# Given a perimeter of 180, how do you find the length and the width of the rectangle of maximum area?

Sep 24, 2016

Given a perimeter of 180, the length and width of the rectangle with maximum area are 45 and 45.

#### Explanation:

Let $x =$ the length and $y =$ the width of the rectangle.
The area of the rectangle $A = x y$

$2 x + 2 y = 180$ because the perimeter is $180$.

Solve for $y$
$2 y = 180 - 2 x$
$y = 90 - x$

Substitute for $y$ in the area equation.
$A = x \left(90 - x\right)$
$A = 90 x - {x}^{2}$

This equation represents a parabola that opens down. The maximum value of the area is at the vertex.

Rewriting the area equation in the form $a {x}^{2} + b x + c$
$A = - {x}^{2} + 90 x \textcolor{w h i t e}{a a a} a = - 1 , b = 90 , c = 0$

The formula for the $x$ coordinate of the vertex is
$x = \frac{- b}{2 a} = \frac{- 90}{2 \cdot - 1} = 45$

The maximum area is found at $x = 45$
and $y = 90 - x = 90 - 45 = 45$

Given a perimeter of 180, the dimensions of the rectangle with maximum area are 45x45.