最新年美赛A题H奖58265390

数学建模美赛2024题目全文共四篇示例，供读者参考第一篇示例：今年的题目是关于气候变化和环境保护的议题。

题目涉及到了全球变暖对气候和环境的影响，以及如何通过有效的政策和措施来减缓这种影响。

参赛者需要结合大量的气象数据、环境数据和经济数据，建立数学模型来分析不同政策对环境的影响，并提出具体的政策建议。

题目要求参赛者首先了解全球变暖的背景和影响，包括气候变化对冰川、海平面和生态系统的影响。

然后需要收集大量的数据，包括气温、降水、二氧化碳排放量等信息，建立数学模型来模拟气候变化的趋势和影响。

在此基础上，参赛者需要分析不同政策对气候和环境的影响，比如减排政策、再生能源政策、森林保护政策等。

最终，他们需要提出具体的政策建议，用数学模型来验证这些政策的有效性和可行性。

这道题目不仅考验参赛者的数学建模能力，还要求他们具备丰富的跨学科知识和分析能力。

参赛者需要深入了解气候变化和环境问题的本质，同时还需要掌握大量的数据处理和模型建立技巧。

他们需要运用数学、统计学、计算机科学等知识，同时还要具备创新思维和团队合作能力。

通过参与这项挑战性的比赛，大学生们不仅可以提升自己的数学建模能力，还可以培养跨学科的综合能力和团队合作精神。

这对于他们未来从事科研、工程或管理等领域的工作都将大有裨益。

这也是一次展示自己才华和创造力的绝佳机会，可以让他们在学术界和工业界获得更多的认可和机会。

2024年美国大学生数学建模竞赛的题目涉及到了气候变化和环境保护这一全球性议题，要求参赛者建立数学模型来分析不同政策对环境的影响，并提出具体的政策建议。

这是一项极具挑战性和实践意义的比赛，将为参赛者提供一个全面发展和展示自己才华的平台。

希望所有参赛者都能在这场比赛中收获满满的成绩和经验！第二篇示例：2024年美国大学生数学建模竞赛（MCM/ICM）是一个全球性的高水平数学建模比赛。

在这个比赛中，参赛队伍需要在72小时内利用自己的数学建模技能解决提出的真实世界问题。

2022年第十二届MathorCup 高校数学建模挑战赛题目A 题 大规模指纹图像检索的模型与实现在生物特征识别领域，指纹作为最具独特性与持久性的生物特征之一，被广泛应用于身份识别。

指纹识别过程分为特征提取和比对两个环节。

其中特征提取环节会提取用于指纹识别的指纹特征，一般国际上最为常见的指纹特征为“细节点”特征，其可视化展示形式如图1中的浅蓝色小圆圈及对外伸出的浅蓝色短线段，短线段用于指示细节点处纹线方向。

细节点一般采用三元存储格式： ，分别表示x 轴像素坐标、y 轴像素坐标及细节点方向。

一般而言：（1）指纹图像坐标体系：左上角为坐标原点，且x 轴方向向右，y 轴方向向下；（2）细节点表达约定：细节点x , y 的位置采用指纹图像坐标系表达，其方向规定：零度方向为x 轴正方向（向右），90度方向为y 轴负方向（向上），180度方向为x 轴负方向（向左），270度方向为y 轴正方向（向下），最大角度为359度。

角度的最小区分单位为1度。

图1 指纹识别原理(,,)x y q在指纹匹配环节，需要对两幅指纹图像的“同一性”进行定量评价，通常采用相似度指标。

常见的两枚指纹之间的相似度评价主要依据每枚指纹图像中各个细节点之间的匹配关系。

如图1所示，相互具有匹配关系的细节点之间用一根跨越两幅图像的红线将其互相连接，用于可视化展示。

在指纹图像匹配环节，常需要考虑如下的情况：考虑到在采集指纹图像时，手指按压图像采集设备的角度、轻重及位置各不相同，因此两幅指纹图像需要做图像的旋转、平移后才能相互对准。

由于手指皮肤较为柔软，通过按压方式采集到的指纹图像会发生一定程度的不规则弹性形变，在图1中会发现两幅指纹图像中，某些相互匹配的细节点在对准时，不能完全“重叠”，有一定幅度的位置及角度的偏差。

这一现象也可以从“跨越两幅图像的红线并不是都平行”现象中观察到。

考虑到手指可能存在临时性蜕皮、褶皱等因素，且空气中的湿度及皮肤表面的干燥程度或粘附在皮肤上的异物等都会导致采集到的指纹图像存1中可以观察到并不是所有的细节点都有对应的红线进行关联。

