A farmer needs to put in a fence in their rectangular field. If we look at the field from above, the cost of the west and east sides are $10/ft, the cost of the south side is $2/ft and the cost of the north side is $7/ft. If we have $700, use optimization to determine the dimensions of the field that will maximize the enclosed area.

Step-by-step explanation:

Short answer: area is maximized when half the cost is spent in each of the orthogonal directions. This means the east and west sides will total $350 at $20 per foot, so will be 17.5 feet. The north and south sides will total $350 at $9 per foot, so will be 38 8/9 feet.

The dimensions that maximize the area are 17.5 ft in the north-south direction by 38 8/9 ft in the east-west direction.

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Long answer: If x represents the length of the north and south sides, and y represents the length of the east and west sides, then the total cost is ...

  10y +10y +2x +7x = 700

  9x +20y = 700

  y = (700 -9x)/20

We want to maximize the area:

  A = xy = x(700 -9x)/20

We can do this by differentiating and setting the derivative to zero:

  dA/dx = 700/20 -9x/10 = 0

  350 -9x = 0 . . . . multiply by 10

  x = 350/9 = 38 8/9

  y = (700 -9(350/9))/20 = 350/20 = 17.5

The north and south sides are 38 8/9 ft long; the east and west sides are 17.5 ft long to maximize the area for the given cost.