美赛2013A题论文(终极布朗尼盘)

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2013Mathematical Contest in Modeling(MCM/ICM)Summary Sheet

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There are plenty of people who are fond of Brownies.Industrial manufacturing of them requires to bake the biggest amount of Brownies in the same batch and also to avoid overcooking of the batter next to the corners of the pan,in order to avoid losses. Overcooking can affect the volumes of batter that are in direct contact with more than one hot surface of the pan,for instance the bottom of it and one of its sides.The corners of the pan are the hottest regions,as three surfaces deliver heat simultaneously(two sides plus the bottom).That’s why overcooking is frequent in the corners.

Round pans distribute heat in a more even way,due to the fact that there are never three surfaces delivering heat to a volume of batter(two at most),.The problem with them is that they cause a big loss of oven space as compared to square ones,even to hexagonal ones.As round pans minimize overcooking and rectangular pans minimize oven-space loss,our model posits a compromise between these two shapes,namely modified rectangular pans having a 90°centred-angle arc instead of being standard rectangles.

In order to clarify which factor will be advantaged by variations of this type of shape(either space saving or heat evenness),we put weights p and(1-p)on the sum of the line lengths and on the sum of the arc lengths.As we consider that the area of the pan is fixed and its width-to-length ratio W/L is given,we propose3equations expressing the effect of having longer or shorter arcs on the pans performances(the longer the arcs,the evener the heat distribution;the shorter the arcs,the better the space in the oven is filled).

We use matlab to solve these equations.Once we give p and W/L two values between0to1, we can get a shape of pan.By changing the weights p by small steps while the ratio W/L is fixed,we can plot the figure in a three-dimentional space layer by layer,so that we can see clearly how the figure changes with weights changing for a given W/L ratio.From these graphs,we can conclude that the space the pan takes increases with increasing weights on the arc factor(we provide some charts to make clear).

In general,if the producer wants to increase the quality of the brownies by avoiding overcooking of volumes placed in critical locations,he needs to decrease p.If he wants to increase the production by filling the oven as much as possible,he needs to increase p. Large p values amount to short arcs,small p values mean long arcs.

Part1.Overview

Obtaining perfectly baked brownies depends much on the shape of the pan used to cook them,as cakes tend to get overcooked at the corners in angular pans.After having visited a few bakeries,we found out that most pans are rectangular with rounded corners.We noticed there were stains showing overcooking in the four corners and,to a less extent,on the sides,which shows that the heat absorbed by the batter had been higher for the volumes placed at the corners and against the faces of the pan.We therefore engaged in finding a pan shape able to prevent overcooking,keeping in mind that bakeries need to fit as many pans as possible into their ovens,in order to take the best advantage of each batch of brownies.

After discussions and some study,we discovered that shapes maximizing the evenness of heat distribution waste oven space.We thus moved to the goal of both preventing overcooking and avoiding waste of oven space in order to define the best suited pan for productive and flawless cooking of brownies.As will be seen,our model tends to make the best compromise between these two constraints.

Part2.Assumptions

Assumption1(Space optimization)

Given a rectangle of fixed width-to-length ratio(W/L)and unknown area,to maximize the number of figures fitting into the rectangle requires the shape of these figures to satisfy the following conditions:

1)As n figures are put together in the rectangle,the area left between figures must be the least possible one,

2)No arcs are necessary,as replacing arcs by angles consumes the same area.

Assumption2(Heat distribution optimization)

The heat each volume unit gets from the pan is proportional to the area unit of the pan it is in contact with.Area units are considered to be the bottom and each one of the faces of the pan.As a volume unit placed in a corner receives heat from3area units of the pan,its heat value equals3units(bottom and2sides).By contrast,a volume unit placed against one of the faces of the pan receives heat from2area units of it(bottom and side),so its heat value

equals2units.For a volume unit placed in the middle of the pan,the heat received comes from a single area unit of the pan(the bottom),thus its heat value equals1unit.[Figure2-1]

Assumption3(Model oven and relevant pan properties)

1.The heating devices in our model oven being evenly distributed,the pan absorbs the same amount of heat value from all directions.

