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?

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

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

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.

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:(

To find the dimensions of each pen that maximize the total pen area, we need to set up an equation based on the available fencing.

Let's assume that the length of each pen is x feet, and the width is y feet.

Since there are six rectangular pens, we have the following relations:

1. The length of each pen, 6x.
2. The width of each pen, y.

The total amount of fencing used will be the sum of the perimeters of all six pens. Each pen has two equal-length sides (length) and two equal-width sides (width), so the total amount of fencing used is:

2(6x) + 2(y) = 12x + 2y

According to the given information, the available fencing is 600 feet. So we have:

12x + 2y = 600

To maximize the total pen area, we need to find the dimensions that maximize the area of a single pen. The area of a rectangle is given by:

Area = length * width
Area = (6x) * (y)
Area = 6xy

Since we want to maximize the area, we need to maximize 6xy while satisfying the fencing constraint.

To proceed, we can use the fencing constraint equation to express y in terms of x:

12x + 2y = 600
2y = 600 - 12x
y = (600 - 12x) / 2
y = 300 - 6x

Substitute this expression for y into the area equation:

Area = 6xy
Area = 6x(300 - 6x)
Area = 1800x - 36x^2

Now we have the area equation in terms of x. To find the maximum area, we can take the derivative of the area equation with respect to x and set it equal to zero:

d(Area)/dx = 1800 - 72x
1800 - 72x = 0
72x = 1800
x = 1800/72
x = 25

Now that we have the value of x, we can substitute it back into the equation for y to find the corresponding value:

y = 300 - 6x
y = 300 - 6(25)
y = 300 - 150
y = 150

So, the dimensions that maximize the total pen area while using 600 feet of fencing are: x = 25 feet and y = 150 feet.

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