# 6 10 pts locate all relative extrema and saddle

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______________________________________________________________________ 6. (10 pts.) Locate all relative extrema and saddle points of the following function. f ( x , y ) x 2 8 y 2 x 2 y Use the second partials test in making your classification. (Fill in the table below after you locate all the critical points.) Crit.Pt. f xx @ c.p. f yy @ c.p. f xy @ c.p. D @ c.p. Conclusion (x,y) 2 - 2 y 16 -2 x 32-32 y -4 x 2 (0,0) 2 16 0 32 Rel. Min. (4,1) 0 16 -8 -64 Saddle Pt. (-4,1) 0 16 8 -64 Saddle Pt. // Since f x ( x , y ) = 2 x (1 - y ) and f y ( x , y ) = 16 y - x 2 , the critical points of f are given by the solutions to the following system: 2 x (1 - y ) = 0 and 16 y - x 2 = 0. The system is plainly equivalent to x = 0 and 16 y = x 2 , or y = 1 and 16 y = x 2 . Solving these systems yields three critical points: (0,0), (4,1), and (-4,1). Now go up and fill in the silly table.

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TEST3/MAC2313 Page 4 of 5 ______________________________________________________________________ 7. (5 pts.) Write down the triple iterated integral in spherical coordinates that would be used to compute the volume of the solid G bounded above by the sphere defined by ρ = 4 and below by the cone defined by φ = π /3. Do not attempt to evaluate the iterated integrals. Ouch! It feels like an easy spherical wedgie. ______________________________________________________________________ 8. (15 pts.) Let f ( x , y ) = x 2 y 2 on the closed unit disk defined by x 2 + y 2 1. Find the absolute extrema and where they occur. Use Lagrange multipliers to analyze the function on the boundary. Do not neglect the interior of the disk in doing your analysis!! // First, we deal with interior points of the disk. Since f ( x , y ) = < 2 xy 2 , 2 yx 2 >, there are infinitely many critical points in the interior of the disk where either component of the pair ( x , y ) is zero. Obviously, f (x,y) = 0 if either x = 0 or y = 0. Now to study f on the boundary using Lagrange multipliers, set g ( x , y ) = x 2 + y 2 - 1. Then ( x , y ) is on the circle defined by x 2 + y 2 = 1 precisely when ( x , y ) satisfies g ( x , y ) = 0. Since g ( x , y ) = <2 x , 2 y >, g ( x , y ) <0, 0> when ( x , y ) is on the circle given by g ( x , y ) = 0. Plainly f and g are smooth enough to satisfy the hypotheses of the Lagrange Multiplier Theorem.

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