Q53. - Algebra 2

Question

WRITING IN MATH

A rectangle is inscribed in an isosceles triangle as shown. Find the dimensions of the inscribed rectangle with maximum area. (Hint: Use similar triangles.)

Step-by-Step Solution

Answer

The dimensions are $5\times 4$ inches 

Step 1. Given information

A figure is given as above

Step 2. Concept used

Make a quadratic equation by assuming one side as $x$. Find the maximum value of it and then find the value of dimensions.

Step 3. Calculation

Let $x=$  the length of the rectangle.

Then $8-\frac{4}{5}x=$ the height of the rectangle.

 $\begin{array}{l}=x\left(8\right)+x\left(-\frac{4}{5}x\right)\\ =8x-\frac{4}{5}{x}^2\\ =-\frac{4}{5}{x}^2+8x\end{array}$

$A\left(x\right)$is a quadratic function with $a=-\frac{4}{5},b=8$, and c=0. Since a<0, the function has a maximum value at the vertex of the graph.

 $\begin{array}{l}=-\frac{8}{2\left(-\frac{4}{5}\right)}\\ =5\end{array}$

This means the maximum area occurs with length 5 inches and height $8-\frac{4}{5}\left(5\right)$  or 4 inches.