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Unformatted text preview: Solutions Padmini April 10, 2008 Note: These solutions have been made in a span of two hours. There may be egregious errors, but well.. I hope it helps. A. = / 3 is a cone and cos = z = 2 is a plane and since takes all possible values its the volume bounded by the cone and plane z = 2. B. f x ( x, y ) = 1 /x 2 + y = 0 and f y ( x, y ) = 1 /y 2 + x and hence, f xx = 2 /x 3 , f yy = 2 /y 3 , f xy = 1 and at (1 , 1) these evaluate to f xx = 2 , f yy = 2 , f xy = 1 and hence D = f xx f yy f 2 xy = 8 > 0 and hence its a saddle point. C. (a) On the circle y 2 = 4 x 2 and hence the function is f = x 2 + 2 x + 2(4 x 2 ) = 8 + 2 x x 2 and hence, f = 2 2 x = 0 or x = 1 , y = 3 and f = 9. Also because on the circle 2 x 2 we put these boundary conditions to see that f = 0 , 8 at these two points. Note: you may also use Lagrange multipliers for this. hence, abs min f = 0 and max f = 9 (b) Here we have to find critical points inside the circle also....
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This note was uploaded on 08/10/2008 for the course MATH 251 taught by Professor Beck during the Spring '08 term at Rutgers.
 Spring '08
 Beck
 Cone

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