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
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
2 answers
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
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