x + 1 Dx x 2 + 3 x 2 + 3 Dx x + 1 x+1 2x 2 + 2x x 2 3 x+1 in .NET framework

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x + 1 Dx x 2 + 3 x 2 + 3 Dx x + 1 x+1 2x 2 + 2x x 2 3 x+1
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x + 1 2x x 2 + 3 1 x+1
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x 2 + 2x 3 x+1
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f (x) is not de ned when (x + 1)2 = 0; that is, when x = 1. But since 1 is not in (0, 3), the only critical numbers that need to be considered are the zeros of x 2 + 2x 3 in (0, 3): x 2 + 2x 3 = 0 (x + 3)(x 1) = 0 x+3=0 x = 3 Thus, 1 is the only critical number in (0, 3). Now construct Table 14-4: 1+3 4 (1)2 + 3 = = =2 1+1 2 2 2+3 3 (0) = =3 f (0) = 0+1 1 9+3 12 (3)2 + 3 f (3) = = = =3 3+1 4 4 f (1) = Thus, the absolute maximum, achieved at 0 and 3, is 3; and the absolute minimum, achieved at 1, is 2. Table 14-4 x 1 0 3 f (x) 2 min 3 max 3 max or x 1 = 0 or x=1
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14.3 Among all pairs of positive real numbers u and v whose sum is 10, which gives the greatest product uv
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Let P = uv. Since u + v = 10, v = 10 u, and so P = u(10 u) = 10u u2 Here, 0 < u < 10. But since P would take the value 0 at u = 0 and u = 10, and 0 is clearly not the absolute maximum of P, we can extend the domain of P to the closed interval [0, 10]. Thus, we must nd the absolute maximum of P = 10u u2 on the closed interval [0, 10]. The derivative dP/du = 10 2u vanishes only at u = 5, and this critical point must yield the maximum. Thus, the absolute maximum is P(5) = 5(10 5) = 5(5) = 25, which is attained for u = 5. When u = 5, v = 10 u = 10 5 = 5. 102 (u v)2 (u + v)2 (u v)2 = , which is 4 4 largest when u v = 0, that is, when u = v. Then 10 = u + v = 2u and, therefore, u = 5. algebra Calculus was not really needed in this problem, for P =
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14.4 An open box is to be made from a rectangular piece of cardboard that is 8 feet by 3 feet by cutting out four equal squares from the corners and then folding up the aps (see Fig. 14-5). What length of the side of a square will yield the box with the largest volume
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Fig. 14-5
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Let x be the side of the square that is removed from each corner. The volume V = lwh, where l, w, and h are the length, width, and height of the box. Now l = 8 2x, w = 3 2x, and h = x, giving V (x) = (8 2x)(3 2x)x = (4x 2 22x + 24)x = 4x 3 22x 2 + 24x The width w must be positive. Hence, 3 2x > 0 or 3 > 2x or 3 >x 2
Furthermore, x > 0. But we also can admit the values x = 0 and x = 3 , which make V = 0 and which, therefore, cannot 2 yield the maximum volume. Thus, we have to maximize V (x) on the interval 0, 3 . Since 2 dV = 12x 2 44x + 24 dx the critical numbers are the solutions of 12x 2 44x + 24 = 0 3x 2 11x + 6 = 0 (3x 2)(x 3) = 0 3x 2 = 0 or x 3 = 0 x=3 x=3 3x = 2 or 2 or x= 3
CHAP. 14]
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