Question

a famer has 120 feet of fencing with whcih to enclose two adjacent rectangular pens as shown. what dimeensions should be used that the enclosed area will be a maximum? what will the area be?

Answer 1

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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