Question

A rancher has 50,000 linear feet of fencing and wants to enclose a rectangular field and then divide it into four equal pastures with three internal fences parallel to one of the rectangular sides. what is the maximum area of each pasture?

Answer 1

Given that a rancher wants to enclose a rectangular field and then divide it into four equal pastures with three internal fences parallel to one of the rectangular sides.

Let the side of the rectangle parallel to the three internal fences be x and the other side of the rectangle be y, then the total length to be enclosed is given by 5x + 2y.

Given that the rancher has 50,000 linear feet of fencing available, then

$5x + 2y = 50,000 \\ \\ 2y = 50,000 - 5x \\ \\ y = \frac{50,000}{2} - \frac{5}{2} x=25,000-2.5 x$

Thus, the dimensions of the rectangular field to be fences is x by (25,000 - 2.5x).

The area of a rectangle is given by length times width.

Thus, the area of the rectangular field is given by

$Area=x(25,000 - 2.5x)=25,000x-2.5x^2$

For the area to be maximum, the derivative of the area with respect to x equals 0.

Thus,

$\frac{d}{dx}(Area)=25,000-5x=0 \\ \\ \Rightarrow5x=25,000 \\ \\ \Rightarrow x= \frac{25,000}{5} =5,000$

Thus, the the maximum area to be enclosed is given by

$Area=25,000(5,000)-2.5(5,000)^2 \\ \\ =125,000,000-2.5(25,000,000) \\ \\ =125,000,000-62,500,000 \\ \\ =62,500,000$

Therefore, the maximum area of each pasture is given by

$\frac{62,500,000}{4} =15,625,000\ square\ feet$