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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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 Fall '07
 SEGUR,HARV

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