Maximum Area of Rectangle
Optimization Problem with Solution

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

Problem

OAB is a triangle whose vertices are given. Find the dimensions of the rectangle with maximum area inscribed in the triangle and with one of its sides on the side OA of the triangle.

triangle for the problem

Solution to Problem:

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

triangle and rectangle for the solution of the problem

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
\[ m_1 = 2 = \frac{y - 0}{x - 0} = \frac{y}{x} \]
Hence the width \(W\) of the rectangle is given by
\[ W = y = 2x \]
The slope \(m_2\) through AB is given by.
\[ m_2 = \frac{12 - 0}{6 - 10} = -3 \]
If the top right vertex of the rectangle has coordinates \((x , y)\) then .
\[ m_2 = -3 = \frac{y - 0}{x - 10} \]
Hence
\[ y = -3x + 30 \]
if we substitute \(x\) by \(x + L\) in the above equation then \(y\) is equal to \(W\) the width of the rectangle
\[ y = W = -3(x + L) + 30 \]
We now equate the expressions of \(W = 2x\) and \(W = -3(x + L) + 30\) to find and expression for \(L\)
\[ 2x = -3(x + L) + 30 \]
Solve the above for \(L\).
\[ L = 10 - \frac{5}{3}x \]
The area \(A\) is given by.
\[ A = W L = 2x \left(10 - \frac{5}{3}x\right) = -\frac{10}{3} x^2 + 20x \]
\(A\) is a quadratic function of \(x\), of the form \(ax^2 + bx + c\), and its leading coefficient \(a = -\frac{10}{3}\) is negative hence it has a maximum value at the critical value of the first derivative \(A'\) of \(A\).
\[ A'(x) = -\frac{20}{3}x + 20 \]
The critical point is found by solving the equation \(A'(x) = 0\). Hence
\[ -\frac{20}{3}x + 20 = 0 \]
The critical point is given by \(x = 3\).
The area \(A\) of the rectangle has a maximum value for \(x = 3\),
\(W = 2x = 6\)
and
\(L = 10 - \frac{5}{3} \times 3 = 5\).

More references on calculus problems