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**Unformatted text preview: **Math 192, Final - Solutions December 6, 2007. 2:00-4:30 1) a) b) f x = 2 x- 2, f y = 2 y so f x = 0 and f y = 0 gives us x = 1 and y = 0 but (1 , 0) is not an interior point so there are no critical points. Let’s now look at the boundary of the region. On x 2 + y 2 = 1 we get f ( x,y ) = 1- 2 x ,- 1 ≤ x ≤ 1, and thus the maximum value of f on that part is f (- 1 , 0) = 3 and the minimum is f (1 , 0) =- 1. On x 2 + 1 4 y 2 = 1 we get f ( x,y ) = x 2 + 4(1- x 2 )- 2 x = 4- 2 x- 3 x 2 . The derivative of 4- 2 x- 3 x 2 is- 2- 6 x so critical value when x = 1 3 and the value of f there is f (- 1 3 , 4 √ 2 3 ) = 1 9 + 32 9 + 2 3 = 39 9 = 4+ 1 3 . We should also check the value of f at the boundary of x 2 + 1 4 y 2 = 1, that is at the points (1 , 0) and (- 1 , 0), but we already checked the value of f there so no need to do it again. From this we see that the absolute minimum of f is f (1 , 0) =- 1 at (1 , 0) and the absolute maximum of f is f (- 1 3 , 4 √ 2 3 ) = 4 + 1 3 at (- 1 3 , 4 √ 2 3 ). 2) Set f ( x,y,z ) = x 2- y 2- z and g ( x,y,z ) = xyz + 30. Then our surfaces S 1 and S 2 are respectively the level surfaces f ( x,y,z ) = 0 and g ( x,y,z ) = 0. Now ∇ f = 2 x i- 2 y j- k , ∇ f (- 3 , 2 , 5) =- 6 i- 4 j- k , ∇ g = yz i + xz j + xy k , ∇ g (- 3 , 2 , 5) = 10 i- 15 j- 6 k so...

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