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APPM 1350 Review Solutions 3

# 12 a farmer with 200 ft of fencing wants to enclose a

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Unformatted text preview: that x > 0 implies x = 1 and x2 S 00 (x) = 2/x3 and so S 00 (1) > 0 which implies a minimum. 12. A farmer with 200 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens? Solution: x y Let x and y be the length of the sides of the rectangular fence. We wish to build pens with walls that are parallel to the side of length x. Thus, we wish to maximize the area A = xy subject to the contstraint on the perimeter, 5x +2y = 200. Solving for y in the constraint condition and substituting ✓ ◆ 200 5x 200x 5x2 into A yields, A(x) = x = and note that y 0 implies x 40 and so we 2 2 200x 5x2 200 10x wish to maximize A(x) = where 0 x 40. Now note that A0 (x) = and so 2 2 x = 20 is a critical point. Now note that A(0) = 0 = A(40) and A(20) =1000, thus the largest possible area is A = 1000 ft2 ....
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