Respuesta :

opudodennis

Answer:

21 ft by 28 ft

Step-by-step explanation:

To maximize the area, see the attached.

Perimeter will be 4l+3w which is equal to the fencing perimeter, given as 168

4l+3w=168

Making l the subject then

4l=168-3w

l=42-¾w

Area of individual land will be lw and substituting l with l=42-¾w

Then

A=lw=(42-¾w)w=42w-¾w²

A=42w-¾w²

Getting the first derivative of the above with respect to w rhen

42-w6/4=0

w6/4=42

w=42*4/6=28

Since

l=42-¾w=42-¾(28)=21

Therefore, maximum dimensions are 21 for l and 28 for w

Ver imagen opudodennis
raphealnwobi

The dimension of the individual coop has is 42 ft. by 21 ft.

Let the length of each chicken coop be l ft and the width of each chicken  coop be w ft. Since the coops are built adjacent to each other, hence the total length = l + l = 2l and the width = w

The perimeter of the coop = 2(length + width) = 2(2l + w) = 4l + 2w

Hence:

168 = 4l + 2w

w + 2l = 84

w = 84 - 2l     (1)

The area (A) = length * width = 2l * w = 2l(84 - 2l)

A = 168l - 4l²

At maximum area, dA/dl = 0, therefore:

dA/dl = 168 - 8l

168 - 8l = 0

8l = 168

l = 21 ft

putting l = 21 ft in equation 1 gives:

w = 84 - 2(21) = 84 - 42 = 42

The dimension of the individual coop has width of 42 ft and length of 21 ft

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