Solutions_to_Sample_Final

# Solutions_to_Sample_Final - blue MATH 32A FINAL EXAM LAST...

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blue MATH 32A FINAL EXAM December 13, 2007 LAST NAME FIRST NAME ID NO. Your TA: To receive credit, you must write your answer in the space provided . DO NOT WRITE BELOW THIS LINE 1 (20 pts) 4 (20 pts) 2 (20 pts) 5 (20 pts) 3 (20 pts) 6 (20 pts) TOTAL FOR WRITTEN PROBLEMS 1

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2 PROBLEM 1 (20 Points) The temperature at position ( x, y, z ) in a room is f ( x, y, z ) = 18 + 2 z + xy o C A woman walks down a spiral staircase in the middle of the room. She holds a thermometer whose position at time t (seconds) is c ( t ) = ( cos t 3 , sin t 3 , 8 t 3 ) ( t in seconds). How fast is temperature reading on the thermometer changing at t = 6 π s. Solution: We have f = ( y, x, 2 ) c ( t ) = (− 1 3 sin t 3 , 1 3 cos t 3 , 1 3 ) At t = 6 π , c (6 π ) = ( x, y, z ) = ( 1 , 0 , 8 2 π ) f = ( 0 , 1 , 2 ) c (6 π ) = ( 0 , 1 3 , 1 3 ) By the Chain Rule for paths, the rate of change of temper- ature is d dt f ( c ( t )) = f · c ( t ) The temperature reading at t = 6 π is changing at the rate f · c (6 π ) = ( 0 , 1 , 2 ) · ( 0 , 1 3 , 1 3 ) = 1 3 2 3 = 1 3 o C/s
3 PROBLEM 2 (20 Points) Find the maximum value of f ( x, y, z ) = xyz , subject to the constraint g ( x, y, z ) = 4 x 2 + y 2 + z 2 = 48 Solution: Use the method of Lagrange multipliers: f = ( yz, xz, xy ) = λ g = λ ( 8 x, 2 y, 2 z ) Case 1: x, y, z are all non-zero. Then yz 8 x = xz 2 y = xy 2 z This yields y 2 = 4 x 2 , z 2 = 4 x 2 Plug into the constraint: 4 x 2 + y 2 + z 2 = 4 x 2 + 4 x 2 + 4 x 2 = 48 x 2 = 4 We obtain x = ± 2, y = ± 2 x = ± 4, z = ± 2 x = ± 4, and f (2 , 4 , 4) = 2(4)(4) = 32 More generally, f ( ± 2 , ± 4 , ± 4) = ± 32. Case 2: At least one of x, y, z is zero. In this case f ( x, y, z ) = xyz = 0. We conclude that the maximum value of f = xyz subject to the constraint is 32.

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4 PROBLEM 3 (20 Points) Let f ( x, y ) = 2 xy 1 6 x 3 y 2 (A) Find the critical points of f ( x, y ). (B) Determine the nature of the critical points (min, max or saddle).
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