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Dmitry_Shevchenko [17]
3 years ago
6

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?

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
4 0

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

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Question:

What is the solution set to the inequality (4x – 3) (2x – 1) ≥ 0?

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