2013年美国大学生数学建模大赛A题 一等奖
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 美国数学建模竞赛题目中文 MCM
2013 竞赛题目
MCM
问题A：
当在矩形平底锅上烘烤食物时，热量集中在四个角，因此食物在四个角上容易烤焦（边上焦的程度较轻一点）。
在圆形平底锅内烘烤时，热量均与地分布于整个外部边缘，食物的边缘部分不会烤焦。
然而，由于大部分烤炉形状为矩形，就利用烤炉中的空间而言，使用圆形平底锅并不是最有效率的做法。
开发一个模型来展现平底锅外边缘热量分布，该平底锅的形状可为矩形或圆形，也可能是其它形状。
假设
1．矩形烤炉长宽比W/L
2．每个平底锅面积为A
3．初始烤炉中的两个架子在空间上平均分布。
开发一模型可用于在下列条件下挑选最佳类型的平底锅（形状）：
1．最大限度提高放入烤炉中的平底锅数量（N）
2．最大限度均匀分布在平底锅上的热量分布（H）
3．最优组合条件（1）和（2），其中指定权值p和（1-p）来解释这些结果如何随W/L和p 的不同值变动。
除了MCM格式化的解决方案，为新的《布朗尼美食家杂志》准备一至两页的广告，表达你的思路。
问题B：水，水，到处都是水
新鲜的水对于世界各地的发展是一个限制性因素。
建立一个数学模型用于决定一个有效、可行且成本效益最大的2013水战略来应对投射出来的2025年水需求[从下列国家中挑选一个国家]，并确认最佳水资源战略。
尤其要注意的是，你的模型必须强调存储量和水的运动；去盐渍及保水。
如果可能的话，使用你的模型来讨论你拟定战略的经济、物理和环境影响。
向政府官员提供一个非技术性意见书，阐述你的方法、可行性和成本，以及你的战略之所以为“最佳水资源战略选择”的原因。
国家：美国，中国，俄国，埃及或沙特阿拉伯。
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。
2013年美赛数模A题答案
中国水资源战略摘要Summary为了确定中国最佳的水资源战略，将中国分为九大流域，首先借助MATLAB建立多项式拟合模型来预测出中国2013年到2025年每年各流域的供水量和需水量，接着在可持续发展的原则指导下建立区域水资源合理配置模型，对每一个流域，采用水资源综合短缺度最小为目标函数, 对地表水、地下水等多种水源统筹考虑, 用权重区别对待工业、农业、生活、生态环境等不同领域的用水需求, ,从而求出各个流域最小的缺水量。
再根据前面的两个模型所预测出来的各流域的缺水量，建立最佳的补水模型解决缺水问题：通过对实际问题的分析，可能的补水方案有两个：方案一是直接从珠江流域调水到缺水的流域，方案二是沿海流域采取海水淡化补水，内陆流域采取直接从珠江流域调水过去，经过分析、计算发现方案二是最佳的。
最后，我们统筹考虑我们所制定的水策略，发现其无论是对经济、社会还是生态环境都将产生重大影响。
In order to determine the best water resources strategy, we divided China into nine basins. Firstly, we established polynomial fitting model with the use of MATLAB to predict the water supply and the water demand of every basin from 2013 to 2025. Secondly, we established the regional water resources rational allocation model under the guidance of the principle of sustainable development. In this model, through taking the minimum comprehensive water shortage degree as objective , surface water , groundwater and other water are considered, and different weightings are used for industrial, agricultural, domestic and ecological water users in order to realize regional water resources rational allocation .In this way can we obtained the minimum amount of water scarcity in every basin. Thirdly, according to the data predicted based on the previous two models, we