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exam2w05sol

exam2w05sol - MATH 215 Winter 2005 SECOND EXAM Problem 1(5...

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MATH 215 – Winter 2005 SECOND EXAM Problem 1. (5+5+10=20 points) Throughout this problem f is the function f ( x, y ) = xye - x - 2 y . (a) Find all critical points of f . (b) A calculation shows that f xx = e - x - 2 y ( - 2 y + xy ) , f xy = e - x - 2 y ( 1 - x - 2 y + 2 xy ) , f yy = e - x - 2 y ( - 4 x + 4 xy ) . Using these formulas, classify the critical points of f . You do not have to compute the second partials yourself. Solution . The partial derivatives are: f x = e - x - 2 y ( y - xy ) , f y = e - x - 2 y ( x - 2 y ) . Since the exponential does not vanish anywhere, the critical points are the solutions to the system y - xy = 0 , x - 2 y = 0 . If y 6 = 0 then the first equation implies that x = 1 . If y = 0 then x must be zero, by the 2nd equation. Therefore, the critical points are: (0 , 0) and (1 , 1 / 2) . The Hessian is e - x - 2 y - 2 y + xy 1 - 2 y - x + 2 xy 1 - 2 y - x + 2 xy - 4 x + 4 xy At (0 , 0) , the Hessian is e 1 / 2 0 1 1 0 and therefore this point is a saddle point. At (1 , 1 / 2) , the Hessian is e - 2 - 1 / 2 0 0 - 2 and therefore this point is a local maximum. / (c) Find the absolute maximum of the function f on the triangle with vertices at (0 , 0), (1 , 0) and (0 , 1 / 2). Hint: There are not any critical points in the interior of the triangle. Solution . There are no critical points in the interior of the region. We must check the boundary, which consists of 3 sides. On each of the two sides that lie on the coordinate axes, the function is identically

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exam2w05sol - MATH 215 Winter 2005 SECOND EXAM Problem 1(5...

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