2019美赛数学建模题目【原创实用版】目录1.2019 美赛数学建模题目概述2.题目分类与解析3.题目解答方法与技巧4.2019 美赛数学建模题目的挑战与启示正文【2019 美赛数学建模题目概述】美国大学生数学建模竞赛（MCM，Mathematical Contest in Modeling）是由美国数学及其应用联合会主办，最高等级的国际性数学建模竞赛，也是世界范围内最具影响力的数学建模竞赛，一般指数学建模竞赛。

2019 年美赛数学建模题目共有 6 道题目，分别是 A、B、C、D、E、F 题，难度各异，涉及多个领域。

【题目分类与解析】A 题：供水管道的设计与优化题目要求根据给定的供水网络，设计一种方案以满足未来一定时间内的用水需求，同时保证供水成本最低。

主要涉及图论、最短路径、网络流等知识。

B 题：无人驾驶汽车的路径规划题目要求设计一种算法，使得无人驾驶汽车能够在繁忙的城市道路上安全、高效地行驶。

主要涉及图论、最短路径、动态规划等知识。

C 题：社交网络分析题目要求分析给定的社交网络数据，挖掘用户的兴趣、行为等信息，并预测未来可能的社交行为。

主要涉及数据挖掘、机器学习、网络分析等知识。

D 题：生物多样性保护策略题目要求根据给定的生态数据，制定一种有效的生物多样性保护策略。

主要涉及生态学、统计学、最优化等知识。

E 题：空气质量预测题目要求预测未来一段时间内某地区的空气质量，并分析影响空气质量的主要因素。

主要涉及时间序列分析、回归分析、机器学习等知识。

F 题：智能温控系统设计题目要求设计一种智能温控系统，使得建筑物在保证舒适度的同时，能耗最低。

主要涉及控制论、最优化、系统分析等知识。

【题目解答方法与技巧】1.熟悉题目要求，明确建模目标。

2.对题目进行深入分析，充分挖掘题目中的信息。

3.结合实际问题，选择合适的数学模型和方法。

4.充分利用计算机软件和编程语言进行求解。

5.结果分析与检验，撰写论文与报告。

美赛2021数模A题论文解法思路美赛2021数模A题解法思路问题:真菌木质纤维分解解法思路：建立不同真菌木质纤维分解数学模型，考虑温度湿度对分解速度的作用。

真菌木质纤维分解数学模型摘要真菌木质纤维分解是本文要解决的数学问题，为了明确真菌木质纤维分解问题，本文针对真菌木质纤维分解问题进行了分析建模,对真菌木质纤维分解问题进行了参考文献研究，建立了真菌木质纤维分解问题的相应模型，推导出真菌木质纤维分解问题的计算公式，编写了真菌木质纤维分解问题的计算程序，经过程序运行，得到真菌木质纤维分解问题程序计算结果。

具体有：对于问题一，这是真菌木质纤维分解问题最重要的问题，根据题目，对问题一进行了分析，参考已有的资料，建立了真菌木质纤维分解问题一的数学模型，推导出问题一的计算公式，编写出真菌木质纤维分解问题一的计算程序。

求出了真菌木质纤维分解问题一的计算结果。

对于问题二，真菌木质纤维分解问题二比问题一复杂的，是真菌木质纤维分解问题的核心，分析的内容多，计算机的东西也多。

在真菌木质纤维分解问题一的基础上，根据真菌木质纤维分解问题，对问题二进行了分析，参考已有的资料，建立了真菌木质纤维分解问题二的数学模型，推导出问题二的计算公式，编写出真菌木质纤维分解问题二的计算程序。

求出了问题二的计算结果，并以图表形式表达结果。

对于问题三，真菌木质纤维分解问题三是问题一和问题二的深入。

在问题一和问题二的基础上，根据真菌木质纤维分解问题，对问题三进行了分析，参考已有的资料，建立了问题三的数学模型，推导出真菌木质纤维分解问题三的计算公式，编写出真菌木质纤维分解问题三的计算程序。

求出了真菌木质纤维分解问题三的计算结果，并以图表形式表达结果，并且进行了分析讨论。

对于问题4，真菌木质纤维分解问题4是问题一、问题二和问题三的扩展。

在问题一、问题二和问题三的基础上，根据真菌木质纤维分解问题，对真菌木质纤维分解问题4进行了分析，参考已有的资料，建立了真菌木质纤维分解数学模型，推导出真菌木质纤维分解问题4的计算公式，编写出问题4的计算程序。

