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

Answer:

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

Step-by-step 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

$L = \dfrac{5000}{3} - \dfrac{2w}{3}$

Recall that:

The area of the rectangle = L×w

$A(w) = ( \dfrac{5000}{3}-\dfrac{2}{3} ) w$

$A(w) = \dfrac{5000}{3}w-\dfrac{2}{3} w^2$

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

$A' (w) = \dfrac{5000}{3} - \dfrac{2}{3} ( 2 w)$

$A' (w) = \dfrac{5000}{3} - \dfrac{4w}{3}$

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

So; $\dfrac{5000}{3} - \dfrac{4w}{3}=0$

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 $L = \dfrac{5000}{3} - \dfrac{2w}{3}$ can now be :

$L = \dfrac{5000}{3} - \dfrac{2(1250)}{3}$

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.