# Difference between revisions of "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 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
Invalid username
Login to AoPS