# How do you find the maximum area of rectangle with 80ft perimeter?

Jan 24, 2016

Algebraically, we can show that the maximum area of the rectangle is achieved when it is a square. In this case, the maximum area is ${\text{400 ft}}^{2}$

#### Explanation:

Since this is not in calculus, I'll provide a non-calculus answer.

We know the rectangle always has a perimeter of $80$, so $2 l + 2 w = 80$, which simplifies to be $l + w = 40$.

We also know that the area of the rectangle is $A = l w$, but we can express this as a function of a single variable.

Use the perimeter expression $l + w = 40$ to say that $l = 40 - w$. Because this will always be true in the rectangle, we can substitute $40 - w$ for $l$ in $A = l w$.

$A = l w$

$A = \left(40 - w\right) w$

$A = - {w}^{2} + 40 w$

This quadratic function can be graphed, and the highest point will be the spot where the rectangle's area is maximized.

graph{-x^2+40x [-5, 45, -145, 460]}

The highest point, the vertex of the parabola, is $\left(20 , 400\right)$. When $w = 20$, we also know that $l = 20$ (which forms a square) and the maximum area of the rectangle (square) is ${\text{400 ft}}^{2}$.