2013年美赛A题
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。
【免费下载】美赛模拟题
2013年美赛模拟题PROBLEM A: Who Moved the Presidential Candidate's Cheese? New media is rapidly becoming an important part of a presidential candidate's media strategy. The use of the new media by most of the candidates had been unprecedentedly frequent and overwhelming in 2008 American president election cycle. It is most believed that U.S President Barack Obama, who had massively used new media as one of his campaign strategies, not only won the election, but also drawn the world's attention and gained himself the reputation as a "Internet President".Requirement 1: Model the effectiveness of the New media for president election.Requirement 2:Verify your model.Requirement 3: Suggestion for the presidential candidate.PROBLEM B: Air Politics in BeijingOne day last month, the reading was so high compared with the standards set by the U.S. Environmental Protection Agency that it was listed as "beyond index." In other words,it had soared right off the chart. But China's own assessment that day, Oct. 9, was that Beijing's air was merely "slightly polluted."NASA:PM2.5 MAPNot even the most fervent propagandist would call thecity's air clean, but the Chinese government made great efforts to improve air quality for the 2008 Olympic Summer Games. Beijing authorities moved huge steelworks out of the capital, switched city dwellers from coal to natural gas heating, raised emissions standards for trucks, and creatednew subway and bus lines. The cost of the cleanup was estimated at $10 billion, not including the investment in mass transit.Three years later, the difference between the Americans and the Chinese is at least in part about what they're measuring. And it highlights the rapid growth in the number of cars in Beijing.Chinese monitoring stations around the capital track large particulates of up to 10 micrometers. The number of those particles has dropped as a result of reforestation programs that lessen the dust storms that blew in from deserts. The Chinese have also been successful in reducing sulfur dioxide emissions by limiting coal heating and imposing stricter emissions standards.The U.S. monitor tracks tinier particles — less than 2.5 micrometers — that physicians say are capable of penetrating human lungs and other organs. Car and truck exhaust is a major source of fine particulate pollution, a particular problem in Beijing, where the number ofregistered cars has skyrocketed from to 5 million from 3.5 million in 2008.The embassy of US in Beijing installed its monitor in 2008 before the Olympics to advise its personnel about air quality, but then decided it should make readings public under diplomatic rules that require that information regarding health and security risks be made available.The measurements drew widespread attention last November, the