2006 iTest Problems/Problem 26

Problem

A rectangle has area $A$ and perimeter $P$. The largest possible value of $\tfrac A{P^2}$ can be expressed as $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Solution

Let $l$ and $w$ be the length and width of the rectangle, respectively. The area of the rectangle is $lw$, and the perimeter of the rectangle is $2l+2w$. We wish to maximize $\frac{lw}{(2l+2w)^2}$.


By the AM-GM Inequality, $\frac{l+w}{2} \ge \sqrt{lw}$, with equality occurring when $l = w$. Multiply both sides by 4 to get $2l+2w \ge 4\sqrt{lw}$.


Because the length and width of a rectangle is positive, $2l+2w \ge 4\sqrt{lw} > 0$. Thus, squaring both sides would not affect the inequality sign, so $(2l+2w)^2 \ge 16lw$. Finally, since $(2l+2w)^2$ is positive, we can divide both sides by $(2l+2w)^2$ and $16$ to get $\frac{lw}{(2l+2w)^2} \le \frac{1}{16}$. The equality case $l = w$ satisfies the equality statement, so the highest possible ratio is $\frac{1}{16}$. Therefore, $m+n = \boxed{17}$.

See Also

2006 iTest (Problems)
Preceded by:
Problem 25
Followed by:
Problem 27
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