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
[tex]L = \dfrac{5000}{3} - \dfrac{2w}{3}[/tex]
Recall that:
The area of the rectangle = L×w
[tex]A(w) = ( \dfrac{5000}{3}-\dfrac{2}{3} ) w[/tex]
[tex]A(w) = \dfrac{5000}{3}w-\dfrac{2}{3} w^2[/tex]
Taking the differentiation of both sides with respect to t; we have:
[tex]A' (w) = \dfrac{5000}{3} - \dfrac{2}{3} ( 2 w)[/tex]
[tex]A' (w) = \dfrac{5000}{3} - \dfrac{4w}{3}[/tex]
Then; we set A'(w) to be equal to zero;
So; [tex]\dfrac{5000}{3} - \dfrac{4w}{3}=0[/tex]
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 [tex]L = \dfrac{5000}{3} - \dfrac{2w}{3}[/tex] can now be :
[tex]L = \dfrac{5000}{3} - \dfrac{2(1250)}{3}[/tex]
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.