first time a reading for fine particulate matter went above 500 micrograms per cubic meter, about seven times the U.S. standard for "acceptable" air quality.The expert suggests that its problem is political as well as technical. Politically, it is hard to revise the standard. After 10 years of saying things are getting better and better, if you reverse that, people will be justifiably angry. Although it is government data that are published in newspapers and broadcast on television, other Chinese media are increasingly citing the U.S. figures. Please Survey data and writing papers.。
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。
历届美国数学建模竞赛赛题(汉语版)
历届美国数学建模竞赛赛题, 1985－2006AMCM1985问题-A 动物群体的管理AMCM1985问题-B 战购物资储备的管理AMCM1986问题-A 水道测量数据AMCM1986问题-B 应急设施的位置AMCM1987问题-A 盐的存贮AMCM1987问题-B 停车场AMCM1988问题-A 确定毒品走私船的位置AMCM1988问题-B 两辆铁路平板车的装货问题AMCM1989问题-A 蠓的分类AMCM1989问题-B 飞机排队AMCM1990问题-A 药物在脑内的分布AMCM1990问题-B 扫雪问题AMCM1991问题-A 估计水塔的水流量AMCM1992问题-A 空中交通控制雷达的功率问题AMCM1992问题-B 应急电力修复系统的修复计划AMCM1993问题-A 加速餐厅剩菜堆肥的生成AMCM1993问题-B 倒煤台的操作方案AMCM1994问题-A 住宅的保温AMCM1994问题-B 计算机网络的最短传输时间AMCM1995问题-A 单一螺旋线AMCM1995问题-B A1uacha Balaclava学院AMCM1996问题-A 噪音场中潜艇的探测AMCM1996问题-B 竞赛评判问题AMCM1997问题-A Velociraptor(疾走龙属)问题AMCM1997问题-B为取得富有成果的讨论怎样搭配与会成员AMCM1998问题-A 磁共振成像扫描仪AMCM1998问题-B 成绩给分的通胀AMCM1999问题-A 大碰撞AMCM1999问题-B “非法”聚会AMCM1999问题- C 大地污染AMCM2000问题-A空间交通管制AMCM2000问题-B: 无线电信道分配AMCM2000问题-C：大象群落的兴衰AMCM2001问题- A: 选择自行车车轮AMCM2001问题-B：逃避飓风怒吼（一场恶风…）AMCM2001问题-C我们的水系-不确定的前景AMCM2002问题-A风和喷水池AMCM2002问题-B航空公司超员订票AMCM2002问题-C蜥蜴问题AMCM2003问题-A: 特技演员AMCM2003问题-C航空行李的扫描对策AMCM2004问题-A：指纹是独一无二的吗？AMCM2004问题-B：更快的快通系统AMCM2004问题-C：安全与否？AMCM2005问题-A：.水灾计划AMCM2005问题-B：TollboothsAMCM2005问题-C：.Nonrenewable ResourcesAMCM2006问题-A：用于灌溉的自动洒水器的安置和移动调度AMCM2006问题-B：通过机场的轮椅AMCM2006问题-C：在与HIV/爱滋病的战斗中的交易AMCM85问题-A 动物群体的管理在一个资源有限，即有限的食物、空间、水等等的环境里发现天然存在的动物群体。
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美国数学竞赛12年级试题A
INSTRUCTIONS1. DO NOT OPEN THIS BOOKLET UNTIL YOUR PROCTOR TELLS YOU.2. This is a twenty-five question multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.3. Mark your answer to each problem on the AMC 12 Answer Form with a #2 pencil. Check the blackened circles for accuracy and erase errors and stray marks completely. Only answers properly marked on the answer form will be graded.4. SCORING: You will receive 6 points for each correct answer, 1.5 points for each problem left unanswered, and 0 points for each incorrect answer.5. No aids are permitted other than scratch paper, graph paper, rulers, compass, protractors, and erasers. No calculators are allowed. No problems on the test will require the use of a calculator.6. Figures are not necessarily drawn to scale.7. Before beginning the test, your proctor will ask you to record certain information on the answer form.8. 