Farmer joe is enclosing a rectangular area on his arm for his chicken with 200 feet of fencing that he recently acquired in a trade. two equal lengths of fencing, of unknown length x ft, will run perpendicular to the side of the barn, and a single length of fencing of unknown length (200-2x) ft will run parallel to the side of the barn. to the nearest square foot, what is the maximum possible area that joe can enclose with his 200ft of fencing

Question

Karson Greer

Mathematics

14 August, 15:09

Farmer joe is enclosing a rectangular area on his arm for his chicken with 200 feet of fencing that he recently acquired in a trade. two equal lengths of fencing, of unknown length x ft, will run perpendicular to the side of the barn, and a single length of fencing of unknown length (200-2x) ft will run parallel to the side of the barn. to the nearest square foot, what is the maximum possible area that joe can enclose with his 200ft of fencing

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Answers (1)

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Cailyn · Answer 1

For this case, the area is given by:

A = x * (200-2x)

Rewriting:

A = 200x-2x ^ 2

Deriving the expression we have:

A ' = 200-4x

Equaling zero we have:

200-4x = 0

We clear x:

4x = 200

x = 200/4

x = 50 feet

Then, the maximum area is:

A (50) = 50 * (200-2 * 50)

A (50) = 5000 feet ^ 2

Answer:

the maximum possible area that can be enclosed with his 200ft of fencing is:

A (50) = 5000 feet ^ 2