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PilotLPTM [1.2K]
2 years ago
6

You have 132 m of fencing with which to make 2 side by side rectangular enclosures against an existing wall. If the enclosures a

re adjacent and of the same depth, what is the maximum area that can be enclosed
Please answer ASAP will give out brainiest
Mathematics
1 answer:
Deffense [45]2 years ago
8 0
The two enclosures will need three equal fences coming out from the wall and meeting another fence running parallel to the wall. If the fences coming out from the wall are x metres long the parallel fence will be (132 - 3x) metres long.
The area A = x(132 - 3x) = 132x - 3x^2
The derivative of A = zero when 132 - 6x = 0 which means the maximum area is when x = 22m
The maximum area = 22 x (132 - 3 x 22) = 1452 m^2

If you don’t know how to find derivatives then you could sketch the graph of y = x(132 - 3x).
This is an inverted parabola (hill) with x intercepts at 0 and 132/3 = 44.
The maximum point (top of the hill) is halfway between 0 and 44 I.e. 22m

Try any other value for x and the area will be smaller.
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