test2sol - Math 263 Test 2 Solutions 1. Use Lagrange...

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Unformatted text preview: Math 263 Test 2 Solutions 1. Use Lagrange multipliers to find the maximum and minimum values of the function f ( x, y ) = 1 x + 1 y subject to the constraint 1 x 2 + 1 y 2 = 1 . Solution: We define the function g ( x, y ) = 1 x 2 + 1 y 2- 1 and using Lagrange multipliers er get that critical points of f subject to the constraint g = 0 satisfy f ( x, y ) = g ( x, y ) and g ( x, y ) = 0 . Now f ( x, y ) = h- 1 x 2 ,- 1 y 2 i and g ( x, y ) = h- 2 x 3 ,- 2 y 3 i . Hence we have to solve the system of equations- 1 x 2 =- 2 x 3 ,- 1 y 2 =- 2 y 3 and 1 x 2 + 1 y 2 = 1 . From the first two equations we get x = 2 = y. Inserting this into the third equation gives x = 2 . Hence the maximum value of f is f ( 2 , 2) = 2 and the minimum value is f (- 2 ,- 2) =- 2 . 1 2. Consider the integral Z 1 Z 1 x 2 x 3 e 2 y 3 dy dx. (a) Sketch the region of integration....
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test2sol - Math 263 Test 2 Solutions 1. Use Lagrange...

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