# hw11 - Granillo Yvette – Homework 11 – Due 3:00 am –...

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Unformatted text preview: Granillo, Yvette – Homework 11 – Due: Nov 11 2005, 3:00 am – Inst: Edward Odell 1 This print-out should have 15 questions. Multiple-choice 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 y-axis 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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hw11 - Granillo Yvette – Homework 11 – Due 3:00 am –...

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