Asked by Jonathan

There is 1116 meters of fencing available to enclose a field. The field is to be divided into two sections by a fence in the middle. What is the maximum area of the field that can be enclosed?

A=L xW
2w+3L=1116
3L=1116-2w
l=1116-2w/3

my final answer came out to 51894

can you check my work please and thank you
Answered by oobleck
correct.
To save some work in the future, know that for this kind of problem, the maximum area is when the fencing is divided equally among lengths and widths. 1116 = 558*2
so since you have two lengths, L = 558/2 = 279
there are 3 widths, so W = 559/3 = 186
279*186 = 51894
Answered by Anonymous
L = (1116-2w) / 3 I think
A = w * (1116-2w) / 3
3 A = 1116 w - 2 w^2
Find vertex of that parabola, I am going to cheat and use calculus
d(3A) /dw = 0 at max = 1116 - 4 w
so
4 w = 1116
w = 279
then L = (1116-2w) / 3 = 186
so
A = 51894
we agree
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