SOLUTION: A farmer would like to enclose a rectangular portion of his field using fences on all four sides. In addition, he must divide this rectangle into three smaller rectangular regions

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Question 966051: A farmer would like to enclose a rectangular portion of his field using fences on all four sides. In addition, he must divide this rectangle into three smaller rectangular regions by building two more fences parallel to one of the rectangle's sides. If he only has 240 meters of fencing material to use, what is the maximum total area he can enclose?
Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
Main outer dimensions of the rectangle, x and y. Two fence portions are of y, and four fence portions are of x, so the total fencing to be 240 meters,
2x%2B4y=240.

The area is A=xy, for the whole outer rectangle.

Simplified, the fence length equation is x%2B2y=120 and then x=120-2y.

Substituting,
A=%28120-2y%29y
But if you just want to find the MAXMIUM for A, then this will be in the exact middle of the roots.

Roots: y%28120-2y%29=A=0
y=0 or 120-2y=0, y=60;

The exact middle of 0 and 60 is 30.
highlight%28y=30%29 for the maximum area.]

x=120-2%2A30
x=120-60
highlight%28x=60%29 for the maximum area.

The maximum area is 60%2A30=highlight%281800%29.