# Calculus 2

posted by on .

A farmer wishes to build a fence for 6 adjacent rectangular pens. If there is 600 feet of fencing available, what are the dimensions of each pen that maximizes total pen area?

I keep getting x=150 but I have been told that is not enough fencing. Can anyone help?

• Calculus 2 - ,

You must give a description of the pens.

Is there a large rectangle with equal partitions parallel to the widths ? (the usual case)

Since you don't say what the x stood for, I have no way of telling what the 150 represents, since "dimension" implies length and width.

• Calculus 2 - ,

That was the only thing the question said :/. It never gave a description of the pens. There is a picture with 6 boxes connected to each other 3 boxes on top and 3 on bottom:

box box box
box box box

other than that, that was all the info I was given:(

• Calculus 2 - ,

In that case, if each pen has width x and height y in the drawing, then

total area is 6xy
Also, 3x+3x+3x+2y+2y+2y+2y = 600, so 9x+8y=600

a = 6xy = 6x(600-9x)/8
= 9/4 x(200-3x)
da/dx = 9/2 (100-3x)
so, da/dx = 0 when x = 100/3

so, each small pen is 100/3 by 75/2

max area = 7500

• Calculus 2 - ,

Ok, then it isn't that bad
Make a sketch,
label the length of each small pen as x and its width y
counting up all the x's and y's, I get
9x + 8y = 600
y = (600 - 9x)/8
where 600-9x > 0
9x < 600
x < 66.67

area = 3x(2y)= 6xy
= 6x(600-9x)/8
= 3600x - (27/4)x^2

This is a parabola which opens downwards, so it has a maximum
the x of the vertex is -b/(2a) = -3600/(-27/2) =266.67
which is beyond our restriction of x

Thus this question has no solution

by Calculus:
d(area)/dx = 3600 - 27x/2
= 0 for a max area
27x/2 = 3600
27x = 7200
x = 266.66..
y = (600 - 9(266.67)/8 which is a negative
no solution

here is a picture of why
http://www.wolframalpha.com/input/?i=plot+y+%3D+3600x+-+%2827%2F4%29x%5E2

• Just ignore my last post -Calculus 2 - ,

What stupid error that was.

my mistake is in
= 6x(600-9x)/8
= 3600x - (27/4)x^2
what garbage that is !!!

Go with STeve