L.We can see how the shape changes with different given values of p and W/L between 0to 1.Thus,our final equations of our model are are::()1010..221)(22<<<<=++−=+=+++LW p t s LWr a r b p pr b a Aab b a r r rb a find ππWe use matlab software to solve these equations andwe use p and W/L to refer to a,b and r （For detailed code please refer to the appendix1）.MA kp k p p a )2222(++−−=πMA kp k p kp b )2222(−−++=πMAkp p k r )1(+−−=In which A is given and ：⎟⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎜⎝⎛=−−−+++−−−++++−−+++=L W k p k kp kp kp p k p k p k p k kp p k kp p k p k M 42644448816228822222222222222ππππππππππPart 5.Sensitivity and stability analysisNext,we analyze the sensitivity and stability of our model by assigning different values to the variables.First of all ，we assume that the area of the pan is fixed as A=10,when the width-to-length ratio W/L=1,we use matlab to calculate the value of a,b and r ：A=10W/L=1Nop a b r S 10.10.14020.1402 1.695412.468020.150.28060.2806 1.607412.217830.20.42140.4214 1.520211.984140.250.56290.5629 1.433511.764250.30.70540.7054 1.347211.558660.350.8490.849 1.261111.365070.40.9940.994 1.175211.185080.45 1.1408 1.1408 1.089311.018490.5 1.2895 1.2895 1.003310.8643100.55 1.4404 1.44040.91710.7217110.6 1.594 1.5940.830310.5924120.65 1.7505 1.75050.742910.4736130.7 1.9103 1.91030.654810.3678140.75 2.0738 2.07380.565810.2746150.8 2.2415 2.24150.475710.1946160.85 2.4139 2.41390.384210.1270170.9 2.5915 2.59150.291110.0724180.95 2.7749 2.77490.196310.03311912.96492.96490.099310.0077In this table ，the last column is the area of the minimum bounding rectangle of the pan S.S=（a+2r)(b+2r).As may be seen in Figure5-1,we multiply p by10to make the graph clearer.For p from 0.1to1step0.05,we plot every figure and pile them up layer by layer in the3-dimentional space.We observe how S changes while changing the value of p,and notice that when p increases,S decreases and tends to10.That is,while p increases from0to1,the area(S) decreases slightly but proportionally to the value of p.Changing the value of W/L shows that the phenomenon is stable for every value of W/L.A=10W/L=0.8Np a b r So10.050.53660.2541 1.708614.515620.10.6740.1807 1.619713.384230.150.81210.037 1.531612.014240.20.95090.1832 1.44411.790050.25 1.09090.33 1.356911.580760.3 1.23220.4778 1.270111.385170.35 1.37520.6268 1.183411.202180.4 1.520.7772 1.096811.032490.45 1.66690.9295 1.010110.8758100.5 1.8163 1.08380.923110.7311110.55 1.9685 1.24050.835710.5990120.6 2.1238 1.40.747810.4803130.65 2.2827 1.56250.659110.3731140.7 2.4455 1.72860.569410.2776150.75 2.6128 1.89870.478710.1968160.8 2.7849 2.07330.386610.1281170.85 2.9626 2.25290.29310.0741180.9 3.1464 2.43810.197510.0331190.95 3.3371 2.62970.110.0089A=10W/L=0.6Np a b r So10.05 1.06250.7733 1.765619.7736 20.1 1.20120.6173 1.672618.0151 30.15 1.33790.4617 1.580516.2982 40.2 1.47570.306 1.489314.6306 50.25 1.61470.1501 1.398613.0032 60.3 1.75530.0064 1.308511.4703 70.35 1.89770.1637 1.218611.2746 80.4 2.04220.3222 1.128911.0940 90.45 2.18910.4821 1.032910.8411 100.5 2.33870.64380.949410.7743 110.55 2.49150.80750.859210.6338 120.6 2.64770.97380.768510.5069 130.65 2.8708 1.14290.677110.5502 140.7 2.9722 1.31540.584910.2937 150.75 3.1414 1.49160.491610.2076 160.8 3.3161 1.67210.396910.1346 170.85 3.4968 1.85750.300810.0784 180.9 3.6842 2.04830.202710.0347 190.95 3.8791 2.24540.102610.0090A=10W/L=0.4Np a b r So10.05 1.883 1.5638 1.930831.1668 20.1 2.0196 1.3825 1.825228.5365 30.15 2.1571 1.2028 1.721426.0149 40.2 2.296 1.0244 1.61923.5881 50.25 2.43640.8469 1.517921.2469 60.3 2.57870.6698 1.417818.9793 70.35 2.72320.4929 