# Use Derivatives to Maximize Area

of A Rectangle with Given Perimeter

A problem to maximize (optimization) the area of a rectangle with a constant perimeter is presented. An interactive applet (you need Java in your computer) is used to understand the problem. Then an analytical method, based on the derivatives of a function and some calculus theorems, is developed in order to find an analytical solution to the problem.

## Problem

You decide to construct a rectangle of perimeter \(400 \, \text{mm}\) and maximum area. Find the length and the width of the rectangle.__Solution to the Problem__

We now look at a solution to this problem using derivatives and other calculus concepts. Let \(x\) ( = distance DC) be the width of the rectangle and \(y\) ( = distance DA) its length, then the area \(A\) of the rectangle may be written: \[ A = xy \] The perimeter may be written as \[ P = 400 = 2x + 2y \] Solve equation \(400 = 2x + 2y\) for \(y\) \[ y = 200 - x \] We now substitute \(y = 200 - x\) into the area \(A = xy\) to obtain: \[ A = x(200 - x) \] Area \(A\) is a function of \(x\). As you change the width \(x\) in the applet, the area \(A\) on the right panel changes. Expand the expression for the area \(A\) and write it as a function of \(x\). \[ A(x) = -x^2 + 200x \] We might consider the domain of function \(A(x)\) as being all values of \(x\) in the closed interval \([0 , 200]\) since \(x \geq 0\) and \(y = 200 - x \geq 0\) (if you solve the second inequality, you obtain \(x \leq 0\)). To find the value of \(x\) that gives an area \(A\) maximum, we need to find the first derivative \(\dfrac{dA}{dx}\) (\(A\) is a function of \(x\)). \[ \dfrac{dA}{dx} = -2x + 200 \] If \(A\) has a maximum value, it happens at \(x\) such that \(\dfrac{dA}{dx} = 0\). At the endpoints of the domain we have \(A(0) = 0\) and \(A(200) = 0\). \[ \dfrac{dA}{dx} = -2x + 200 = 0 \] Solve the above equation for \(x\). \[ x = 100 \] \(\dfrac{dA}{dx}\) has one zero at \(x = 100\). The second derivative \(\dfrac{d^2A}{dx^2} = -2\) is negative (see calculus theorem on using the first and second derivative to determine extrema of functions). The value of the area \(A\) at \(x = 100\) is equal to \(10000 \, \text{mm}^2\) and it is the largest (maximum). So if you select a rectangle of width \(x = 100 \, \text{mm}\) and length \(y = 200 - x = 200 - 100 = 100 \, \text{mm}\) (it is a square!), you obtain a rectangle with maximum area equal to \(10000 \, \text{mm}^2\).

### Exercises

1 - Solve the same problem as above but with the perimeter equal to \(500 \, \text{mm}\).__Solution to the above exercise__

Width \(x = 125 \, \text{mm}\) and length \(y = 125 \, \text{mm}\).