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Unformatted text preview: blue MATH 32A Final Exam March 17, 2008 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) 5 (20 pts) 2 (20 pts) 6 (20 pts) 3 (20 pts) 7 (20 pts) 4 (20 pts) TOTAL 2 PROBLEM 1 (20 Points) Let function f ( x, y ) = x 4 x 2 y 2 . (A) Find all critical points of f ( x, y ) and classify each one as a maximum, minimum, or saddle point. (B) Find the absolute maximum and minimum of f ( x, y ) on the domain x 2 + y 2 ≤ 4. Answer : 3 Solution: (A) First solve for the critical points: f x = 4 x 3 2 x = 2 x (2 x 2 1) ⇒ x = 0 , ± 1 √ 2 f y = 2 y = 0 ⇒ y = 0 There are three critical points: (0 , 0) ( 1 √ 2 , 0) , ( 1 √ 2 , 0) The discriminant is D ( x, y ) = f xx f yy f 2 xy = (12 x 2 2)( 2) 0 = 4 24 x 2 We see that D (0 , 0) = 4 > 0 so (0 , 0) is a local min or max. In fact, it is a local max since f x x (0 , 0) = 2 < 0. We have D ( ± 1 √ 2 , 0) = 4 24( 1 2 ) = 8 < Therefore the other two critical points are saddle points. (B) We evaluate f ( x, y ) at the critical points: f (0 , 0) = 0 , f ( ± 1 √ 2 , 0) = 1 4 1 2 0 = 1 4 Next, we test f ( x, y ) on the circle x 2 + y 2 = 4. We find that on the circle f ( x, y ) = x 4 x 2 y 2 = x 4 4 (for 2 ≤ x ≤ 2) Since x 4 is positive, we may conclude immediately that the maximum of f on the circle is f (2 , 0) = f ( 2 , 0) = 2 4 4 = 12 and the minimum value is f (0 , 0) = 4. Conclusion: the max of f ( x, y ) on the disk x 2 + y 2 ≤ 4 is 12 and the min is 4 (since the values at the critical points lie between 4 and 12). 4 PROBLEM 2 (20 Points) Suppose that f ( x, y ) is a function of x and y , and that x = u 2 v 2 , y = uv . Use the table of values ∂f ∂x and ∂f ∂y to compute ∂f ∂u ( u,v )=(1 , 1) and ∂f ∂v ( u,v )=(1 , 0) x y ∂f ∂x ( x, y ) ∂f ∂y ( x, y ) 23 1 5 1 114 1 1 3 2 Answer : ∂f ∂u ( u,v )=(1 , 1) Answer : ∂f ∂v ( u,v )=(1 , 0) 5 Solution: (A) When ( u, v ) = (1 , 1), we have x = u 2 v 2 = 0 , y = uv = 1 The table gives the values ∂f ∂x ( x,y )=(0 , 1) = 1 , ∂f ∂y ( x,y )=(0 , 1) = 4 By the Chain Rule: ∂f ∂u = ∂f ∂x ∂x ∂u + ∂f ∂y ∂y ∂u = ∂f ∂x (2 u ) + ∂f ∂y ( v ) Evaluate at ( u, v ) = (1 , 1) and ( x, y ) = (0 , 1): ∂f ∂u ( u,v )=(1 , 1) = ( 1)(2) + ( 4)(1) = 6 (B) When ( u, v ) = (1 , 0), we have x = u 2 v 2 = 1 , y = uv = 0 The table gives the values ∂f ∂x ( x,y )=(1 , 0) = 5 , ∂f ∂y ( x,y )=(1 , 0) = 1 By the Chain Rule: ∂f ∂v = ∂f ∂x ∂x ∂v + ∂f ∂y ∂y ∂v = ∂f ∂x ( 2 v ) + ∂f ∂y ( u ) Evaluate at ( u, v ) = (1...
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This test prep was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.
 Winter '08
 GANGliu
 Math

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