Arectangular area is to be enclosed and divided into thirds. the family has $1040 to spend for the fencing material. the outside fence costs $10 per foot installed, and the dividers cost $20 per foot installed. what are the dimensions that will maximize the area enclosed? (the answer contains a fraction.)

  26 m long by 8 2/3 m wide (dividers are 8 2/3 m)

Step-by-step explanation:

Let x represent the length of the side of the rectangular area parallel to the partitions. Then the cost of the ends and partitions is ...

  10x +20x +20x +10x = 60x

The cost of the other two sides of the rectangle will be (1040 -60x), so those sides will have length ...

  (1/2)(1040 -60x)/10 = 52 -3x

The area of the rectangle is the product of its two dimensions ...

  area = x(52 -3x)

This is a downward-opening parabola that has zeros at x=0 and x=52/3. The vertex of the parabola will be halfway between these values, at ...

  x = (0 +52/3)/2 = 26/3

  52 -3(26/3) = 26 . . . . other side dimension

The area is maximized when the enclosure side parallel to the partitions is 8 2/3 meters, and the other side is 26 meters.

Comment on this problem

By making the cost of the partitions double, the problem effectively becomes one of maximizing the areas of three separate enclosures. The area of each of those is maximum when each is a square, $1040/(3 enclosures×4 sides/enclosure×$10/m) = 8 2/3 m/side. When these are placed side-by-side, the rectangle is 8 2/3 m by 3×(8 2/3 m), or 8 2/3 m by 26 m.

Comment on this type of problem

This sort of problem has the solution that the area of the rectangle is maximized for the cost when half the cost is spent in each of the orthogonal directions of the sides of the rectangle. This is true for a variety of geometries, including ones where all or part of a side or partition may be missing.

72 is the answer to this problem

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