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Society for Industrial and Applied Math (SIAM) 1.Square ABCD has side length 10.Point E is on BC ,and the area of ABE is 40.What is BE ?(A)4(B)5(C)6(D)7(E)8E 2.A softball team played ten games,scoring 1,2,3,4,5,6,7,8,9,and 10runs.They lost by one run in exactly five games.In each of their other games,they scored twice as many runs as their opponent.How many total runs did their opponents score?(A)35(B)40(C)45(D)50(E)553.A flower bouquet contains pink roses,red roses,pink carnations,and red carna-tions.One third of the pink flowers are roses,three fourths of the red flowers are carnations,and six tenths of the flowers are pink.What percent of the flowers are carnations?(A)15(B)30(C)40(D)60(E)704.What is the value of 22014+2201222014−22012?(A)−1(B)1(C)53(D)2013(E)240245.Tom,Dorothy,and Sammy went on a vacation and agreed to split the costs evenly.During their trip Tom paid $105,Dorothy paid $125,and Sammy paid $175.In order to share the costs equally,Tom gave Sammy t dollars,and Dorothy gave Sammy d dollars.What is t −d ?(A)15(B)20(C)25(D)30(E)356.In a recent basketball game,Shenille attempted only three-point shots and two-point shots.She was successful on20%of her three-point shots and30%of her two-point shots.Shenille attempted30shots.How many points did she score?(A)12(B)18(C)24(D)30(E)367.The sequence S1,S2,S3,...,S10has the property that every term beginningwith the third is the sum of the previous two.That is,S n=S n−2+S n−1for n≥3.Suppose that S9=110and S7=42.What is S4?(A)4(B)6(C)10(D)12(E)168.Given that x and y are distinct nonzero real numbers such that x+2x =y+2y,what is xy?(A)14(B)12(C)1(D)2(E)49.In ABC,AB=AC=28and BC=20.Points D,E,and F are on sides AB,BC,and AC,respectively,such that DE and EF are parallel to AC and AB, respectively.What is the perimeter of parallelogram ADEF?(A)48(B)52(C)56(D)60(E)7210.Let S be the set of positive integers n for which1n has the repeating decimalrepresentation0.ab=0.ababab...,with a and b different digits.What is the sum of the elements of S?(A)11(B)44(C)110(D)143(E)15511.Triangle ABC is equilateral with AB =1.Points E and G are on AC and points D and F are on AB such that both DE and F G are parallel to BC .Furthermore,triangle ADE and trapezoids DF GE and F BCG all have the same perimeter.What is DE +F G ?(A)1(B)32(C)2113(D)138(E)5312.The angles in a particular triangle are in arithmetic progression,and the side lengths are 4,5,and x .The sum of the possible values of x equals a +√b +√c ,where a ,b ,and c are positive integers.What is a +b +c ?(A)36(B)38(C)40(D)42(E)4413.Let points A =(0,0),B =(1,2),C =(3,3),and D =(4,0).QuadrilateralABCD is cut into equal area pieces by a line passing through A .This line intersects CD at point (pq ,r s ),where these fractions are in lowest terms.What is p +q +r +s ?(A)54(B)58(C)62(D)70(E)7514.The sequencelog 12162,log 12x,log 12y,log 12z,log 121250is an arithmetic progression.What is x ?