Chapter 7.7, Problem 6P

To determine

Tofind: thegreatest possible area of the rectangle.

Answer to Problem 6P

The greatest possible area of the rectangle is ${\text{1250m}}^2$ .

Explanation of Solution

Given information:

The three sides sum is $100\text{m}$ .

Calculation: to find the greatest area, let $x$ represents the sides of the fence be of length then the long side of length is $100-2x$ .

So the area of the rectangle is,

$\begin{array}{l}a=x\left(100-2x\right)\\ a=100x-2x^2\text{ [multiply]}\end{array}$

To find the greatest area , first differentiate and then equal to zero,

$\begin{array}{l}a=100x-2x^2\\ a\text{'}=100-4x\text{ [differentiate]}\\ 100-4x=0\text{ [equal to zero]}\\ x=25\text{ [simplify]}\end{array}$

Now, substitute in the area function,

$\begin{array}{l}a=100\left(25\right)-2{\left(25\right)}^2\text{ [substitute]}\\ a=2500-1250\text{ [multiply]}\\ a=1250\text{ [subtract]}\end{array}$

Thus, the greatest possible area is ${\text{1250m}}^2$ .