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Until now we have been able to nd the absolute maxima and minima of differentiable functions only on closed intervals (see Section 14.2). The following result often enables us to handle cases where the function is de ned on a half-open interval, open interval, in nite interval, or the set of all real numbers. Remember that, in general, there is no guarantee that a function has an absolute maximum or an absolute minimum on such domains. Theorem 24.1: Let f be a continuous function on an interval I , with a single relative extremum within I . Then this relative extremum is also an absolute extremum on I . Intuitive Argument: Refer to Fig. 24-1. Suppose that f has a relative maximum at c and no other relative extremum inside I . Take any other number d in I . The curve moves downward on both sides of c. Hence, if the value f (d) were greater than f (c), then, at some point u between c and d, the curve would have to change direction and start moving upward. But then f would have a relative minimum at u, contradicting our assumption. The result for a relative minimum follows by applying to f the result just obtained for a relative maximum.
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For a rigorous proof, see Problem 24.20.
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Fig. 24-1
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(a) Find the shortest distance from the point P(1, 0) to the parabola x = y2 [see Fig. 24-2(a)].
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Fig. 24-2
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The distance from an arbitrary point Q(x, y) on the parabola to the point P(1,0) is, by (2.1), u= = = (x 1)2 + y2 (x 1)2 + x [y2 = x at Q] x2 x + 1
x 2 2x + 1 + x =
But minimizing u is equivalent to minimizing u2 F(x) = x 2 x + 1 on the interval [0, + ) (the value of x is restricted by the fact that x = y2 0). F (x) = 2x 1 The only critical number is the solution of F (x) = 2x 1 = 0 or x= 1 2 F (x) = 2
Now F ( 1 ) = 2 > 0. So by the second-derivative test, the function F has a relative minimum at x = 1 , Theorem 24.1 implies 2 2 that this is an absolute minimum. When x = 1 , 2 1 y2 = x = 2 and 1 2 2 1 y = = = 2 2 2 2 2/2) and ( 1 , 2/2). 2
Thus, the points on the parabola closest to (1,0) are ( 1 , 2
(b) An open box (that is, a box without a top) is to be constructed with a square base [see Fig. 24-2(b)] and is required to have a volume of 48 cubic inches. The bottom of the box costs 3 cents per square inch, whereas the sides cost 2 cents per square inch. Find the dimensions that will minimize the cost of the box. Let x be the side of the square bottom, and let h be the height. Then the cost of the bottom is 3x 2 and the cost of each of the four sides is 2xh, giving a total cost of C = 3x 2 + 4(2xh) = 3x 2 + 8xh The volume is V = 48 = x 2 h. Hence, h = 48/x 2 and C = 3x 2 + 8x 48 x2 = 3x 2 + 384 = 3x 2 + 384x 1 x
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[CHAP. 24
which is to be minimized on (0, + ). Now 384 dC = 6x 384x 2 = 6x 2 dx x and so the critical numbers are the solutions of 6x 384 =0 x2 384 6x = 2 x x 3 = 64 x=4 Now 768 d2C = 6 ( 2)384x 3 = 6 + 3 > 0 dx 2 x
for all positive x; in particular, for x = 4. By the second-derivative test, C has a relative minimum at x = 4. But since 4 is the only positive critical number and C is continuous on (0, + ), Theorem 24.1 tells us that C has an absolute minimum at x = 4. When x = 4, 48 48 h= 2 = =3 16 x So, the side of the base should be 4 inches and the height 3 inches.
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