Question

A rectangular area is to be enclosed and divided into thirds. The family has $760 to spend for the fencing material. The outside fence costs $10 per running foot installed, and the dividers cost $20 per running foot installed. What are the dimensions that will maximize the area enclosed?

Answer 1

`x=(19)/(3)\ feet\\y=19\ feet`

Explanation:

Derivatives and Optimization

We can find where a function f has points of maxima/minima by using the first derivative criteria, i.e. f'=0 and evaluate its critical points.

Let's say the rectangular area has dimensions x (height) and y (width) and we are going to split the width into thirds. The area of that rectangle is

`A=xy`

The perimeter of that rectangle is the length of the outside fence

`P_o=2x+2y`

It costs $10 per feet, so the cost of the outside fence is

`C_o=10(2x+2y)=20x+20y`

The dividers cost $20 per foot, and we have assumed we split the width, so each divider has a length of x feet. The cost of the internal dividers is

`C_i=20(2x)=40x`

The total cost of the fencing is

`C=20x+20y+40x=60x+20y=20(3x+y)`

We know there is a limit of $760 to spend for the fencing material, so

`20(3x+y)=760`

Reducing

`3x+y=38`

Solving for y

`y=38-3x`

The area can be now expressed in terms of x alone

`A=xy=x(38-3x)=38x-3x^2`

To find the critical point, and possible maxima/minima, we set A'=0

`A'=38-6x=0`

We find:

`x=(19)/(3)`

Since

`y=38-3x=19`

The second derivative is

A''=-6

This means the area is maximum at the critical point

The dimensions that maximize the area are:

`x=(19)/(3)\ feet\\y=19\ feet`