a. The function that gives the area A(x) of the playground(in square meters) in terms of x is A(x) = -2x² + 240x
b. The length of each side(x) running perpendicular to the school building is 60 cm.
c. The maximum total area that the playground can have is 120 × 60 = 7200 cm²
The 2 sides perpendicular to the school building = x
Let
The side parallel to the side school building = y
Therefore,
The perimeter for the 3 sides = 2x + y
240 = 2x + y
y = 240 - 2x
Area = xy
a.
The function that gives the area A(x) of the playground(in square meters) in terms of x can be calculated below:
- Area = x(240 - 2x)
- A(x) = 240x - 2x²
- A(x) = -2x² + 240x
b.
The sides x that gives the maximum area that the playground have can be calculated as follows:
This is a parabola facing downward because the leading coefficient is less than zero.
The maximum point are (h, k).
h = - b / 2a (this gives the maximizing point)
b = 240
a = -2
h = - 240 / 2 × - 2
h = -240 / -4
h = 60
Therefore,
x(side length) = 60 meter
c.
The maximum area that the playground can have can be calculated below:
The 2 sides perpendicular to the building playground is 60 meters each. The third will be 240 - 60(2) = 240 - 120 = 120 cm.
The maximum area = 120 × 60 = 7200 cm²
Therefore,
- The function that gives the area A(x) of the playground(in square meters) in terms of x is A(x) = -2x² + 240x
- The length of each side(x) running perpendicular to the school building is 60 cm.
- The maximum total area that the playground can have is 120 × 60 = 7200 cm².
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