CPM Homework Help : APCALC Problem 7-124

What are the dimensions of the right triangle with hypotenuse of length $13$ that has a maximum area?

Hint:

$\text{Earlier, you took the derivative }\frac{dA}{dt}\text{. Now use the same process but find }\frac{dA}{da}\text{ instead and solve from there.}$

Solution:

$\left. \begin{array}{l}{ a ^ { 2 } + b ^ { 2 } = 13 ^ { 2 } \rightarrow b = \sqrt { 13 ^ { 2 } - a ^ { 2 } } }\\{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A = \frac { a b } { 2 } = \frac { 1 } { 2 } a \sqrt { 13 ^ { 2 } - a ^ { 2 } } }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ { \frac { d A } { d a } = \frac { 1 } { 2 } \sqrt { 13 ^ { 2 } - a ^ { 2 } } + \frac { 1 } { 2 } a ( \frac { 1 } { 2 } ) ( 13 - a ^ { 2 } ) ^ { - 1 / 2 } \cdot ( 2 a ) = 0 }\end{array} \right.$

$\left. \begin{array}{l}{ ( 13 ^ { 2 } - a ^ { 2 } ) - a ^ { 2 } = 0 \text { multiply each side by } 2 \sqrt { 13 ^ { 2 } - a ^ { 2 } } }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ { 13 ^ { 2 } = 2 a ^ { 2 } }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ { a = \frac { 13 } { \sqrt { 2 } } }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ { \approx 9.192 }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ { b = \sqrt { 13 ^ { 2 } - ( \frac { 13 } { \sqrt { 2 } } ) ^ { 2 } } = a \approx 9.192 }\end{array} \right.$

$\left. \begin{array}{l}{ a ^ { 2 } + b ^ { 2 } = 13 ^ { 2 } \rightarrow b = \sqrt { 13 ^ { 2 } - a ^ { 2 } } }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ { A = \frac { a b } { 2 } = \frac { 1 } { 2 } a \sqrt { 13 ^ { 2 } - a ^ { 2 } } }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ { \frac { d A } { d a } = \frac { 1 } { 2 } \sqrt { 13 ^ { 2 } - a ^ { 2 } } + \frac { 1 } { 2 } a ( \frac { 1 } { 2 } ) ( 13 - a ^ { 2 } ) ^ { - 1 / 2 } \cdot ( 2 a ) = 0 }\end{array} \right.$