M408K Hm. 11 Solutions

# M408K Hm. 11 Solutions - Mahon Kevin Homework 11 Due 3:00...

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Mahon, Kevin – 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 – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points IF f is a di±erentiable Function such that f 0 ( x ) = ( x 2 - 4) 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 = 2 2. at both x = - 2 , 2 3. only at x = - 4 4. only at x = - 2 correct 5. only at x = 4 6. at both x = - 4 , 4 Explanation: By the ²irst Derivative test, f will have (i) a local minimum at x 0 iF f 0 ( x 0 ) = 0 and iF the sign oF f 0 ( x ) changes From negative to positive as x passes through x 0 ; (ii) a local maximum at x 0 iF f 0 ( x 0 ) = 0 and iF the sign oF f 0 ( x ) changes From positive to negative as x passes through x 0 . Now the only solutions oF f 0 ( x ) = ( x 2 - 4) g ( x ) = ( x - 2)( x + 2) g ( x ) = 0 occur at x = - 2 , 2. On the other hand, the sign chart -∞ - + - - 2 2 For f 0 ( x ) shows that the sign oF f 0 ( x ) changes From negative to positive only at x = - 2. Consequently, f has a local minimum only at x = - 2 . 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. 3 cm 2. 20 cm 3. 10 cm 4. 10 π cm correct 5. 30 π 2 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 .

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Mahon, Kevin – Homework 11 – Due: Nov 11 2005, 3:00 am – Inst: Edward Odell 2 003 (part 1 of 1) 10 points A rectangle is inscribed between the y -axis and the parabola y 2 = 3 - x as shown in Determine the maximum possible area, A max , of the rectangle. 1. A max = 5 sq. units 2. A max = 7 sq. units 3. A max = 6 sq. units 4. A max = 3 sq. units 5. A max = 4 sq. units correct 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 = 6 y - 2 y 3 . DiFerentiating A ( y ) with respect to y we see that A 0 ( y ) = 6 - 6 y 2 . The critical points of A are thus the solutions of 6 - 6 y 2 = 0 , i . e ., y = - 1 , 1 ; the solution y = - 1 can obviously be disre- garded for practical reasons. Substituting for y = 1 in A ( x ) we get A max = 4 sq. units . keywords: Stewart5e, 004 (part 1 of 1) 10 points What is the minimum value of the product P = xy when y = 2 + 5 x ? 1. minimum value = - 3 20 2. minimum value = 1 5 3. minimum value = - 1 4 4. minimum value = 1 4 5. minimum value = - 1 5 correct 6. minimum value = 3 20 Explanation: When y = 2 + 5 x , P = xy = x (2 + 5 x ) = 2 x + 5 x 2 . Now the minimum value of P will occur at a critical point of P , i.e. , at the solution, x 0 , of P 0 ( x ) = 2 + 10 x = 0 , in other words when x 0 = - 1 5 . Since P 00 ( x 0 ) = 10 > 0 ,
Mahon, Kevin – Homework 11 – Due: Nov 11 2005, 3:00 am – Inst: Edward Odell 3 we thus see that minimum value = - 1 5 . keywords: Stewart5e, 005 (part 1 of 1) 10 points A 24 00 × 24 00 square sheet of metal is made into an open box by cutting out a square at each corner and then folding up the four sides.

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M408K Hm. 11 Solutions - Mahon Kevin Homework 11 Due 3:00...

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