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Prove that the square has the smallest perimeter of all rectangles of a given area

Prove that for a fixed area, the rectangle with the minimal perimeter is a square.


Proof. Let x and y denote the sides of the rectangle. If the area is fixed, then xy is a constant, say xy = A. The perimeter of the rectangle is then a function P(x,y) = 2x + 2y. But, since

    \[ xy = A \quad \implies y = \frac{A}{x}, \]

we can write P as a function of x alone,

    \[ P(x) = 2x + 2 \frac{A}{x}. \]

So, to find the value of x at which P(x) is minimal we take the derivative,

    \[  P(x) &= 2x + 2\frac{A}{x} \quad \implies \quad P'(x) = 2 - \frac{2A}{x^2}. \]

This has a zero at x = \sqrt{A} and we have

    \begin{align*}  P'(x) &< 0 & \text{when} && x &< \sqrt{A} \\  P'(x) &> 0 & \text{when} && x &> \sqrt{A}. \end{align*}

Therefore, P(x) is decreasing for x < \sqrt{A} and increasing for x > \sqrt{A}. Hence, P(x) has its minimum at x = \sqrt{A}. Since x = \sqrt{A} implies y = \sqrt{A} as well, we have that the perimeter is minimal when x = y, i.e., when the rectangle is a square. \qquad \blacksquare

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