A right triangle has a perimeter of 38. What are the lengths of each side if the area contained within the triangle is to be maximized?
A) , ,
B) 15, 15, 8
C) , ,
D) , ,
E) 17, 17, 4

Question

A right triangle has a perimeter of 38. What are the lengths of each side if the area contained within the triangle is to be maximized?
A) , , 
B) 15, 15, 8 
C) , , 
D) , , 
E) 17, 17, 4

Quizplus · Accepted Answer

Let the sides of the right triangle be $a$, $b$, and $c$, where $c$ is the hypotenuse. Without loss of generality, assume that $a\leq b$. Then we have $c = \sqrt{a^2+b^2}$. Since the perimeter is given as 38, we have $a+b+c = 38$.
We want to maximize the area of the triangle, which is $\frac{1}{2}ab$. Using the fact that $c = \sqrt{a^2+b^2}$ and $a+b+c = 38$, we can eliminate $c$ to get $a+b+\sqrt{a^2+b^2} = 38$. Squaring both sides gives $2ab+2a\sqrt{a^2+b^2}+2b\sqrt{a^2+b^2}+a^2+b^2 = 38^2$. Rearranging gives $$2ab = 38^2-(2a+2b)\sqrt{a^2+b^2}-(a^2+b^2).$$ Since $a$ and $b$ are positive, $ab$ is maximized when $\sqrt{a^2+b^2}$ is minimized. This occurs when $a$ and $b$ are as close as possible, which is when $a=b$. (To prove this, you can use the AM-GM inequality: $\sqrt{a^2+b^2} = \sqrt{2}\sqrt{\frac{a^2+b^2}{2}} \geq \sqrt{2}\sqrt{ab} = \sqrt{a^2+b^2}$, with equality holding only when $a=b$.) 
So we have $a=b=\frac{19-c}{2}$, and from $a+b+c=38$ we get $c=19$. Therefore, the sides of the triangle are $\frac{18}{\sqrt{2}}$, $\frac{18}{\sqrt{2}}$, and $19$. 
Therefore, the answer is (C).

A Right Triangle Has a Perimeter of 38