2.The pan being evenly heated,every point of the pan shares the same temperature.

3.Only the pan provides the heat,as we ignore the heat in the air surrounding it.

4.Only the shape of the pan will be taken into consideration.

Part3.Goals of the model

As the design of the pan is meant to fit the requirements of energy-saving while perfect cooking of brownies(no burned or uncooked pastry),the following modelling goals have been set:

(1)Maximize number of pans that can fit in the oven(N),

(2)Maximize the evenness of heat distribution(H)for a pan.

In order to achieve an optimal balance between goals(1)and(2),putting weights on them [p and(1-p)],will allow to decide which one should be given more attention.

Part4.Model

4.1Discussion of objective1:how to maximize the filling of the oven According to assumption1,we have first identified rectangles and regular hexagons as the best shapes for figures:whereas both of them meet the requirement of minimizing the space left between figures,rectangles not only satisfy this constraint but also minimize the lost space within the oven when the figures are in place,due to the fact that the W/L ratio of rectangles is more consistent with the W/L ratio of a rectangular oven than the equivalent ratio of hexagons.

From this perspective,our final choice is thus a rectangle.

4.2Discussion of objective2:how to maximize the evenness of heat distribution

As to the constraint to maximize the evenness of the heat distribution,we interpret it as the fact that the difference between the quantity of heat every volume unit of the batter receives is the lowest possible.As this difference increases with the number of faces of the pan, volume-type ensuring the most even heat distribution should be a cylinder.From a two-dimensional perspective,the ideal shape is thus a circle.

circle..

From this perspective,our final choice is thus a circle

4.3Putting perspectives together

In order to combine both shapes,we replaced the four angles of the rectangle with four arcs with a central angle of90degrees,as shown in figure4-1below.

If the central angle of the arc were greater than 90degrees (figure 4-2),the lines and the arc could not be put together in a way that satisfies assumption 1.Conversely,if the central angle of the arc was less than 90degrees (figure 4-3)space would be wasted as the figure would get bigger because the area A is

fixed.

4.4Building the model

In order to satisfy 4.1and 4.2,rectangle and circle can be combined and enhanced with weights p and (1-p),so as to obtain the shape model illustrated by figure 4-1.The model is expressed via the following equations:

(Each pan must have an area of A)

()22............................................................................r r a b ab A π+++=①

(Weights p and (1-p)are assigned to the perimeter of lines and arcs so as to allow decisions about the most critical factor)(This shape takes as much space as the same sized rectangle.According to Assumption 1,we have equation ③)

We assume that p and W/L are independent variables,a,b and r being dependent ones.We solve the equations by expressing a,b and r with p and W/L.We can see how the shape changes with different given values of p and W/L between 0to 1.Thus,our final equations of our model are are:

:()1

01

0.

.221)(22<<<<=++?=+=+++L

W p t s L

W

r a r b p p

r b a A

ab b a r r r

b a find ππWe use matlab software to solve these equations and

we use p and W/L to refer to a,b and r （For detailed code please refer to the appendix1）.

M

A kp k p p a )

2222(++??=πM

A kp k p kp b )

2222(??++=πM

A

kp p k r )

1(+??=

In which A is given and ：

???

??

??

?????????=???+++???++++??+++=L W k p k kp kp kp p k p k p k p k kp p k kp p k p k M 4264444881622882

2222222222222ππππππππππPart 5.Sensitivity and stability analysis

Next,we analyze the sensitivity and stability of our model by assigning different values to the variables.First of all ，we assume that the area of the pan is fixed as A=10,when the width-to-length ratio W/L=1,we use matlab to calculate the value of a,b and r ：

A=10W/L=1No

p a b r S 10.10.14020.1402 1.695412.468020.150.28060.2806 1.607412.217830.20.42140.4214 1.520211.984140.250.56290.5629 1.433511.764250.30.70540.7054 1.347211.558660.350.8490.849 1.261111.365070.40.9940.994 1.175211.185080.45 1.1408 1.1408 1.089311.018490.5 1.2895 1.2895 1.003310.8643100.55 1.4404 1.44040.91710.7217110.6 1.594 1.5940.830310.5924120.65 1.7505 1.75050.742910.4736130.7 1.9103 1.91030.654810.3678140.75 2.0738 2.07380.565810.2746150.8 2.2415 2.24150.475710.1946160.85 2.4139 2.41390.384210.1270170.9 2.5915 2.59150.291110.0724180.95 2.7749 2.77490.196310.033119

1

2.9649

2.9649

0.0993

10.0077

In this table ，the last column is the area of the minimum bounding rectangle of the pan S.S=（a+2r)(b+2r).