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。

2024美赛a题完整思路
2024美赛A题是一道数学建模题目，通常会涉及到一些实际问题，需要进行数学建模和分析。

由于我无法透露具体的比赛题目内容，但我可以给你一些一般性的思路，希望能够帮助你更好地理解
和解决这类题目。

首先，解决数学建模题目通常需要以下几个步骤，问题理解、
建立数学模型、求解模型、对结果进行分析和验证。

在解决这类题
目时，首先需要仔细阅读题目，确保充分理解问题的背景和要求。

然后，根据问题的特点和要求，建立相应的数学模型，这可能涉及
到微积分、概率统计、线性代数等数学知识。

接下来，需要对建立
的模型进行求解，可能需要使用数值计算、优化算法等方法。

最后，对求解结果进行分析和验证，确保结果符合实际情况，并能够给出
合理的结论。

在解决具体的数学建模题目时，可能会涉及到不同的数学知识
和方法。

例如，可能需要进行数据分析和处理，使用统计方法对数
据进行分析；可能需要进行优化建模，使用线性规划或整数规划等
方法进行优化求解；可能需要进行动态建模，使用微分方程或差分
方程描述系统的动态变化等等。

在解决数学建模题目时，需要注意以下几点，首先，要对问题进行合理的简化和抽象，将实际问题转化为数学模型；其次，要合理选择适当的数学方法和工具，确保能够有效地解决问题；最后，要对结果进行合理的解释和分析，确保结果符合实际情况，并能够给出合理的结论。

总的来说，解决数学建模题目需要综合运用数学知识、建模能力和计算能力，希望以上的一般性思路能够帮助你更好地理解和解决这类题目。

11A 设计单板滑雪场摘要本文研究的是单板滑雪轨道的设计问题。

首先，我们考虑使运动员的腾空 高度尽量达到最大，建立以腾空高度为目标函数的模型， 针对问题一， 要使滑雪者的垂直距离最长，通过对滑雪者运动过程的分析， 我们选取滑雪者运动的一个周期进行研究， P 平台通过 F 平台再到到 P 平台的 从 另一端，即从 A—B—C—D(如图 2)。

首先， 结合给定的滑雪路线 A—B—C—D 运用物理学知识对滑雪过程进行分 析，建立以最大腾空高度为目标函数，滑雪道曲率半径 r 、倾斜角  、底部平台 宽度 d 以及运动员出槽时与槽边缘的夹角  为自变量的微分方程， 根据搜索相关 资料得知，滑雪道长度、曲率半径、倾斜角及底部平台宽度的设计都有一定的参 考范围限制， 我们结合 U 型单板滑雪道设计数据。

其次， 对于滑雪道长度的求解， 我们是根据运动员的平均腾空次数及每次腾空时所行距离 l 求解得出，而腾空 次数的确定是取 5 次腾空为标准。

关键词： 受力分析 能量守恒 微分方程一、问题重述单板滑雪场地主要由 Flat 平台、过渡区、垂直区、Platform 平台、入口坡组 成。

运动员的滑雪技巧、身体素质，滑雪场地的坡度、深度、宽度、长度等成为 影响运动员成绩的诸多因素。

在单板滑雪比赛中，当滑雪运动员最大限度地产生 垂直腾空后就能够做出各种动作，那么应该如何考虑哪些权衡因素，从而设计出 比较优化的单板滑雪场地。

确定一个滑雪场的形状，使得滑雪选手垂直腾空高度最大化。

“垂直腾空” 最大化就是指最大的垂直腾空距离在半管边缘以上的距离。

定制形状时要优化其他可能的要求， 如空气中的最大扭曲等。

综合各种条件， 选择最优的滑雪场模型。

初步观察得到U型管道的合理设计有利于运动员水平的发挥，因此我们从设 计U型槽的坡度，弧长，场地长度等角度入手。

需要建立数学模型解决的问题： （1）设计出滑雪道的形状（现在一般为半圆柱管内侧面） ，以使得滑雪者的垂 直距离（滞空时间）最长。

2024数学建模美赛a题全文共四篇示例，供读者参考第一篇示例：2024年数学建模美赛A题的题目是一个挑战性的问题，需要参赛选手在短时间内进行思考和分析，然后给出一个合理的解决方案。

