acob wanted to buy some ice cream, but did not know what flavor to get so he asked Caitlyn to pick for him. The shop has 4 flavo
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I've attached a plot of one such cross-section (orange) over the region in the x-y plane (blue), including the bounding curves (red). (I've set 
for this example.)
The length of each cross section (the side lying in the base) has length determined by the horizontal distance
between the y-axis 
and the curve 
. In terms of 
, this distance is 
. The height of each cross section is twice the value of , so the area of each rectangular cross section should be 
.
This means the volume would be given by the integral
The length of each cross section (the side lying in the base) has length determined by the horizontal distance
This means the volume would be given by the integral
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.
Disclaimer : The missing figure for the question is attached below.
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