As may be seen in Figure5-1,we multiply p by10to make the graph clearer.For p from 0.1to1step0.05,we plot every figure and pile them up layer by layer in the3-dimentional space.We observe how S changes while changing the value of p,and notice that when p increases,S decreases and tends to10.That is,while p increases from0to1,the area(S) decreases slightly but proportionally to the value of p.

Changing the value of W/L shows that the phenomenon is stable for every value of W/L.

A=10W/L=0.8

N

p a b r S

o

10.050.53660.2541 1.708614.5156

20.10.6740.1807 1.619713.3842

30.150.81210.037 1.531612.0142

40.20.95090.1832 1.44411.7900

50.25 1.09090.33 1.356911.5807

60.3 1.23220.4778 1.270111.3851

70.35 1.37520.6268 1.183411.2021

80.4 1.520.7772 1.096811.0324

90.45 1.66690.9295 1.010110.8758

100.5 1.8163 1.08380.923110.7311

110.55 1.9685 1.24050.835710.5990

120.6 2.1238 1.40.747810.4803

130.65 2.2827 1.56250.659110.3731

140.7 2.4455 1.72860.569410.2776

150.75 2.6128 1.89870.478710.1968

160.8 2.7849 2.07330.386610.1281

170.85 2.9626 2.25290.29310.0741

180.9 3.1464 2.43810.197510.0331

190.95 3.3371 2.62970.110.0089

A=10W/L=0.6

N

p a b r S

o

10.05 1.06250.7733 1.765619.7736 20.1 1.20120.6173 1.672618.0151 30.15 1.33790.4617 1.580516.2982 40.2 1.47570.306 1.489314.6306 50.25 1.61470.1501 1.398613.0032 60.3 1.75530.0064 1.308511.4703 70.35 1.89770.1637 1.218611.2746 80.4 2.04220.3222 1.128911.0940 90.45 2.18910.4821 1.032910.8411 100.5 2.33870.64380.949410.7743 110.55 2.49150.80750.859210.6338 120.6 2.64770.97380.768510.5069 130.65 2.8708 1.14290.677110.5502 140.7 2.9722 1.31540.584910.2937 150.75 3.1414 1.49160.491610.2076 160.8 3.3161 1.67210.396910.1346 170.85 3.4968 1.85750.300810.0784 180.9 3.6842 2.04830.202710.0347 190.95 3.8791 2.24540.102610.0090

A=10W/L=0.4

N

p a b r S

o

10.05 1.883 1.5638 1.930831.1668 20.1 2.0196 1.3825 1.825228.5365 30.15 2.1571 1.2028 1.721426.0149 40.2 2.296 1.0244 1.61923.5881 50.25 2.43640.8469 1.517921.2469 60.3 2.57870.6698 1.417818.9793 70.35 2.72320.4929 1.318416.7752 80.4 2.87020.3156 1.219714.6279 90.45 3.02010.1376 1.121412.5278 100.5 3.17330.0414 1.023310.8992 110.55 3.33020.2220.925110.7348 120.6 3.49120.40450.826710.5871 130.65 3.65680.58490.727810.4319 140.7 3.82760.77720.628210.3388 150.75 4.00420.96850.527610.2387 160.8 4.1871 1.16390.425810.1555 170.85 4.3773 1.36390.322510.0893 180.9 4.5755 1.56940.217310.0402 190.95 4.7827 1.78110.1110.0109

