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.
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.

The slope m1 of the line through OB is given by
m1 = (12  0) / (6  0) = 2

Let (x,y) be the coordinates of the top left vertex of the rectangle. Hence
m1 = 2 = (y  0) / (x  0) = y / x

Hence the width W of the rectangle is give by
W = y = 2x

The slope m2 through AB is given by.
m2 = (12  0) / (6  10) = 3

If the top right vertex of the rectangle has coordinates (x , y) then .
m2 = 3 = (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  (5/3) x

The area A is given by.
A = W L = 2x (10  (5/3) x) = (10 / 3) x^{ 2} + 20 x

A is a quadratic function of x, of the form ax^{ 2} + bx + c, and its leading coefficient a = 10 / 3 is negative hence it has a maximum value at the critical value of the first derivative A' of A.
A'(x) = (20/3) x + 20

The critical point is found by solving the equation A'(x) = 0. Hence
(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  (5/3) 3 = 5.