Wednesday
March 29, 2017

Post a New Question

Posted by 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!!!!!!!!!!

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question