can establish the optimal replenishment model to solve the problem of water shortage. We identified two possible replenishment program based on the analysis of the actual problems. One is to transfer the water of the Pearl River to basins where lack of water resources, another is to transfer the water of the Pearl River to inland basins directly while we meet the water shortage of coastal basins by desalination. After analysis and calculation, we find second program is the best. Finally, we find the water strategy we developed has a significant impact on the economic, social and ecological environment after we considered the models we established.关键字：水策略多项式拟合模型区域水资源合理配置模型补水模型Keywords：Water strategythe Polynomial fitting modelThe Regional water resources rational allocation modelthe Replenishment model§1.问题重述Problem restatement水是生命之源, 是人类生存和发展不可替代的资源, 是经济、社会可持续发展的基础。
2013数学建模美赛题目(中文版)[1]
![2013数学建模美赛题目(中文版)[1]](https://img.taocdn.com/s3/m/2a43be292f60ddccda38a018.png)
2013 ICM问题problem A:当用矩形平底锅高温加热物品时，热量一般集中于4个角落，因而在角落的物品会被焙烧过度（较小程度在角落的物品一部分会被焙烧过度）。
当用一个圆锅加热物品时，热量是均匀分布在整个外缘，因而物品不会在边缘被焙烧过度。
然而，大多数烤箱是长方形的，而圆型的锅被认为效率低的。
建立一个模型以显示不同形状如矩形圆形或者其他介于两者之间的形状的锅在整个外缘的热量分布。
假设1长方形烤箱的宽/长=W/L ；2 每个锅的面积是确定的常熟A；3 最初，烤箱里的烤架两两之间间隔均匀。
建立一个模型，该模型可用于在下列条件之下选择最佳形状的锅：1烤箱中，锅数量（N）最大；2均匀分布的热量（H）最大的锅；3 优化组合条件1和条件2，以比重p和（1-p）的不同分配来说明结果与W/L 和p的不同值的关系。
problem B :对世界来说，新鲜的水资源是限制发展的制约因素。
对2013年建立一个确实有效的，可行的和具有成本效益的水资源战略数学模型，以满足2025年[从下面的列表选择一个国家]预计的用水需求,并确定最佳水资源战略。
尤其是，你的数学模型必须解决水的存储，运动，盐碱化和保护等问题。
如果可能的话，用你的模型，探讨经济，物理和环境对于你的战略的影响。
提供一个非技术性的文件，向政府领导介绍你的方法，介绍其可行性和成本，以及为什么它是“最好的的水战略选择。
”国家有：美国，中国，俄罗斯，埃及，沙特阿拉伯3.网络建模的地球的健康背景:社会是感兴趣的发展和使用模型来预测生物和环境卫生条件我们的星球。
许多科学研究认为越来越多的压力在地球的环境和生物吗系统,但是有很少的全球模型来测试这些索赔。
由联合国支持的年生态系统评估综合报告》显示,近三分之二的地球的维持生命的生态系统——包括干净的水,纯净的空气,和稳定的气候-正在退化,被不可持续的使用。
人类是归咎于很多这次的损坏。
不断飙升的要求食品、新鲜水、燃料和木材有贡献到戏剧性的环境变化,从森林砍伐,空气,土地和水的污染。
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。
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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。
当7273.0=LW，5977.0p =时，此时的6n =。
