Posted by Michael on Monday, March 26, 2012 at 4:31pm.
I don't understand the two by two arrangement.
What you need to do is write an expression for the total amount of fence (outside+inside lengths)=900
it will be in terms of fractions of length, width.
Then you want to maximize area:
Area= Outsidelength*width.
Now in the fence length needed equation, solve for w. Put that in the area equation.
Take the derivative dA/dl, set to zero, and solve for l.
Might as well go for the general nxn case, then we can plug in what we want.
If there are n^2 small pens, with width x and length y, with length parallel to the long fence, then the total fence used is
n(n+1)x + n^2y = 900
y = (900-n(n+1)x)/n^2
A = n^2 xy
= n^2 x (900-n(n+1)x)/n^2
= x(900-n(n+1)x)
= 900x - n(n+1)x^2
dA/dx = 900 - 2n(n+1)x
max at x = 900/[2n(n+1)]
x = 450/[n(n+1)]
y = 450/n^2
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