(A)125√3(B)270(C)162√5(D)434(E)225√615.Rabbits Peter and Pauline have three offspring—Flopsie,Mopsie,and Cotton-tail.These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child.It is not required that every store gets a rabbit.In how many different ways can this be done?(A)96(B)108(C)156(D)204(E)37216.A ,B ,and C are three piles of rocks.The mean weight of the rocks in A is 40pounds,the mean weight of the rocks in B is 50pounds,the mean weight of the rocks in the combined piles A and B is 43pounds,and the mean weight of the rocks in the combined piles A and C is 44pounds.What is the greatest possible integer value for the mean in pounds of the rocks in the combined pilesB andC ?(A)55(B)56(C)57(D)58(E)5917.A group of 12pirates agree to divide a treasure chest of gold coins among themselves as follows.The k th pirate to take a share takes k 12of the coins that remain in the chest.The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins.How many coins does the 12th pirate receive?(A)720(B)1296(C)1728(D)1925(E)385018.Six spheres of radius 1are positioned so that their centers are at the vertices of a regular hexagon of side length 2.The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon.An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere.What is the radius of this eighth sphere?(A)√2(B)32(C)53(D)√3(E)219.In ABC ,AB =86,and AC =97.A circle with center A and radius AB intersects BC at points B and X .Moreover BX and CX have integer lengths.What is BC ?(A)11(B)28(C)33(D)61(E)7220.Let S be the set {1,2,3,...,19}.For a,b ∈S ,define a b to mean that either 0<a −b ≤9or b −a >9.How many ordered triples (x,y,z )of elements of S have the property that x y ,y z ,and z x ?(A)810(B)855(C)900(D)950(E)98821.ConsiderA =log (2013+log (2012+log (2011+log(···+log (3+log 2)···)))).Which of the following intervals contains A ?(A)(log 2016,log 2017)(B)(log 2017,log 2018)(C)(log 2018,log 2019)(D)(log 2019,log 2020)(E)(log 2020,log 2021)22.A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10with no leading zeros.A 6-digit palindrome n is chosen uniformly at random.What is the probability that n 11is also a palindrome?(A)825(B)33100(C)720(D)925(E)113023.ABCD is a square of side length √3+1.Point P is on AC such that AP =√2.The square region bounded by ABCD is rotated 90◦counterclockwise with center P ,sweeping out a region whose area is 1c (aπ+b ),where a ,b ,and c are positive integers and gcd(a,b,c )=1.What is a +b +c ?(A)15(B)17(C)19(D)21(E)2324.Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon.What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?(A)553715(B)443572(C)111143(D)81104(E)22328625.Let f :C →C be defined by f (z )=z 2+iz +1.How many complex numbersz are there such that Im(z )>0and both the real and the imaginary parts of f (z )are integers with absolute value at most 10?