Difference between revisions of "2006 iTest Problems/Problem 26"

(Solution to Problem 26 -- Ratio Maximization ft. Rectangle)
 
m (See Also)
 
Line 16: Line 16:
 
{{iTest box|year=2006|num-b=25|num-a=27|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
 
{{iTest box|year=2006|num-b=25|num-a=27|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
  
[[Category:Introductory Algebra Problems]]
+
[[Category:Intermediate Algebra Problems]]

Latest revision as of 19:40, 1 December 2018

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