这个题目涉及到了数学建模、数据分析和计算机编程等多个领域，需要选手具备较强的逻辑思维能力和解决问题的能力。

题目要求参赛选手利用给定的数据集，对某个特定问题进行建模和分析，然后给出解决方案。

选手需要根据现有的数据集进行数据清洗和预处理，然后利用统计学和数学建模的方法对数据进行分析和建模，最终提供一个可行的解决方案。

在解题过程中，选手需要运用各种数学工具和编程语言来处理数据和进行计算，例如Python、R语言等。

选手还需要结合实际问题的背景知识和专业知识，对数据进行合理的解释和分析。

在解题过程中，选手需要注意数据的质量和可靠性，同时还需要对模型的准确性和稳定性进行评估。

最终，选手需要给出一个详细的报告，说明解决问题的方法和步骤，以及给出相关的结论和建议。

参加数学建模比赛可以锻炼选手的团队合作能力和解决问题的能力，同时也能够提高选手的数学建模和数据分析能力。

希望参赛选手在比赛中能够充分发挥自己的潜力，充分展现出自己的优势和才华，最终取得优异的成绩。

【字数不足，正在努力补充中……】第二篇示例：2024数学建模美赛a题分析数学建模是一门涵盖数学、计算机科学和工程等多学科知识的综合性学科，应用广泛，涉及领域广泛。

每年举办的数学建模比赛更是为广大热爱数学和挑战智力的学生提供了一个展示自己才华的舞台。

今天我们就来分析一下2024年数学建模美赛的a题。

让我们来看一下2024年数学建模美赛a题的具体问题描述：根据指定信息，设计出最佳的实体投资组合。

实体投资组合包括个人、公司、政府、银行等单位所投资的资金和资产，投资的目的是为了获得更高的回报率。

在实际投资中，投资者需要根据市场行情、经济形势等因素来选择不同的投资产品，以实现最大化的利润。

我们需要通过收集数据来分析市场行情和经济形势，以确定合适的投资产品。

2020美赛A题翻译：向北迁徙全球海洋温度影响某些海洋生物的栖息地质量。

当温度变化太大以至于无法继续生长时，这些物种便开始寻找其他更适合其现在和将来的生活和生殖成功的栖息地。

其中一个明显的例子就是美国缅因州的龙虾种群，该种群正缓慢地向北迁徙到加拿大，那里较低的海洋温度提供了更合适的栖息地。

这种地理上的种群迁移会严重破坏依赖海洋生物物种稳定性的公司的生计。

您的团队已被苏格兰北大西洋渔业管理协会聘为顾问。

如果全球海洋温度升高，该财团希望更好地了解与苏格兰鲱鱼和鲭鱼从其目前在苏格兰附近的栖息地迁徙有关的问题。

这两种鱼类是苏格兰渔业的重要经济来源。

鲱鱼和鲭鱼种群位置的变化可能使以苏格兰为基地的小型捕捞公司在经济上造成不确定风险，后者使用没有船上制冷的渔船来捕捞鲜鱼并将其运送到苏格兰渔港的市场。

要求1.建立一个数学模型，以识别未来50年内这两种鱼类最可能的位置，假设水温将发生足够的变化以导致种群移动。

2.根据海水温度变化的速度，使用您的模型预测最佳情况、最坏情况和最有可能经过的时间，直到这些种群距离小渔业公司太远以至于如果小渔业公司继续在其当前位置外作业将一无所获。

3.根据您的预测分析，这些小型捕捞公司是否应该改变其经营方式？•a.如果是，请使用您的模型为小型捕捞公司识别和评估实用且经济上有吸引力的策略。

您的策略应考虑但不限于现实的选择，包括：o将部分或全部捕捞公司的资产从苏格兰港口的当前位置迁移到两个鱼类种群都迁徙的附近；o使用一定比例的小型渔船，这些渔船可以在没有陆上支持的情况下运行一段时间，同时仍确保渔获物的新鲜度和高质量。

o您的团队可以识别和模拟的其他可能的选项。

•b.如果您的团队拒绝进行任何更改，请根据建模结果来说明拒绝的原因，因为建模结果与您的团队所做的假设有关。

4.使用您的模型来解决:如果有一部分渔业移至另一个国家的领海时您的建议受到的影响。

5.除了技术报告外，还要为 Hook Line and Sinker 杂志准备一份长达两页的文章，以帮助渔民了解问题的严重性以及您提出的解决方案将如何改善他们的未来的业务前景。