How do you find the maximum area of a rectangle inscribed in a right triangle?

To maximize the area of a rectangle inscribed in a right triangle, use similar triangles or area partitioning to express the rectangle's dimensions as a function of a single variable. Then, find the first derivative of the area function, set it to zero to find critical points, and apply the first derivative test to confirm the maximum.

Maximum Area of Rectangle in a Right Triangle – Calculus Optimization Problem with Solution

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

In what follows, \( C \) represents the constant of integration where applicable.

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.

Rectangle inscribed in right triangle problem — Figure 1. Rectangle inscribed in a right triangle

Show 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:

\[ \frac{1}{2} \times 40 \times 30 = 600 \]

The area of the right triangle may also be calculated as the sum of the areas of triangle BEC and triangle BEA (partitioned by the rectangle). 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 \( x \):

\[ y = \frac{600 - 15x}{20} \]

Substitute \( y \) into \( A \) to obtain area as a function of \( x \):

\[ A(x) = \frac{x(600 - 15x)}{20} \]

An expansion of \( A(x) \) shows that \( A(x) \) is a quadratic function with a negative leading coefficient, and therefore has a maximum value:

\[ A(x) = -\frac{3}{4}x^2 + 30x \]

The graph of \( A(x) \) as a function of \( x \) is shown below. \( A(x) \) has a maximum value for \( x = 20 \), as verified analytically below:

Rectangle inscribed in right triangle solution graph — Figure 2. Graph of area function \( A(x) \)

We now calculate the first derivative of \( A(x) \):

\[ A'(x) = -\frac{3}{2}x + 30 \]

Set \( A'(x) = 0 \) and solve for \( x \):

\[ -\frac{3}{2}x + 30 = 0 \implies x = 20 \]

It is easy to check that \( A'(x) \) is positive for \( x < 20 \) and negative for \( x > 20 \), confirming that \( A(x) \) has a maximum at \( x = 20 \).

Table of sign of the derivative and variations of the function — Figure 3. Table of signs for the first derivative \( A'(x) \)

The maximum area is given by \( A(20) \):

\[ A(20) = -\frac{3}{4} \times 20^2 + 30 \times 20 = 300 \]

We now find \( y \) using the area formula:

\[ A = 300 = x \times y \implies y = \frac{300}{20} = 15 \]

Conclusion: The dimensions of the rectangle that maximize its area are \( x = 20 \) and \( y = 15 \).

Problem with Solution

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