Final_Fall_2007

# Final_Fall_2007 - Math 192 Final Solutions December 6 2007...

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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 f ( - 3 , 2 , 5) × ∇ g ( - 3 , 2 , 5) = 9 i - 46 j + 130 k . Thus the tangent line to the curve of intersection of the two surfaces at ( - 3 , 2 ,

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