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Supplementary Problems
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24.4 A rectangular eld is to be fenced in so that the resulting area is 100 square meters. Find the dimensions of that eld (if any) for which the perimeter is: (a) a maximum; (b) a minimum. 24.5 Find the point(s) on the parabola 2x = y2 closest to the point (1, 0). 24.6 Find the point(s) on the hyperbola x 2 y2 = 2 closest to the point (0, 1). 24.7 A closed box with a square base is to contain 252 cubic feet. The bottom costs \$5 per square foot, the top costs \$2 per square foot, and the sides cost \$3 per square foot. Find the dimensions that will minimize the cost. 24.8 Find the absolute maxima and minima (if any) of f (x) = x2 + 4 on the interval [0, 2). x 2
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24.9 A printed page must contain 60 square centimeters of printed material. There are to be margins of 5 centimeters on either side, and margins of 3 centimeters each on the top and the bottom. How long should the printed lines be in order to minimize the amount of paper used 24.10 A farmer wishes to fence in a rectangular eld of 10 000 square feet. The north south fences will cost \$1.50 per foot, whereas the east west fences will cost \$6.00 per foot. Find the dimensions of the eld that will minimize the cost.
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24.11 (a) Sketch the graph of y =
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(b) Find the point on the graph where the tangent line has the greatest slope. 24.12 (a) Find the dimensions of the closed cylindrical can [see Fig. 24-4(a)] that will have a capacity of k volume units and use the minimum amount of material. Find the ratio of the height h to the radius r of the top and bottom. (The volume is V = r 2 h, and the lateral surface area is S = 2 rh.) (b) If the bottom and the top of the can have to be cut from square pieces of metal and the rest of these squares is wasted [see Fig. 24-4(b)], nd the dimensions that will minimize the amount of material used, and nd the ratio of the height to the radius.
Fig. 24-4
Fig. 24-5
24.13 A thin-walled cone-shaped cup is to hold 36 cubic inches of water when full. What dimensions will minimize the amount of material needed for the cup (The volume is V = 1 r 2 h and the surface area is A = rs; see Fig. 24-5.) 3 24.14 (a) Find the absolute extrema on [0, + ) (if they exist) of f (x) = x . (b) Sketch the graph of f . (x 2 + 1)3/2
24.15 A rectangular bin, open at the top, is required to contain a volume of 128 cubic meters. If the bottom is to be a square, at a cost of \$2 per square meter, whereas the sides cost \$0.50 per square meter, what dimensions will minimize the cost 24.16 The selling price P of an item is 100 0.02x dollars, where x is the number of items produced per day. If the cost C of producing and selling x items is 40x + 15 000 dollars per day, how many items should be produced and sold every day in order to maximize the pro t 24.17 Consider all lines through the point (1, 3) and intersecting the positive x-axis at A(x, 0) and the positive y-axis at B(0, y) (see Fig. 24-6). Find the line that makes the area of BOA a minimum.
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Fig. 24-6
k 24.18 Consider the function f (x) = 1 x 2 + . (a) For what value of k will f have a relative minimum at x = 2 (b) For the value 2 x of k found in part (a), sketch the graph of f . (c) For what value(s) of k will f have an absolute minimum 24.19 Find the point(s) on the graph of 3x 2 + 10xy + 3y2 = 9 closest to the origin. [Hint: Minimize x 2 + y2 , making use of implicit differentiation.] 24.20 Fill in the gaps in the following proof of Theorem 24.1. Assume that f is continuous on an interval I and that f has a relative maximum on I at c. Assume, to the contrary, that d = c is a point in I with f (d) > f (c). On the closed interval I with endpoints c and d, f has an absolute minimum at some point u. Clearly, u = d. Also, u cannot lie in the interior of I . (Otherwise, f would have a relative minimum on I at u, contradicting the hypothesis that c is the only point in I at which f has a relative extremum.) Hence, u = c. This would imply that f (x) = f (c) in a subinterval of I and that f would have relative extrema at all points of that subinterval. This would contradict the hypothesis that c is the unique point in I at which f has a relative extremum. (A similar argument would show that f has an absolute minimum at c when it is assumed that f has a relative minimum at c.) 24.21 Prove the following theorem, similar to Theorem 24.1: If the graph of f is concave upward (downward) over an interval I , then any relative minimum (maximum) of f in I is an absolute minimum (maximum) on I . [Hint: Consider the relationship of the graph of f to the tangent line at the relative extremum.) 24.22 Find the absolute extrema (if any) of f (x) = x 2/5 1 x 7/5 on ( 1, 1]. 7