Maximum Area of Rectangle in a Right Triangle - Problem with Solution

Maximize the area of a rectangle inscribed in a right triangle using the first derivative. The problem and its solution are presented

Problem with Solution

BDEF is a rectangle inscribed in the right triangle ABC whose side lengths are 40 and 30. Find the dimensions of the rectangle BDEF so that its area is maximum.

Solution to Problem:

Let the length BF of the rectangle be $$y$$ and the width BD be $$x$$. The area of the right triangle is given by $$\frac{1}{2} \times 40 \times 30 = 600$$. But the area of the right triangle may also be calculated as the sum of the areas of triangle BEC and BEA. Hence
$600 = \frac{1}{2} \times 40 \times y + \frac{1}{2} \times 30 \times x$
Let $$A$$ be the area of the rectangle. Hence
$A = y \times x$
We now use the first equation to express $$y$$ in terms of $$x$$ as follows
$y = \frac{600 - 15x}{20}$
Substitute in $$A$$ to obtain
$A(x) = \frac{x(600 - 15x)}{20}$
The graph of $$A(x)$$ as a function of $$x$$ is shown below. $$A(x)$$ has a maximum value for $$x = 20$$. This will be shown analytically as well

An expansion of $$A(x)$$ shows that $$A(x)$$ is a quadratic function with negative leading coefficient and therefore has a maximum value.
$A(x) = -\frac{3}{4}x^2 + 30x$
We now calculate the first derivative of $$A$$.
$A'(x) = -\frac{3}{2}x + 30$
Set $$A'(x) = 0$$ and solve for $$x$$.
$x = 20$
It is easy to check that $$A'(x)$$ is positive for $$x \lt 20$$ and negative for $$x > 20$$ and therefore $$A(x)$$ has a maximum at $$x = 20$$.

The maximum area is given by $$A(20)$$
$A(20) = -\frac{3}{4} \times 20^2 + 30 \times 20 = 300$
We now find $$y$$ as follows
$A = 300 = x \times y , \quad y = \frac{300}{20} = 15$
The dimensions of the rectangle that make its area maximum are $$x = 20$$ and $$y = 15$$.