SOLUTION: Find the rectangle of perimeter l which has the maximum area

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Question 1087451: Find the rectangle of perimeter l which has the maximum area
Found 2 solutions by Alan3354, math_helper:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the rectangle of perimeter l which has the maximum area
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The max area of a rectangle of given perimeter is a square.
You can believe that, or I can prove it.
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P = 1 = 4s --- s = side length
s = 1/4

Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
Max area is when the rectangle is a square: L=W= +highlight%281%2F4%29+
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Proof: 2L + 2W = 1 —> +W=%281-2L%29%2F2+
A = LW = +L%281-2L%29%2F2++=+%281%2F2%29%28L-2L%5E2%29+
Take derivative of A with respect to L:
dA/dL = +%281%2F2%29%281-4L%29+
Set that to 0 and solve for L, while noting that d%5E2A%2FdL%5E2+=+-2+ —> Concave down so our answer will give us a MAX:
+%281%2F2%29%281-4L%29+=+0+
+++4L+=+1+
++L+=+1%2F4+ —> +W+=+1%2F4+

Since L=W we have a square.