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Practice2

# Practice2 - ft 2 and the two ends \$3 ft 2 What is the shape...

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CME 100 Fall 2010 Vector Calculus for Engineers Page 1 of 2 Prof. Eric Darve Practice problems 2 For each question, choose the right answer. 1. f ( x , y ) = x exp( y ). Find the rate of of change at (2,0) in the direction from (2,0) to (1 / 2,2). (a) 2 (b) 3 / 2 (c) 1 (d) 5 / 2 2. f ( x , y ) = x exp( y ). Find the direction of maximum rate of change at (2,0). (a) (1 / 5 , - 2 / 5) (b) (2 / 5 , 1 / 5) (c) (2 / 5 , - 1 / 5) (d) (1 / 5 , 2 / 5) 3. f ( x , y ) = x exp( y ). What is the maximum rate of change at (2,0)? (a) 3 (b) 2 (c) 5 (d) 2 4. Use the chain rule to ﬁnd dz / dt : z = q 1 + x 2 + y 2 , x = ln t , y = cos t at t = π/ 2 (a) L = - (1 ) ln( π/ 2) / p 1 + ln( π/ 2) 2 (b) L = (2 ) ln( π/ 2) / p 1 + ln( π/ 2) 2 (c) L = - (2 ) ln( π/ 2) / p 1 + ln( π/ 2) 2 (d) L = (1 ) ln( π/ 2) / p 1 + ln( π/ 2) 2 5. Consider a box with volume 48 ft 3 . The front and back cost \$1 / ft 2 , the top and bottom \$2

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Unformatted text preview: / ft 2 , and the two ends \$3 / ft 2 . What is the shape of the box that minimizes cost? Position the box so that z = corresponds to the bottom and y = 0 corresponds to the front. (a) x = 6, y = 2, z = 4 (b) x = 2, y = 6, z = 4 (c) x = 4, y = 2, z = 6 (d) x = 2, y = 4, z = 6 2/2 CME 100, Fall 2010 6. Consider f ( x , y , z ) = x 2 + y 2 + z 2 with the constraint x 4 + y 4 + z 4 = 1. Find the maximum and minimum of f using the method of Lagrange multipliers. (a) Maximum 2, minimum 1 / 2 (b) Maximum √ 3, minimum 1 (c) Maximum √ 3, minimum 1 / 2 (d) Maximum 2, minimum 1...
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Practice2 - ft 2 and the two ends \$3 ft 2 What is the shape...

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