math10c fall 2009 practice final

# math10c fall 2009 practice final - SOLUTIONS Problem 1 Find...

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SOLUTIONS Problem 1. Find the critical points of the function f ( x, y ) = 2 x 3 - 3 x 2 y - 12 x 2 - 3 y 2 and determine their type i.e. local min/local max/saddle point. Are there any global min/max? Solution: Partial derivatives f x = 6 x 2 - 6 xy - 24 x, f y = - 3 x 2 - 6 y. To find the critical points, we solve f x = 0 = x 2 - xy - 4 x = 0 = x ( x - y - 4) = 0 = x = 0 or x - y - 4 = 0 f y = 0 = x 2 + 2 y = 0 . When x = 0 we find y = 0 from the second equation. In the second case, we solve the system below by substitution x - y - 4 = 0 , x 2 + 2 y = 0 = x 2 + 2 x - 8 = 0 = x = 2 or x = - 4 = y = - 2 or y = - 8 . The three critical points are (0 , 0) , (2 , - 2) , ( - 4 , - 8) . To find the nature of the critical points, we apply the second derivative test. We have A = f xx = 12 x - 6 y - 24 , B = f xy = - 6 x, C = f yy = - 6 . At the point (0 , 0) we have f xx = - 24 , f xy = 0 , f yy = - 6 = AC - B 2 = ( - 24)( - 6) - 0 > 0 = (0 , 0) is local max . Similarly, we find (2 , - 2) is a saddle point since AC - B 2 = (12)( - 6) - ( - 12) 2 = < 0 and ( - 4 , - 8) is saddle since AC - B 2 = ( - 24)( - 6) - (24) 2 < 0 . The function has no global min since lim y →∞ ,x =0 f ( x, y ) = -∞ and similarly there is no global maximum since lim x →∞ ,y =0 f ( x, y ) = . 1

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Problem 2. Determine the global max and min of the function f ( x, y ) = x 2 - 2 x + 2 y 2 - 2 y + 2 xy over the compact region - 1 x 1 , 0 y 2 . Solution: We look for the critical points in the interior: f = (2 x - 2 + 2 y, 4 y - 2 + 2 x ) = (0 , 0) = 2 x - 2 + 2 y = 4 y - 2 + 2 x = 0 = y = 0 , x = 1 . However, the point (1 , 0) is not in the interior so we discard it for now. We check the boundary. There are four lines to be considered: the line x = - 1 : f ( - 1 , y ) = 3 + 2 y 2 - 4 y. The critical points of this function of y are found by setting the derivative to zero: ∂y (3 + 2 y 2 - 4 y ) = 0 = 4 y - 4 = 0 = y = 1 with f ( - 1 , 1) = 1 . the line x = 1 : f (1 , y ) = 2 y 2 - 1 . Computing the derivative and setting it to 0 we find the critical point y = 0 . The corre- sponding point (1 , 0) is one of the corners, and we will consider it separately below. the line y = 0 : f ( x, 0) = x 2 - 2 x. Computing the derivative and setting it to 0 we find 2 x - 2 = 0 = x = 1 . This gives the corner (1 , 0) as before. the line y = 2 : f ( x, 2) = x 2 + 2 x + 4 with critical point x = - 1 which is again a corner. Finally, we check the four corners ( - 1 , 0) , (1 , 0) , ( - 1 , 2) , (1 , 2) . The values of the function f are f ( - 1 , 0) = 3 , f (1 , 0) = - 1 , f ( - 1 , 2) = 3 , f (1 , 2) = 7 . From the boxed values we select the lowest and the highest to find the global min and global max. We conclude that global minimum occurs at (1 , 0) global maximum occurs at (1 , 2) . Problem 3. Using Lagrange multipliers, optimize the function f ( x, y ) = x 2 + ( y + 1) 2
subject to the constraint 2 x 2 + ( y - 1) 2 18 . Solution: We check for the critical points in the interior f x = 2 x, f y = 2( y + 1) = (0 , - 1) is a critical point . The second derivative test f xx = 2 , f yy = 2 , f xy = 0 shows this a local minimum with f (0 , - 1) = 0 . We check the boundary g ( x, y ) = 2 x 2 + ( y - 1) 2 = 18 via Lagrange multipliers. We compute f = (2 x, 2( y + 1)) = λ g = λ (4 x, 2( y - 1)) .

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