Contents1 Analysis32 Model Assumptions3 3 Modeling and solving 33.1 Definition ...................................................................................................... 4 3.2 Model 1 ............................................................................................................. 4 3.3 Model2 ............................................................................................................ 10 3.4 Model3 ........................................................................................................................... 11 3.5 Model4 ........................................................................................................................... 13 4 References 15 5 Appendix 151.问题分析Analysis本文讨论了在有限的烤箱内，不同形状烤盘的外部边缘的热量的分布问题。
当烤箱内部预热到一定时间时，烤箱内温度达到一个均衡值。
由于预热的一段时间很短，我们假设在烤箱的工作时间，炉内热量分布是均匀的。
因此烤箱内的气体可以看成为温度不变的流体。
烤盘的每一条边都可以看成无限大平板在一维时的情况。
可以建立半无限大平板在第三类边界条件下的一维非稳态导热函数，并结合多维非稳态导热的乘积解法，可以得到多边形烤盘在二维的热量分布。
然后模拟出多边形烤盘热量分布的图像，通过观察，得到各种形状烤盘所受到的热量分布情况。
问题二：讨论烤箱所能够容纳烤盘数最多的情况。
实际上也就是讨论多边形在L W ⋅ 区域内的平铺问题。
在这里，我们假设L W +为定值。
一方面当LW分别为不同值时，多边形的平铺区域面积会有不同的值。
另一方面，多边形在区域L W ⋅的烤盘数量N 会随着多边形边数的变化而变化。
因此，平铺数量N 会随着LW和边数n 的变化而变化。
讨论烤盘平均热量最大的情况，实际上也就是讨论非烤焦区域面积占总区域面积比例的问题。
我们认为烤焦区域面积为温度出现重叠的区域面积。
一方面，当LW分别为不同值时，热量平均分布H 会有不同的值。
另一方面，多边形在区域L W ⋅的热量平均分布会随着多边形边数的变化而变化。
因此，热量平均分布H 会随着LW和边数n 的变化而变化。
结合以上相关结论，我们可以得到边数n 会随着热量平均分布H 和L W和数量N 变化而变化。
通过作无量纲化处理，数量N 和平均热量分布H 的权重分别为p 和()p -1，所以边数n 会随着LW和P 的变化而变化。
2.模型假设Model Assumptions1. 忽略不同食材，烘焙时间长短等因素对蛋糕成熟的影响；2. 当烤箱工作时，烤箱内的温度为定值；3. 假设烤箱内传热主要为导热传热。
3. 不同形状烤盘热量分布模型3.1烤盘，烤箱的定义本文考虑的烤箱的结构简图（Figure 1）：Figure 1 烤箱结构图本文忽略盘烤的高度，仅考虑烤盘在二维空间内的导热问题,如图2所示：Figure 2 烤盘形态图图3.2模型建立模型解决烤盘形态转变过程中所有烤盘形状热量分布的问题。
当只考虑烤盘的一条边时，此时烤盘相当于半无限大平板。
在一维非稳态传热过程中烤盘内的温度。
坐标分布如图3所示：Figure 3 半无限大平板加热过程中的温度分析由上图可知，烤盘厚度为δ时烤盘的加热情况：第一阶段step1：当烤制时间),0(2ττ∈时，空气流体不断的向烤盘内部导热，但是烤盘仍然有部分处于初始温度，未开始加热。
当2ττ=时，空气流体对烤盘的热量正好传到烤盘的内边缘；第二阶段step2：当),(42τττ∈时，空气流体对整个烤盘加热的一段时间； 第三阶段step3：当4ττ>时，烤盘的温度到达新的稳定状态。
烤盘的加热过程的微分方程[1]为：)1(22ττ∂∂=∂∂t a t其中，t 为烤盘的温度，0t 为烤盘的初始温度，f t 为空气流体的温度，且0t t f >。
f h 为空气流体与烤盘间的对流换热系数，且为常数。
τ为加热时间，δ为烤盘边缘的厚度，α为热量传输系数（或导热系数）。
定解条件：0=τ，δ≤≤x 0，0t t =0>τ，0=x ，00=∂∂=x xt （对称性）0>τ，δ=x ，()δλ===∂∂x f f x tt h xt -0引入过余温度:t t f -=θ。
