Question

A farmer with 5000 feet of fencing wants to enclose a rectangular field and then divide it into two plots by adding a fence in the middle parallel to one of the sides. Using the function A(x) , analytically find the WIDTH of the field that would maximize its area. (Round your answer to the nearest whole number)

Answer 1

The required width of the field that would maximize the area is = 1250 feet

Explanation:

Given that:

The total fencing length = 5000 ft

Let consider w to be the width and L to be the length.

Then; the perimeter of the rectangular field by assuming a parallel direction is:

P = 3L + 2w

⇒ 3L + 2w = 5000

3L = 5000 - 2w

`L = (5000)/(3) - (2w)/(3)`

Recall that:

The area of the rectangle = L×w

`A(w) = ( (5000)/(3)-(2)/(3) ) w`

`A(w) = (5000)/(3)w-(2)/(3) w^2`

Taking the differentiation of both sides with respect to t; we have:

`A' (w) = (5000)/(3) - (2)/(3) ( 2 w)`

`A' (w) = (5000)/(3) - (4w)/(3)`

Then; we set A'(w) to be equal to zero;

So;
`(5000)/(3) - (4w)/(3)=0`

5000 = 4w

w = 5000/4

w = 1250

Thus; the required width of the field that would maximize the area is = 1250 feet

Also, the length
`L = (5000)/(3) - (2w)/(3)` can now be :

`L = (5000)/(3) - (2(1250))/(3)`

L = (5000 -2500)/3

L = 2500/3 feet

Suppose, the farmer divides the plot parallel to the width; Then 2500/3 feet = 833.33 feet and the length L = 1250 feet.