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24.1 A farmer must fence in a rectangular eld with one side along a stream; no fence is needed on that side. If the area must be 1800 square meters and the fencing cost $2 per meter, what dimensions will minimize the cost
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Let x and y be the lengths of the sides parallel and perpendicular to the stream, respectively. Then the cost C is C = 2(x + 2y) = 2x + 4y But 1800 = xy, or x = 1800/y, so that C=2 1800 y + 4y = 3600 + 4y = 3600y 1 + 4y y
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3600 dC = 3600y 2 + 4 = 2 + 4 dy y
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CHAP. 24]
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We wish to minimize C(y) for y > 0. So we look for positive critical numbers, 3600 +4=0 y2 3600 4= 2 y 3600 = 900 y2 = 4 y = +30
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d 7200 d2C ( 3600y 2 + 4) = 7200y 3 = = , which is positive at y = +30. Thus, by the second-derivative dy dy2 y3 test, C has a relative minimum at y = 30. Since y = 30 is the only positive critical number, there can be no other relative extremum in the interval (0, + ). Therefore, C has an absolute minimum at y = 30, by Theorem 24.1. When y = 30 meters, 1800 1800 x= = = 60 meters y 30 Now
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24.2 If c1, c2 , . . ., cn are the results of n measurements of an unknown quantity, a method for estimating the value of that quantity is to nd the number x that minimizes the function f (x) = (x c1 )2 + (x c2 )2 + + (x cn )2 This method is called the least-squares principle. Find the value of x determined by the least-squares principle. f (x) = 2(x c1 ) + 2(x c2 ) + + 2(x cn )
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To nd the critical numbers, 2(x c1 ) + 2(x c2 ) + + 2(x cn ) = 0 (x c1 ) + (x c2 ) + + (x cn ) = 0 nx (c1 + c2 + + cn ) = 0 nx = c1 + c2 + + cn c + c2 + + cn x= 1 n As f (x) = 2 + 2 + + 2 = 2n > 0, we have, by the second-derivative test, a relative minimum of f at the unique critical number. By Theorem 24.1, this relative minimum is also an absolute minimum on the set of all real x. Thus, the least-squares principle prescribes the average of the n measurements.
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24.3 Let f (x) =
4x 2 3 for 0 x < 1. Find the absolute extrema, if any, of f on [0, 1), x 1 f (x) = (x 1)Dx (4x 2 3) (4x 2 3)Dx (x 1) (x 1)(8x) (4x 2 3)(1) = (x 1)2 (x 1)2 2 8x 4x 2 + 3 2 8x + 3 8x 4x (2x 3)(2x 1) = = = 2 2 (x 1) (x 1) (x 1)2
(2x 3)(2x 1) =0 (x 1)2 (2x 3)(2x 1) = 0 2x 3 = 0 x= 3 2 or or 2x 1 = 0 x= 1 2
To nd the critical numbers, set f (x) = 0,
So the only critical number in [0,1) is x = 1 . 2
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[CHAP. 24
Let us apply the rst-derivative test (Theorem 23.4), f (x) =
3 1 2(x 3 ) 2(x 1 ) (2x 3)(2x 1) 2 2 = 4(x 2 )(x 2 ) = 2 2 2 (x 1) (x 1) (x 1)
For x immediately to the left of 1 , x 1 < 0 and x 3 < 0, and so, f (x) > 0. For x immediately to the right of 2 2 2 1 , x 1 > 0 and x 3 < 0 and so, f (x) < 0. Thus, we have the case {+, }, and f has a relative maximum at x = 1 . 2 2 2 2 (The second-derivative test could have been used instead.) The function f has no absolute minimum on [0, 1). Its graph has 4x 2 3 = (see Fig. 24-3).1 the line x = 1 as a vertical asymptote, since lim f (x) = lim x 1 x 1 x 1
Fig. 24-3
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