# Math

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24 m of fencing are available to enclose a play area. what would the maximum area be if the fencing only went on 3 sides because the wall is used as the 4th side.

• Math -

l + 2w = perimeter
l + 2w = 24
l = 24-2w

Area = l times w

(24-2w)w = Area.
24w -2w^2 = Area.

If you are doing this using calculus, you would take the derivative of the area and set it equal to zero to find w.

• Math -

let the side parallel to the wall by y m
let the other two sides be x m each

so we have y + 2x = 24 or y = 24-2x
area = xy
= x(24-2x
= -2x^2 + 24x
let's complete the square to find the vertex of this parabola
= -2(x^2 - 12x + 36 - 36)
= -2( (x-6)^2 - 36)
= -2(x-6)^2 + 72

the vertex is (6,72)
So the maximumum area is 72 ^2 , when x = 6 and y = 12

or

by calculus
d(area) = -4x + 24 = 0 for a max area
x = 6
then y = 24 - 12 = 12
max area = 6(12) = 72

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