Among all rectangles of given area, show that the square has the least perimeter.

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Let the sides of the rectangle be given by x and y. Then the area is `A=xy` and the perimeter is `P=2x+2y` .

`A` is a given constant. Then `x=A/y` .

`P=2(A/y)+2y`

We wish to minimize the function, so take the first derivative and find the extrema:

`(dP)/(dy)=(-2A)/y^2+2` The critical points occur when the derivative doesn't exist, or when the derivative is zero. The derivative fails if y=0, but this doesn't make sense for the problem. Then setting the derivative equal to zero we get:

`(2A)/y^2=2`

`2y^2=2A`

`y^2=A`

Thus the minimum occurs when `y=sqrt(A)` . Since `x=A/y` we have `x=A/sqrt(A)=sqrt(A)` so x=y and the rectangle must be a square.

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