Question

A farmer wants to enclose a rectangular field along a river on three sides. If 4,800 feet of fencing is to be used, what dimensions will maximize the enclosed area

Answer 1

We know that:
\[
2l+2w=4800
\]And we wish to maximize:
\[
lw=A
\]So, we write one in terms of the other:
\[
l+w=2400\implies\\
w=2400-l\implies\\
A=l(2400-l)
\]To find the maximum, we can simply find the vertex, but I'll do it a slightly different way here:
\[
A=2400l-l^2\implies\\
\frac{d}{dl}A=2400-2l\implies\\
\]A is a maximum when A'=0, so:
\[
0=2400-2l\implies\\
l=1200\\
w=2400-l=2400-1200=1200
\]Therefore the maximum is, when \(l=1200\) and \(w=1200\)

Of course, this problem can be simplified to finding the largest square, but that won't always be the case for these kinds of problems.

Answer 2

one small thing, i think the correct equation is 2w + l = 4800 for perimeter, but your answer is correct

Answer 3

err actually i think length should be.. 2400

Answer 4

Yes, you are correct.
Wow, I feel stupid, for not noticing that... but thank you for the correction!

Answer 5

I'll re-trace this problem quickly, to correct my mistake:
\[l+2w=4800\\
l=4800−2w\\
lw=A\]
Which means:
\[A=w(4800−2w)=4800w−2w^2\implies\\
A′=4800−4w\]Gives a maxima at:\[
4w=4800\implies\\
w=1200
\]Which means the length is:\[
l=4800−2(1200)=2400
\]Thus:
\[
l=2400\\
w=1200\]
Jeez, I screwed up twice.. need to get some sleep! G'night!

Answer 6

Thanks, GN