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Here we have to prove that the square has the greatest area for the same perimeter when a rectangle is considered.
So the perimeter P can be taken to be a constant P.
Use a variable x for the length. So the width is P/2 - x.
The area A = ( P/2 -x) *x.
Now to maximize the area, for a variable x, we have to find dA/dx.
dA/dx= P/2 - 2x.
Equate this to zero P/2 - 2x =0
=> x= P/4
So the length is P/4 and the width is P/2 - x = P/4.
Therefore for a given perimeter of a rectangle the area is the largest if the shape is that of a square with each side equal to the perimeter divided by 4.
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