Two chicken coops are to be built adjacent to one another from 168 ft of fencing. What dimensions should be used to maximize the area of an individual coop?

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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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

step-by-step explanation:

step 1: solve one of the equations for one of the variables. let's solve the first equation for y:

step 2: substitute that equation into the other equation, and solve for x.

step 3: substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.