Calculus
posted by Kirsten .
A farmer wishes to enclose a long rectangular pin that will then be divided into four separated, adjacent pins (all 4 pins are in a row) for different breeds of farm animals. He has 1100 feet of fencing available for use. What should the dimensions of the pin be so that he is able to maximize area? What is the maximum area?

Let the width of each of the smaller pens be x, and let the length of each of the smaller pens be y
So the total of all fences = 8x + 5y
8x + 5y = 1100
y = 220  8x/5
Area = 4xy
= 4x(220  8x/5)
= 880x  32x^2/5
d(Area)/dx = 880  64x/5 = 0 for a max of area
64x/5 = 880
64x = 4400
x = 68.75
y = 220  8(68.5)/5 = 110
So each of the small pens should be 68.75 by 110
check my arithmetic
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