# Calculus

posted by
**Carla** on
.

Consider the following problem: A farmer with 810 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total are of the four pens?

your drawing should look like this

......................

. . . .

. . . .

. . . .

. . . .

. . . .

......................

then let the width of the smaller rectangle be x, let its length be y

so the amount of fencing would be

8x + 5y

so 8x+5y=810

y=(810-8x)/5

Area = 4xy = 4x(810-8x)/5

= 648x - 32x^2 /5

d(Area)/dx = 648 - (64/5)x = 0 for a max of Area

solve to get x=50.625

and subst to get y=81

for a max area of 4(50.625)(81)=16402.5

sorry, my diagram did not come out the way I hoped it would.

It is a large rectangle cut into four smaller ones.

Thank you!!!!!!!!!!