Suppose you have 76 feet of fencing to enclose a rectangular dog pen. The function A=38x-x^2, where x = width, gives you the area of the dog pen per square feet. What width gives you the maximum area? What is the maximum area? Round to the nearest tenth if necessary.

Question

Huey

Mathematics

8 May, 04:03

Suppose you have 76 feet of fencing to enclose a rectangular dog pen. The function A=38x-x^2, where x = width, gives you the area of the dog pen per square feet. What width gives you the maximum area? What is the maximum area? Round to the nearest tenth if necessary.

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Answers (1)

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Amelie Lane · Answer 1

If A=38x-x^2 then

dA/dx=38-2x

d2A/dx2=-2

Since the acceleration, d2A/dx2 is a constant negative, when velocity, dA/dx=0, it will be an absolute maximum for A (x)

dA/dx=0 only when 38=2x, x=19

A (19) = 38 (19) - 19^2

A (19) = 722-361

A (19) = 361 ft^2

So the maximum possible area is 361 ft^2

(This will always be true as the maximum possible area enclosed by a given amount of material will always be a perfect square ...)