Mathswatch Unit 13-Relative Frequencey
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The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
Let the area be y .
Area = (base) × (height)
Base = 2x
Height = h
Let the area of the rectangular pens be y .
∴ y = 2xh
Perimeter of all the fencing = 4x+3h
∴ 4x+3h = 120
now we solve for h
3h = 120-4x
h = 40 - 4/3 x
Now we will substitute this value in the above first equation:
y = 2xh
or, y = 2x (40 - 4/3 x)
or, y = 80x - 8/3 x²
Now for the maximum area we have to find the first order differentiation of y
now,
dy /dx = 80 - 16/3 x
At dy/dx = 0 we get the value of x for which y is maximum.
80 - 16/3 x = 0
or, - 16/3 x = -80
or, x = 15 feet
Hence height = 40 - 4/3 x = 40 - 20 = 20feet
Maximum area = 2xh = 2×15×40 = 1200 square feet
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
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