Applied Optimization: Maximizing Area Given Fixed Fencing

조회 수: 2 (최근 30일)
Mona Mona
Mona Mona 2019년 5월 1일
댓글: Walter Roberson 2019년 5월 3일
You have a ranch house 56 feet wide, and wish to enclose a rectangular area in the back with 500 ft of fencing, using the house as part of one side (see figure below). What dimensions will maximize the area?
fence_6_9.jpg
Solve this problem by defining variables L as the length of the field and W as the width on either side of the house (we can assume the house is centered on that side without affecting our solution), then follow these steps:
  1. Write the area A as a function of L and W.
  2. Write an equation eq you know relating L and W.
  3. Eliminate L by solving eq for L (we could solve for W, but L is slightly easier).
  4. Substitute this solution for L in your function A.
  5. Find the critical value of A (you should only get one).
  6. Use the Second Derivative Test to determine if your critical value is a maximum or a minimum.
  7. Answer the question being asked in the problem.
What is the solution of this Matlab
syms L W
% Step 1
A=; % Write A as a symbolic expression, not a function
% Step 2
eq=; % Remember to use == for the equal sign in your equation!
% Step 3
elimL=solve();
% Step 4
AofW=subs()
% Step 5
dA=diff()
cv=solve()
% Step 6
ddA=diff()
SDT=subs()
% Step 7-find the value of the other variable by substituting into eq and solving
eqalt=subs();
Lans=solve();
% DO NOT CHANGE CODE ON REMAINING LINES-displays answer with explanation
Wans=cv;
disp('Desired length is')
disp(Lans)
disp(' ')
disp('Desired width is')
disp(Wans)
  댓글 수: 5
jahanzaib ahmad
jahanzaib ahmad 2019년 5월 3일
where is the figure ?
Walter Roberson
Walter Roberson 2019년 5월 3일
The figure is the box-on-a-box

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