(A)399(B)401(C)413(D)431(E)441WRITE TO US!Correspondence about the problems and solutions for this AMC 12and orders for publications should be addressed to:American Mathematics CompetitionsUniversity of Nebraska, P .O. Box 81606Lincoln, NE 68501-1606Phone 402-472-2257 | Fax 402-472-6087 | amcinfo@The problems and solutions for this AMC 12 were prepared by the MAA’s Committee on theAMC 10 and AMC 12 under the direction of AMC 12 Subcommittee Chair:Prof. Bernardo M. Abrego2013 AIMEThe 31st annual AIME will be held on Thursday, March 14, with the alternate on Wednesday, April 3. It is a 15-question, 3-hour, integer-answer exam. You will be invited to participate only if you score 120 or above or finish in the top 2.5% of the AMC 10, or if you score 100 or above or finish in the top 5% of the AMC 12. T op-scoring students on the AMC 10/12/AIME will be selected to take the 42nd Annual USA Mathematical Olympiad (USAMO) on April 30-May 1, 2013. The best way to prepare for the AIME and USAMO is to study previous exams. Copies may be ordered as indicated below.PUBLICATIONSA complete listing of current publications, with ordering instructions, is at our web site: American Mathematics Competitions。
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最终的布朗尼锅摘要关键字：目录引言题目背景近年来，电烤箱普遍采用远红外加热技术，使电烤箱的技术含量增加，耗能降低，深受广大用户的欢迎。
利用红外线加热物体，就是利用辐射波长与物体接收波长一致时，物体吸收大量的红外能，从而加剧物体内部的分子运动，使之加热升温。
加热时间短，能耗低，使用方便。
但是，当我们使用矩形烤盘烘烤食物时，热传导方程，加上一些边界条件，导致方形烤盘热量集中在的四个角上，因此四个角上的物体会因过度受热（以及在较小程度的边缘处）而变焦。
如果用圆形烤盘，热量会平均分布在整个外围边缘，在外围的物体就不会过度受热。
然而，由于大多数的烤箱都是矩形的，所以用圆形的烤盘就不能较好的利用烤箱的空间。
给烘烤食物的朋友带来了很大的不便。
为什么角部的食物肉容易烤焦，以及选择哪种形状的烤盘，，这是令人很费解的问题。
电烤箱工作原理电烤箱利用电热元件所发出的辐射热来烘烤食品，利用它我们可以制作烤鸡、烤鸭、烘烤面包、糕点等。
根据烘烤食品的不同需要，电烤箱的温度一般可在50-250℃范围内调节。
电烤箱主要由箱体、电热元件、调温器、定时器和功率调节开关等构成。
其箱体主要由外壳、中隔层、内胆组成三层结构，在内胆的前后边上形成卷边，以隔断腔体空气；在外层腔体中充填绝缘的膨胀珍珠岩制品，使外壳温度大大减低；同时在门的下面安装弹簧结构，使门始终压紧在门框上，使之有较好的密封性。
电烤箱的加热方式可分为面火（上加热器加热）、底火（下加热器加热）和上下同时加热三种。
电烤箱技术参数温度范围室温-200℃（300℃）温度稳定度±0.5℃温度分布均匀度±2℃（特佳）排气烟道叶片式设计可调出风量符号和定义l：多边形边长L: 多边形周长k: 周长与面积的比G：单位圆的周长C: 单位椭圆周长a: 椭圆的长半轴b: 椭圆的短半轴假设1、烤箱内温度同一层表分布均匀且稳定2、烤箱内风扇使空气及时流通3、假设烤盘之间相互不影响4、假设各层之间相互不影响5、假设烤盘的深度影响忽略不计6、假设烤箱内垂直分布的热辐射场为递增针对问题1问题1的说明针对问题1，考虑同一层烤架上温度稳定且分布均匀，我们提出了两个模型。
从两个方面验证了烤盘拐角处食物易焦的原理，并画出了不同形状的烤盘外边缘热量分布图。
模型I的提出为了更好的说明为什么角部的食物容易烤焦，我们建立了（从长方形到圆形）图形边数与顶角热量聚集程度之间关系的数学模型。
首先，假设各个图形面积都相同，且为单位1；为通过对比从长方形到圆形边角角度的变化，来引出热量在顶角处聚集程度的变化，最后通过烤盘外边缘处热量的分布，说明了在烤盘角部的食物容易烤糊的原理。
此外，考虑到方形烤盘容易使食物变焦，而圆形烤盘不易使之变焦，会想到食物与空气的接触面积和周长有关。
我们证明了相同底面截面积条件下，长方形的周长比正方形、圆及其他多边形周长大的事实。
因此，我们提出了矩形烤盘四角中食物变焦原因是周长和面积的比值太大的理论模型。
若使食物不易变焦，就要想办法使之与空气的接触面积减小，即减小横截面的周长。
这从另一个方面说明了此现象出现的原因。
模型I的建立与求解当我们用矩形烤盘烘烤食物时，由于热量集中在烤盘的四个角上，因此四个角上的物体会因过度受热（以及在较小程度的边缘处）而变焦。
如果用圆形烤盘，热量会平均分布在整个外围边缘，在外围的物体就不会过度受热。
然而，由于大多数的烤箱都是矩形的，所以用圆形的烤盘就不能较好的利用烤箱的空间。
给烘烤食物的朋友带来很大的不便。
因此建立了一个定面积求周长的模型来分析热量在圆形和矩形及它们之间的其它形状的烤盘外边缘的分布。
由于食物接收到的热量与食物和空气接触的面积有关，比较小的部位和边缘易烤焦，现在建立的模型，使烤盘的周长L与面积S的比最小，说明食物接收的热量较均匀。
考虑到烤盘的形状较多为规则多边形、椭圆，对单位面积的烤盘形状对热量的影响讨论如下：1、单位面积的正多边形：设多边形边长为l,共有n条边，将正对变形分成n个等腰三角形。
