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Unformatted text preview: MATH 16B - SPRING 2009 - FINAL EXAM REVIEW PROBLEM GUIDE Problem 1 Let f ( x,y ) = 3 x 2 + 6 xy + 4 y 2 + 4 y . Find all maxima, minima, and saddle points of f ( x,y ). Solution. At a maximum, minimum, or saddle point, the partial derivatives ∂f ∂x and ∂f ∂y are 0. That means we must have ∂f ∂x = 6 x + 6 y = 0 (1) ∂f ∂y = 6 x + 8 y + 4 = 0 (2) We solve these equations for x and y . From (1), we get y =- x . Substituting for y in (2), we get 2 y + 4 = 0, so y =- 2, whence x =- (- 2) = 2. Therefore, the function has at most one maximum, minimum, or saddle, and any such point is at (2 ,- 2). To determine what type of point, we have found, we use the second derivative test. We compute the three second partial derivatives: ∂ 2 f ∂x 2 ( x,y ) = 6 ∂ 2 f ∂y 2 ( x,y ) = 8 ∂ 2 f ∂x∂y ( x,y ) = 6 . Therefore, D (2 ,- 2) = ∂ 2 f ∂x 2 (2 ,- 2) · ∂ 2 f ∂y 2- ∂ 2 f ∂x∂y (2 ,- 2) 2 = 6 · 8- 6 2 = 12 > . Since D (2 ,- 2) > 0 and ∂ 2 f ∂x 2 (2 ,- 2) = 6 > 0, the point is a local minimum. Important points. • Maximum and minimum values of a function occur where the partial derivatives are 0. • To determine the nature of a point where the partials are 0, use the second derivative test (the statement of the test is on page 392 of the textbook). Problem 2 Let R = { ( x,y ) | ≤ y ≤ ln( x ) , 1 ≤ x ≤ e } . Compute Z Z R y dx dy Solution. We compute the double integral by integrating first with respect to y and then with respect to x . Z Z R y dx dy = Z e 1 Z ln( x ) y dy dx = Z e 1 Z ln( x ) y dy dx = Z e 1 y 2 2 ln( x ) dx = Z e 1 ln( x ) 2 2 dx. At this point, we must integrate by parts. We’ll compute an indefinite integral first, and then use that to compute the definite integral. Let f 1 ( x ) = (ln x ) 2 and g 1 ( x ) = 1 2 . Then f 1 ( x ) = 2 ln x x and g 1 ( x ) has antiderivative G 1 ( x ) = x 2 . Integrating by parts thus gives (3) Z (ln x ) 2 2 dx = x 2 (ln x ) 2- Z 2ln x x · x 2 dx = x 2 (ln x ) 2- Z ln x dx. 1 2 MATH 16B - SPRING 2009 - FINAL EXAM REVIEW PROBLEM GUIDE We integrate by parts again. Let f 2 ( x ) = ln x and g 2 ( x ) = 1. Then f 2 ( x ) = 1 /x and g 2 ( x ) has antiderivative G 2 ( x ) = x . Therefore, Z ln x dx = x ln x- Z 1 x · x dx = x ln x- Z 1 dx = x ln x- x + C. Substituting into (3) now yields Z (ln x ) 2 2 dx = x 2 (ln x ) 2- ( x ln x- x + C ) = x 2 (ln x ) 2- x ln x + x- C. Therefore, Z Z R y dx dy = Z (ln x ) 2 2 dx = h x 2 (ln x ) 2- x ln x + x i e 1 = e 2 · ln e- e ln e + e- 1 2 (ln1) 2- 1 · ln1 + 1 = e 2- 1 Important points. • Compute the double integral as an iterated integral, that is by integrating with respect to x and y separately. • If one or both bounds for x are functions of y , integrate with respect to x first. If one or both bounds for y are functions of x , integrate with respect to y first. If all the bounds are constants, you can choose either order....
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