A=10W/L=0.2

N

p a b r S

o

10.05 3.7351 3.3153 2.538973.9676 20.1 3.8622 3.0321 2.377967.1161 30.15 3.9921 2.7539 2.223660.7722 40.2 4.1251 2.2453 2.075252.9276 50.25 4.2615 2.2386 1.931849.5810 60.3 4.4017 1.988 1.792744.5153 70.35 4.546 1.7425 1.657339.7518 80.4 4.6948 1.501 1.52535.2466 90.45 4.8487 1.2625 1.395230.9605 100.5 5.008 1.0263 1.267426.8607 110.55 5.17340.7913 1.141222.9169 120.6 5.34540.5568 1.016219.1026 130.65 5.52490.32190.891815.3880 140.7 5.71250.08570.767611.7478 150.75 5.90920.15270.643210.3552 160.8 6.11590.39430.518110.2311 170.85 6.3340.640.391710.1309 180.9 6.56480.8910.263710.0596 190.95 6.8098 1.14860.133310.0145

Part6.C onclusion

In general,if the producer wants to increase the quality of the brownies by avoiding over-cooking of volumes placed in critical locations,he needs to decrease p.If he wants to incre ase the production by filling the oven as much as possible,he needs to increase https://www.360docs.net/doc/5b16201542.html,rge p v alues amount to short arcs,small p values mean long arcs.

Improvement of the subsequent models:

One of our tasks in the future is to consider the stability of the pan structures after our current model and add relevant details to it,such as heat conduction.

We still need to test our model in order to ensure that it is reliable and reasonable.

Part7.Strengths and Weaknesses of the Model

?The model describes the pan in detail,coordinating the relevant physical quantities.

?The numerical computations are relatively precise.

?The results illustrated by the graphical representations of the model are generated by numerical computations.

?The model is simple and practical and thus easy to implement.

?Arbitrary results may be obtained from the model according to specific circumstances(the values of W/L and p are different under different conditions).

?The model does not allow to calculate the number of pans in an oven,but it can select the best type of them.

?The model does not take into account detailed heat transfer and loss of energy.

eferen n ce

ces s

Part9.R efere

X.Y.Huang,C.Y.Peng and J.H.Teng2005English expressions of the definitions and theorems in advanced mathematics.ISBN7-118-03773-7

Dilip P Patil2010Introduction to Algebraic Geometry and Commutative Algebra

M.Z.Arslanov1998Continued fractions in optimal cutting of a rectangular sheet into equal small rectangles

Z.F.Zhao and Y.Y.Ma2011The Method of Requesting the Limit of Function of Many Variables Journal of Yichun College2011(8)

Appendix1:Code

Code of the model

clear all

clc

syms a b r A k p

A=10;k=0.2;s=0.05:0.05:1;

for ii=1:length(s)

[a,b,r]=solve('pi*r*r+2*r*(a+b)+a*b=A','(a+b)*(1-p)=pi*r*p','(b+2*r)/(a+2*r)=k', 'a,b,r');

p=s(ii);

a=eval(a);b=eval(b);r=eval(r);

for i=1:length(a)

if a(i)>0

x=a(i);

end

end

a=x

for i=1:length(b)

if b(i)>0

y=b(i);

end

end

b=y

for i=1:length(r)

if r(i)>0

z=r(i);

end

end

r=z

rectA(a+2*r,b+2*r,p)

hold on

circle(r,1,r+a,1,b+r,p)

circle(r,1,a+r,-1,r,p)

circle(r,-1,r,-1,r,p)

circle(r,-1,r,1,b+r,p)

grid on

End

Code of circle drawing

function circle(R,a1,a2,b1,b2,n) alpha=0:pi/50:pi/2;

[0,2*pi]%R=2;

x=a1*R*cos(alpha)+a2;

y=b1*R*sin(alpha)+b2;

z(1:26)=10*n;

plot3(x,y,z,'-');

%axis equal

Code of rectangle drawing

function rectA(w,h,n);

X=[0,w,w,0,0];

Y=[0,0,h,h,0];

d(1:5)=10*n;

%axis([-36-36-210])

plot3(X,Y,d,'r');

Appendix2:Advertising sheet