有(1)(2)式可得每个三角形的面积为：其中单位面积求多边形周长的公式为：4tan()n L nπ=(1.1)可得计算边长与面积比的关系式：(1.2)2、单位面积的圆：对单位面积的各多边形烤盘求（从四边形到圆之间的正多边形）的周长与面积比的MATLAB 程序作图如下：501001502002503.53.553.63.653.73.753.83.853.93.954 0501001502002503.53.553.63.653.73.753.83.853.93.954周长/面积图1：各多边形的周长 图2：各多边形及圆周长与面积的比值 周长与面积的比值结果分析：当多边形的边长越多时，周长无限接近于圆。
即相同面积的烤盘，圆形时周长最短。
周长与面积比最小，烤盘上的热量越均匀，实物越不易烤焦。
3、单位面积的椭圆：已知椭圆周长和面积的计算公式：S ab π=(1.3)24()C ab a b π=+-(1.4)单位面积椭圆周长的计算公式： 22221x y a b+= (1.5)1S ab π==(1.6)将(1.7)、(1.8)式代人式(1.9)求偏倒并令之等于零得： 20.75(2)b ππ=≈-(1.10)120.432a bπππ-==≈ (1.11)由于a b ≥显然不符合逻辑。
所以当10.56a b π==≈时，椭圆演变为圆，此时仍可说明圆是使食物不易被烤焦的最佳选择，但考虑到圆形烤盘和椭圆烤盘在一般矩形烤架内空间利用率不大，所以我们想到了圆角矩形。
4、圆角矩形：考虑到矩形在烤箱中的利用率比圆形高，但由于其形状的原因对食物的烘烤出的质量的影响我们可以考虑结合矩形与圆的优缺，避免各子的缺点，可已将二者结合将矩形的四角变化成1/4圆弧从而可以很好的利用二者各自的优点，从而达到最优。
模型II 的提出在模型I 中，说明了食物在烤盘角处变焦的原因，但从本质上讲，食物通过烤箱加热管释放的辐射能加热，再通过热传导使整个食物变熟。
因此食物在烤箱中接收辐射热的多少决定了食物的成熟度。
因此，研究烤盘周边的热量分布成为了必要。
我们采用无约束优化模拟烤盘内热量的分布。
模型II 的建立与求解 圆我们采用了两种优化方法来模拟烤盘内热量的分布：最速下降法和牛顿法。
最速下降法又称为梯度法，它是以负梯度方向的极小化算法，相邻两次的搜索方向是相互直交的寻找目标函数法 。
牛顿法利用目标函数()f x 在迭代点k x 处的Taylor 展开式作为模型函数，并利用这个二次模型函数的极小点序列去逼近目标函数的极小点。
设无约束目标函数为周长与面积的比值2()2()(4)x y x y A f Aπ+---=(1.12)其中1A =，1:n f R R →一阶连续可微。
基本思想是从当前k x 出发，取函数()f x 在点k x 处下降最快的方向作为我们的搜索方向k p 。
由()f x 的Taylor 展开式知()()()()k k k k Tk k f x f x tp t f x p tp σ-+=-∇+(1.13)略去t 的高阶无穷小项目不计，可见取()k k p f x =-∇时，函数值下降的最多。
于是，我们建立了最速下降法和牛顿法的模型来模拟矩形烤盘。
结果与分析：MAYLAB 试验程序运行结果如图3： 102030405060102030405020406080100120x 12 x 1+...-1/8388608 (241620187839069 x 1 x 2-241620187839069)1/2x 2图3 烤盘温度场的分布 由图3知，边角的食物在加热过程中会接收较多的辐射热，因此从食物接收热量多少的本质上揭示了边角食物易焦的原因。
模型的优缺点：当今尚无标准的椭圆周长计算公式是基础科学中的遗憾之一，现在科学中所使用的椭圆周长都是近似值，这是科学的遗憾之一，也是模型一的缺点之一。
对问题2的说明广告表 优缺点 参考文献文献：/view/e3d7d21dc5da50e2524d7f80.html 人教版圆的面积ppt.pdf/view/e179e54fc850ad02de8041db.html圆的面积教学课件.pdf/view/f4ef9e0c4a7302768e99391c.html热辐射计算公式《精通MATLAB科学计算》王正林龚纯何倩编著电子工业出版社Publishing House of Electronics Industry/view/08ec1c48767f5acfa0c7cd05.html 电烤箱工作原理以及构造建立一个模型分析热量在不同形状烤盘外围边缘的分布- 矩形、圆形以及之间的过渡形状假设下列条件1. 烤箱的宽、长比是W/L,并且是矩形的2. 每一个烤盘的面积都是A3. 初始有烤箱两层烤架，并且在空间里平均分布分别建立一个模型在以下条件中选择最优的烤盘形状1. 最大化烤箱内能容纳的烤盘数量2. 最大化每个烤盘的平均受热量3. 以权重p和(1-p)综合考虑条件(1)和(2)，指出结果是如何随W/L和p变化除了你的解答，准备一页或两页广告宣传你的烤箱，强调你的设计与结果PROBLEM A: The Ultimate Brownie PanWhen baking in a rectangular pan heat is concentrated in the 4 cornersand the product gets overcooked at the corners (and to a lesser extent atthe edges). In a round pan the heat is distributed evenly over the entireouter edge and the product is not overcooked at the edges. However,since most ovens are rectangular in shape using round pans is notefficient with respect to using the space in an oven.Develop a model to show the distribution of heat across the outer edgeof a pan for pans of different shapes - rectangular to circular and othershapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular inshape.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.。