## Calculus: Early Transcendentals 8th Edition

(a) Let $x$ be the width of the rectangle and let $y$ be the length of the rectangle. Note that $x \gt 0$ and $y \gt 0$ We can express $y$ in terms of the area $A$ and $x$: $A = xy$ $y = \frac{A}{x}$ We can write an expression for the perimeter $P$: $P = 2x+2y = 2x+\frac{2A}{x}$ We can find the point where $P'(x) = 0$: $P(x) = 2x+\frac{2A}{x}$ $P'(x) = 2-\frac{2A}{x^2} = 0$ $2=\frac{2A}{x^2}$ $x^2 = A$ $x = \sqrt{A}$ When $0 \lt x \lt \sqrt{A}$, then $P'(x) \lt 0$ When $x \gt \sqrt{A}$, then $P'(x) \gt 0$ Thus, $x=\sqrt{A}$ is the point where $P$ is a minimum. We can find $y$: $y = \frac{A}{x} = \frac{A}{\sqrt{A}} = \sqrt{A}$ Since $x=y$, the rectangle is a square. Therefore, for all rectangles with a given area, the one with the smallest perimeter is a square. (b) Let $x$ be the width of the rectangle and let $y$ be the length of the rectangle. Note that $x \gt 0$ and $y \gt 0$ We can express $y$ in terms of the perimeter $P$ and $x$: $P = 2x+2y$ $y = \frac{P}{2}- x$ We can write an expression for the area $A$: $A = xy = \frac{P}{2}x- x^2$ We can find the point where $A'(x) = 0$: $A(x) = \frac{P}{2}x- x^2$ $A'(x) = \frac{P}{2}- 2x = 0$ $2x = \frac{P}{2}$ $x = \frac{P}{4}$ When $0 \lt x \lt \frac{P}{4}$, then $A'(x) \gt 0$ When $x \gt \frac{P}{4}$, then $A'(x) \lt 0$ Thus, $x=\frac{P}{4}$ is the point where $A$ is a maximum. We can find $y$: $y = \frac{P}{2}-x = \frac{P}{2}-\frac{P}{4} = \frac{P}{4}$ Since $x=y$, the rectangle is a square. Therefore, for all rectangles with a given perimeter, the one with the greatest area is a square.