Calculus 2
posted by Henry .
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 
Reiny
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 
Henry
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 
Steve
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(6009x)/8
= 9/4 x(2003x)
da/dx = 9/2 (1003x)
so, da/dx = 0 when x = 100/3
so, each small pen is 100/3 by 75/2
max area = 7500 
Calculus 2 
Reiny
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 6009x > 0
9x < 600
x < 66.67
area = 3x(2y)= 6xy
= 6x(6009x)/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 
Reiny
What stupid error that was.
my mistake is in
= 6x(6009x)/8
= 3600x  (27/4)x^2
what garbage that is !!!
Go with STeve
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