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Mathematics
Textbook & Expert-Verified
A farmer has 360 feet of fencing to make three identical adjacent rectangular pens. What dimensions of each pen will maximize the total enclosed area?

Answer 1

Let x = width of each pen (across the row) and y = depth (the common length).
Three adjacent pens require 4 vertical fences of length y and two horizontal runs each of length 3x, so the fencing constraint is
6x + 4y = 360.

Total area A = 3xy. Use the constraint to eliminate y:
y = (360 − 6x)/4 = 90 − 1.5x,
so A(x) = 3x(90 − 1.5x) = 270x − 4.5x^2.

Maximize A: A'(x) = 270 − 9x = 0 ⇒ x = 30 ft. Then y = 90 − 1.5(30) = 45 ft.

Thus each pen should be 30 ft by 45 ft. (Each pen area = 1,350 ft^2, total area = 4,050 ft^2.)