Question

See attachment.

Answer 1

I get L=30 and W=15.

Answer 2

So, area = 30*15 = 450 SQ FT.

Answer 3

Here is how I did it.

Perimeter = 2W + 2L. But, one side is covered by the barn. So, 60 FT of fence covers the other three sides. So, 60 = W + 2L.

Area = W*L = W(60-W)/2 = 30W - W^2/2.

Maximum area is when differential of that with respect to W is zero:
d/dW (30W - W^2/2) = 30 - W = 0 => W = 30

Because W + 2L = 60, L has to be 15.

Answer 4

so it looks like this?

Answer 5

?

Answer 6

Yes.

Answer 7

kk THX ^_^

Answer 8

One line solution using Mathematica:
Input: Maximize[ { L w, 2 w + L == 60 }, { L, w} ]
output: { 450, { L ->30, w ->15 } }