在此定解条件下微分方程解的结果为：()()()()()）（2cos sin cos sin 2--10022∑∞=+⎥⎦⎤⎢⎣⎡==n n n n n n f f x s e t t t t a nδβδβδβδβδβθθδτδβ 式中的n β是下列超越方程的根，称为特征值。
()δβδβn in B =tan () 3,2,1=n从上式看出解得结果可表示为：()())3(,,-,-,000⎪⎭⎫ ⎝⎛==δτθτθx B F f t t x t t x i f f从上述的结果可知，烤盘的加热过程函数是一个无穷级数，计算工作量较大。
但对比计算表明，当傅里叶系数2.0o >F 时，采用该级数的第一项与采用完整的级数计算平板中心温度的差别小1%.0。
这样的误差在计算中是被允许的，因而当此2.0o >F 后可以采用以下简化结果：()()()()()⎥⎦⎤⎢⎣⎡+==δδβδβδβδβδβθθδτδβx e t t t t i a f f ncos cos sin sin 2--22111100 (4) 其中特征值n β() ,3,2,1=n 的值与i B 有关。
从上式可知得当2.0o >F 以后平板中的任意一点的过余温度()τθ,x 与平板中心的过余温度()()τθτθm x =,之比为：()⎥⎦⎤⎢⎣⎡=δδβθθx m 1cos (5) 非稳态导热的这一阶段就是所谓的导热正规状况或充分发展阶段。
确认正规状况阶段的存在具有重要的意义，因为本文计算中关心的非稳态导热过程常常处于正规状况阶段，此时的计算可以采用上述的简化公式。
为了便于计算，人们广泛采用按分析解的级数第一项而绘制的一些线算图（诺曼图）。
其中用以确定温度分布的线算图称为海斯勒（Heasler ）图。
以无限大平板为例，它首先根据等式（4）中给出的0m θθ随O F 及i B 变化的曲线（此时0x=δ），然后再根据等式（5）确定m θθ的值。
于是平板中任意一点的0θθ值便为：mm 0θθθθθθ= (6) 无限大平板的0mθθ和mθθ的计算图[2]如图4和图5所示：Figure 4 无限大平板中心无量纲温度图Figure 5无限大平板的mθθ曲线图3.3模型求解设烤盘密度3/10.26m kg =ρ，比热容)./(904C kg J c =，导热率)/(120C m W ⋅=λ，对流换热系数)/(1002C m W h ⋅=，烤盘的宽度m 5.0=δ，烤箱内的温度C t f 200=。
当时间s 10=τ时，根据图4和图5和等式(6)得到若干大平板的温度和大平板距离的散点数据，拟合出大平板的温度和大平板距离的曲线如图6所示：Figure 6 大平板的温度和大平板距离的拟合曲线3.4四边形烤盘情况烤盘形状为四边形的受热情况：Figure 7 烤盘形状为四边形的受热图四边形的烤盘可以看做成由四个半无限大平板所围成的，根据多维非稳态导热的乘积解法可以得出如下结果：()()()()()board0board 0board 0board 004321--y,--,--y,--,--,,δδδδτττττθ⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫⎝⎛ff f f f f fff f t t t t t t t x t t t t t t t t x t tt t y x (7) 图像如图8所示：Figure 8 四边形烤盘的热量分布图3.5 五边形烤盘情况烤盘形状为五边形的受热情况：Figure 9烤盘形状为五边形的受热图五边形的烤盘可以看做成由五个半无限大平板所围成的，根据多维非稳态导热的乘积解法可以得出如下结果：()()()()board0board 0board 00521--,--,--,--,,δδδττττθ⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫ ⎝⎛ff f f ff f f t t t x t t t t y t t t t x t t t t y x (8) 图像如图10所示：Figure 10五边形烤盘的热量分布图3.6 多边形烤盘情况烤盘形状为n 边形的受热情况：：Figure 11烤盘形状为n 边形的受热图n 边形的烤盘可以看做成由n 个半无限大平板所围成的，根据多维非稳态导热的乘积解法可以得出如下结果：()()()()board0board 0board 00n 21--,--,--,--,,δδδττττθ⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛=⎪⎪⎭⎫⎝⎛ff f f fff f t t t x t t t t y t t t t x t tt t y x (9) 图像如图12所示：Figure 12多边形烤盘的热量分布图4烤盘数量最优模型当用相同多的材料做成烤箱时，存在以下等式：ont C W L =+式中，L 为烤箱的长度，W 为烤箱的宽度，ont C 为常数。