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? 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: Write the area A as a function of L and W. Write an equation eq you know relating L and W. Eliminate L by solving eq for L (we could solve for W, but L is slightly easier). Substitute this solution for L in your function A. Find the critical value of A (you should only get one). Use the Second Derivative Test to determine if your critical value is a maximum or a minimum. 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)

Applied Optimization: Maximizing Area Given Fixed Fencing

Mona Mona

2019 5 1

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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?

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:

Write the area A as a function of L and W.
Write an equation eq you know relating L and W.
Eliminate L by solving eq for L (we could solve for W, but L is slightly easier).
Substitute this solution for L in your function A.
Find the critical value of A (you should only get one).
Use the Second Derivative Test to determine if your critical value is a maximum or a minimum.
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)