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Unformatted text preview: Granillo, Yvette – Homework 11 – Due: Nov 11 2005, 3:00 am – Inst: Edward Odell 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points If f is a differentiable function such that f ( x ) = ( x 2 9) g ( x ) , where g ( x ) < 0 for all x , at which value(s) of x does f have a local minimum? 1. only at x = 3 correct 2. only at x = 3 3. only at x = 9 4. at both x = 9 , 9 5. only at x = 9 6. at both x = 3 , 3 Explanation: By the First Derivative test, f will have (i) a local minimum at x if f ( x ) = 0 and if the sign of f ( x ) changes from negative to positive as x passes through x ; (ii) a local maximum at x if f ( x ) = 0 and if the sign of f ( x ) changes from positive to negative as x passes through x . Now the only solutions of f ( x ) = ( x 2 9) g ( x ) = ( x 3)( x + 3) g ( x ) = 0 occur at x = 3 , 3. On the other hand, the sign chart∞ + 3 3 ∞ for f ( x ) shows that the sign of f ( x ) changes from negative to positive only at x = 3. Consequently, f has a local minimum only at x = 3 . 002 (part 1 of 1) 10 points Consider all right circular cylinders for which the sum of the height and the circumference is 30 centimeters. What is the radius of the one with maxi mum volume? 1. 10 cm 2. 10 π cm correct 3. 20 cm 4. 30 π 2 cm 5. 3 cm Explanation: h r S = h + 2 πr = 30 h = 30 2 πr V = πr 2 h = πr 2 (30 2 πr ) = 30 πr 2 2 π 2 r 3 dV dr = 60 πr 6 π 2 r 2 dV dr = 0 ⇒ 6 πr (10 πr ) = 0 r = 10 π 10/ π + [ V' Thus a maximum volume occurs when r = 10 π . 003 (part 1 of 1) 10 points Granillo, Yvette – Homework 11 – Due: Nov 11 2005, 3:00 am – Inst: Edward Odell 2 A rectangle is inscribed between the yaxis and the parabola y 2 = 12 x as shown in Determine the maximum possible area, A max , of the rectangle. 1. A max = 35 sq. units 2. A max = 32 sq. units correct 3. A max = 33 sq. units 4. A max = 31 sq. units 5. A max = 34 sq. units Explanation: Let ( x, y ) be the coordinates of the upper right corner of the rectangle. The area of the rectangle is then given by A ( y ) = 2 xy = 24 y 2 y 3 . Differentiating A ( y ) with respect to y we see that A ( y ) = 24 6 y 2 . The critical points of A are thus the solutions of 24 6 y 2 = 0 , i . e ., y = 2 , 2 ; the solution y = 2 can obviously be disre garded for practical reasons. Substituting for y = 2 in A ( x ) we get A max = 32 sq. units . keywords: Stewart5e, 004 (part 1 of 1) 10 points What is the maximum value of the product P = xy when y = 1 4 x ? 1. maximum value = 5 16 2. maximum value = 3 16 3. maximum value = 1 16 correct 4. maximum value = 3 16 5. maximum value = 5 16 6. maximum value = 1 16 Explanation: When y = 1 4 x , P = xy = x (1 4 x ) = x 4 x 2 ....
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 Spring '08
 schultz
 Critical Point, Optimization, amax, maximum value, Edward Odell

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