1.318416.7752 80.4 2.87020.3156 1.219714.6279 90.45 3.02010.1376 1.121412.5278 100.5 3.17330.0414 1.023310.8992 110.55 3.33020.2220.925110.7348 120.6 3.49120.40450.826710.5871 130.65 3.65680.58490.727810.4319 140.7 3.82760.77720.628210.3388 150.75 4.00420.96850.527610.2387 160.8 4.1871 1.16390.425810.1555 170.85 4.3773 1.36390.322510.0893 180.9 4.5755 1.56940.217310.0402 190.95 4.7827 1.78110.1110.0109A=10W/L=0.2Np a b r So10.05 3.7351 3.3153 2.538973.9676 20.1 3.8622 3.0321 2.377967.1161 30.15 3.9921 2.7539 2.223660.7722 40.2 4.1251 2.2453 2.075252.9276 50.25 4.2615 2.2386 1.931849.5810 60.3 4.4017 1.988 1.792744.5153 70.35 4.546 1.7425 1.657339.7518 80.4 4.6948 1.501 1.52535.2466 90.45 4.8487 1.2625 1.395230.9605 100.5 5.008 1.0263 1.267426.8607 110.55 5.17340.7913 1.141222.9169 120.6 5.34540.5568 1.016219.1026 130.65 5.52490.32190.891815.3880 140.7 5.71250.08570.767611.7478 150.75 5.90920.15270.643210.3552 160.8 6.11590.39430.518110.2311 170.85 6.3340.640.391710.1309 180.9 6.56480.8910.263710.0596 190.95 6.8098 1.14860.133310.0145Part6.C onclusionIn general,if the producer wants to increase the quality of the brownies by avoiding over-cooking of volumes placed in critical locations,he needs to decrease p.If he wants to incre ase the production by filling the oven as much as possible,he needs to increase rge p v alues amount to short arcs,small p values mean long arcs.Improvement of the subsequent models:One of our tasks in the future is to consider the stability of the pan structures after our current model and add relevant details to it,such as heat conduction.We still need to test our model in order to ensure that it is reliable and reasonable.Part7.Strengths and Weaknesses of the Model•The model describes the pan in detail,coordinating the relevant physical quantities.•The numerical computations are relatively precise.•The results illustrated by the graphical representations of the model are generated by numerical computations.•The model is simple and practical and thus easy to implement.•Arbitrary results may be obtained from the model according to specific circumstances(the values of W/L and p are different under different conditions).•The model does not allow to calculate the number of pans in an oven,but it can select the best type of them.•The model does not take into account detailed heat transfer and loss of energy.eferen n ceces sPart9.R efereX.Y.Huang,C.Y.Peng and J.H.Teng2005English expressions of the definitions and theorems in advanced mathematics.ISBN7-118-03773-7Dilip P Patil2010Introduction to Algebraic Geometry and Commutative AlgebraM.Z.Arslanov1998Continued fractions in optimal cutting of a rectangular sheet into equal small rectanglesZ.F.Zhao and Y.Y.Ma2011The Method of Requesting the Limit of Function of Many Variables Journal of Yichun College2011(8)Appendix1:CodeCode of the modelclear allclcsyms a b r A k pA=10;k=0.2;s=0.05:0.05:1;for ii=1:length(s)[a,b,r]=solve('pi*r*r+2*r*(a+b)+a*b=A','(a+b)*(1-p)=pi*r*p','(b+2*r)/(a+2*r)=k', 'a,b,r');p=s(ii);a=eval(a);b=eval(b);r=eval(r);for i=1:length(a)if a(i)>0x=a(i);endenda=xfor i=1:length(b)if b(i)>0y=b(i);endendb=yfor i=1:length(r)if r(i)>0z=r(i);endendr=zrectA(a+2*r,b+2*r,p)hold oncircle(r,1,r+a,1,b+r,p)circle(r,1,a+r,-1,r,p)circle(r,-1,r,-1,r,p)circle(r,-1,r,1,b+r,p)grid onEndCode of circle drawingfunction circle(R,a1,a2,b1,b2,n) alpha=0:pi/50:pi/2;[0,2*pi]%R=2;x=a1*R*cos(alpha)+a2;y=b1*R*sin(alpha)+b2;z(1:26)=10*n;plot3(x,y,z,'-');%axis equalCode of rectangle drawingfunction rectA(w,h,n);X=[0,w,w,0,0];Y=[0,0,h,h,0];d(1:5)=10*n;%axis([-36-36-210])plot3(X,Y,d,'r');Appendix2:Advertising sheet。