Answer
(a) For all rectangles with a given area, the one with the smallest perimeter is a square.
(b) For all rectangles with a given perimeter, the one with the greatest area is a square.
Work Step by Step
(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.
