SOLUTION: The perimeter of a rectangle is 32m. Find the dimensions for which the diagonal is as short as possible

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Question 312445: The perimeter of a rectangle is 32m. Find the dimensions for which the diagonal is as short as possible
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
P=2%2A%28L%2BW%29=32
1.L%2BW=16
The diagonal is then,
2.D%5E2=L%5E2%2BW%5E2
From the first equation,
L%2BW=16
L=16-W
Substituting into the second equation,
D%5E2=%2816-W%29%5E2%2BW%5E2
D%5E2=%28256-32W%2BW%5E2%29%2BW%5E2
D%5E2=2W%5E2-32W%2B256
Now the distance squared is a function only of W.
Take the derivative wrt W and set it equal to zero to find the minimum.
2%2AD%2A%28dD%2FdW%29=4W-32=0
Since D cannot be zero (since L and W are both not equal to zero), then dD%2FdW must be zero when 4W-32=0.
4W-32=0
4W=32
W=8
Then from above,
L=16-8=8
The minimum diagonal is formed by an 8m x 8m square.