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Unformatted text preview: Solution to practice problems for midterm 3 1. If the rectangle has dimension x and y , then its perimeter is 2 x + 2 y = 100. So y = 50 x . Thus, the area is A = xy = x (50 x ) = 50 x x 2 , where 0 ≤ x ≤ 50. We wish to maximize the function A ( x ). Therefoe, we first need to find all critical points of A ( x ). Notice that A ′ ( x ) = 50 2 x = 0 only at x = 25. The area A (25) = 25(50 25) = 625, while A (0) = A (50 = 0. Clearly, the maximum area occurs at x = 25 and y = 50 x = 25, and the maximum area is A (25) = 625. 2. Let x > 0 and f ( x ) = x + 1 x . We wish to minimize f ( x ). Notice that f ′ ( x ) = 1 1 x 2 = 0 implies 1 x 2 = 1. Hence x 2 = 1. We have two critical points, x = 1 and x = 1. However, since we only consider positive numbers x > 0, here we only take on critical point x = 1. Use the first derivative test, notice that f ′ ( x ) < 0 for 0 < x < 1 and f ′ ( x ) > 0 for x > 1, so f has an absolute minimum at x = 1. At this point, f (1) = 2. 3. Let b be the length of the base of the box and h be the height. The surface area is 1200 = b 2 +4 hb . Solve it for h , we have h = 1200 − b 2 4 b . The volume is V = b...
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 Fall '08
 PAGANO
 Math, Calculus

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