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Please help with this problem.

Please help with this problem.-example-1

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Answer:

150 ft by 100 ft

Explanation:

You want the dimensions of the maximum rectangular area that can be enclosed using 600 ft of fence when there is one partition in the area, parallel to the shortest side.

Dimensions

Let x represent the length of the longest side. Then the length of the short side is ...

(600 -2x)/3

Area

And the total enclosed area is ...

A = LW

A = x(200 -2/3x)

Maximum

The area equation defines a parabola that opens downward. Its x-intercepts are x=0 and x=200/(2/3)=300. The maximum area is had when x is halfway between these values, at x = (0 +300)/2 = 150.

The short side is ...

(600 -2(150))/3 = 300/3 = 100

The dimensions that maximize the enclosed area are 150 ft by 100 ft.

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Additional comment

The generic solution to this sort of problem is that the cost of fence is equal in the two orthogonal directions. Here, the cost is measured by the length, so the long sides total half the perimeter, and the short sides (and partition) split the remaining half of the perimeter fence.

In those cases where part of the enclosure boundary is something other than the fence, the half-and-half split remains true. For example, if one side is a "river," then half the fence is parallel to the river, and the other half is perpendicular to the river.

Similarly, if fence one direction (or on one side) has a different cost, the total cost is split evenly in the two orthogonal directions.

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User Karthik AMR
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