Maximum Area of Rectangle
Optimization Problem with Solution

Calculus Optimization Tutorial and Step-by-Step Solution

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

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

Problem

OAB is a triangle with given vertices. Find the dimensions of the rectangle with maximum area inscribed in the triangle such that one of its sides lies on side OA of the triangle.

Triangle for the problem
Figure 1. Triangle OAB
Show Solution to Problem

In the figure below, a rectangle with width \( W \) and length \( L \) is inscribed inside the triangle, with its top vertices lying on the sides of the triangle. We first find a formula for the area of the rectangle in terms of \( x \) only.

Triangle and rectangle for the solution
Figure 2. Inscribed rectangle inside the triangle

The slope \( m_1 \) of the line through OB is given by:

\[ m_1 = \frac{12 - 0}{6 - 0} = 2 \]

Let \( (x, y) \) be the coordinates of the top-left vertex of the rectangle. Hence:

\[ m_1 = 2 = \frac{y - 0}{x - 0} = \frac{y}{x} \]

Therefore, the width \( W \) of the rectangle is given by:

\[ W = y = 2x \]

The slope \( m_2 \) of the line through AB (from \( (6, 12) \) to \( (10, 0) \)) is given by:

\[ m_2 = \frac{0 - 12}{10 - 6} = \frac{-12}{4} = -3 \]

If the top-right vertex of the rectangle has coordinates \( (x + L, y) \), it lies on the line AB passing through \( (10, 0) \):

\[ m_2 = -3 = \frac{y - 0}{(x + L) - 10} \]

Solving for \( y \):

\[ y = -3(x + L - 10) = -3(x + L) + 30 \]

Since \( y = W = 2x \), we equate both expressions for \( y \):

\[ 2x = -3(x + L) + 30 \]

Solving for \( L \):

\[ 2x = -3x - 3L + 30 \implies 3L = 30 - 5x \implies L = 10 - \frac{5}{3}x \]

The area \( A \) of the rectangle is given by:

\[ A = W \cdot L = 2x \left(10 - \frac{5}{3}x\right) = -\frac{10}{3}x^2 + 20x \]

\( A \) is a quadratic function of \( x \) of the form \( ax^2 + bx + c \). Since its leading coefficient \( a = -\dfrac{10}{3} \) is negative, it has a maximum value at the critical point of its first derivative \( A'(x) \).

Computing the first derivative:

\[ A'(x) = -\frac{20}{3}x + 20 \]

Setting \( A'(x) = 0 \) to find critical points:

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

Thus, the area \( A \) of the rectangle has a maximum value at \( x = 3 \), with dimensions:

  • \( W = 2x = 2(3) = 6 \)
  • \( L = 10 - \frac{5}{3}(3) = 10 - 5 = 5 \)

The maximum area is \( A = 6 \times 5 = 30 \).

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