2013美国大学生数学建模竞赛题目和翻译和A题图解
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
1.热力学模型：
1.1热力学内容分析：主要考虑热传导问题，热辐射。
更深入还需考虑：蛋糕和盘子之间的热量传递，烤盘之间热量传递的量化分析，参数计算等等。
1.2过度形状的如何优化选择
1.3偏微分方程的解法（这里要考虑到过度形状的选择），主要软件有annsy，pde等等。
对偏微分方程数值解的误差分析，收敛性分析证明等等。
1.4均匀化程度的量化指标，以及如何改进软件实现各种过度形状的并行计算。
2.平面装箱问题（这里的处理需要联系到热力学模型的烤盘间热量传递分析）：
2.1.矩形的装箱处理（这里估计需要简化，如何实现一般的横竖放置处理是一个难点）
2.2.椭圆形的装箱处理
2.3,过度形状处理
3.组合权重模型主要是解决两个核心问题：
3.1考虑权重后的一般评价函数的确定（考虑非线性处理）
3.2不同烤箱形状对最终结果影响的一般分析。
4.其他
4.1关于2和3的灵敏度分析（注意到关于烤箱形状，N未必一定是稳定的。
因此如何对其作出一般的灵敏度考量是一个关键问题）
4.2广告的处理（主要问题是广告的一般性主旨是宣传，但是又要加上模型的结果但是不能直接标出参数，这两个之间的权衡也十分重要）。
2013mcm题目
2013 Contest ProblemsMCM PROBLEMSPROBLEM A: The Ultimate Brownie PanWhen 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 with respect 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 valuesof 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.PROBLEM B: Water, Water, EverywhereFresh water is the limiting constraint for development in much of the world. Build a mathematical model for determining an effective, feasible, and cost-efficient water strategy for 2013 to meet the projected water needs of [pick one country from the list below] in 2025, and identify the best water strategy. In particular, your mathematical model must address storage andmovement; de-salinization; and conservation. If possible, use your model to discuss the economic, physical, and environmental implications of your strategy. Provide a non-technical position paper to governmental leadership outlining your approach, its feasibility and costs, and why it is the “best water strategy choice.”Countries: United States, China, Russia, Egypt, or Saudi Arabia。
2013年数学建模美赛题目中文翻译_共4页
Problem c:背景：社会正致力于运用和开发模型来预测地球的生物和环境情况。
很多科学研究总结了逐渐增长的地球环境和生物系统压力，但很少有人用全球范围的模型来检测这些观点。
联合国发表的千年生态系统评估综合报告发现：近三分之二的地球生命支持生态系统——包括净水，洁净的空气，稳定的气候——正在因非可持续性使用而逐渐衰减。
其中大部分破坏归咎于人类行为。
暴增的对于食物，淡水，燃料，木材的需求导致了剧烈的环境变化；从森林砍伐到空气，土壤和水污染。
尽管已存在大量关于局部习惯和地区因素的研究，目前的模型还不能告知决定人他们的局部策略是如何影响整个地球的健康的。
许多模型忽略了复杂的全球因素，这些模型无法判断重大政策的长期影响。
尽管科学家们意识到巨大环境和生物系统中存在的复杂关系和交叉作用，当前的模型通常忽略这些管理或限定了系统间的影响。
系统的复杂性体现在多元交互（多个元素的相关性），反馈，突发行为，即将发生的状态变化或触发点。
最近的自然杂志中一篇由22位国际知名科学家撰写的题为“迫近地球生物圈的状态变化”的文章讨论了许多有关科学模型对于预测行星健康系统潜在状态变化的重要性与必需性。
文章提供了两种具体定性的模型，并寻求更好的预测模型：1）通过在全球模型中加入相关系统的复杂性（包括局部情况对全球系统的影响，反之亦然）来优化生物状态预测。
2）辨别不同因素在产生非健康全球状态变化中的作用并展示如何运用有效的生态系统管理来预防或限制这些即将发生的状态变化。
研究最终归结于问题：我们是否能利用全球健康的局部或地区性组成部分预测潜在状态变化来帮助决策者制定基于对全球健康状况潜在影响的，有效的策略。
尽管有越来越多的警示信号出现，没人知道地球是否确实在接近全球性的转折点（极端状态），这种极端的状态是否是不可避免的。
自然杂志等研究指出了地球生态系统中的一些重要工作元素。
（例如：局部因素，全球变化，多维元素与关系，变化的时间与空间范围）。
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。
美国大学生数学建模竞赛试题AB题中文
A 题热水澡一个人进入浴缸洗澡放松。
浴缸的热水由一个水龙头放出。
然而浴缸不是一个可以水疗泡澡的缸，没有辅助加热系统和循环喷头，仅仅就是一个简单的盛水容器。
过一会，水温就会显著下降。
因此必须从热水龙头里面反复放水以加热水温。
浴缸的设计就是当水达到浴缸的最大容量，多余的水就会通过一个溢流口流出。
做一个有关浴缸水温的模型，从时间和地点两个方面来确定在浴缸中泡澡的人能采用的最佳策略，从而泡澡过程中能保持水温并在不浪费太多水的情况下使水温尽量接近最初的水温。
用你的模型来确定你的策略多大程度上依赖于浴缸的形状和容量，浴缸中的人的体型/体重/体温，以及这个人在浴缸中做出的动作。
如果这个人在最开始放水的时候加入了泡泡浴添加剂，这将会对你的模型结果有什么影响？要求提交一页MCM的总结，此外你的报告必须包括一页给浴缸用户看的非技术性的解释，其中描述了你的策略并解释了在泡澡过程中为什么保持平均的水温会非常困难。
B题太空垃圾地球轨道周围的小碎片的数量受到越来越多的关注。
据估计，目前大约有超过50万片太空碎片被视为是宇宙飞行器的潜在威胁并受到跟踪，这些碎片也叫轨道碎片。
2009年2月10号俄罗斯卫星科斯莫斯-2251与美国卫星iridium-33相撞的时候，这个问题在新闻媒体上就愈发受到广泛讨论。
已经提出了一些方法来清除这些碎片。
这些方法包括小型太空水流喷射器和高能量激光来瞄准具体的碎片，还有大型卫星来清扫碎片等等。
这些碎片数量和大小不一，有油漆脱离的碎片，也有废弃的卫星。
碎片高速转动使得定位清除变得困难。
建一个随时间变化的模型来确定一个最佳选择或组合的选择提供给一家私人公司让它以此为商业机遇来解决太空碎片问题。
你的模型应该包括对成本、风险、收益的定量和/或定性分析以及其他重要因素的分析。
你的模型应该既能够评估单个的选择也能够评估组合的选择，且能够探讨一些重要的”what if ”情景。
用你的模型来确定是否存在这样的机会，在经济上很有吸引力；或是根本不可能有这样的机会。
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。
2013年美赛数模A题翻译
2013 Contest ProblemsMCM PROBLEMSPROBLEM A: The Ultimate Brownie PanWhen 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 with respect 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 valuesof 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.终极蛋糕烤箱当用矩形烤盘烘焙时，热量集中在4个角落，在角落的食物容易被烤糊（边上的热量较少），在一个圆形烤盘上，热量均匀的分布在盘的整个边缘，而且分布在边缘的食物不会被过度加热。