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Unformatted text preview: Math 2220 Problem Set 4 Solutions Spring 2010 Section 4.3 : 4. Use Lagrange multipliers to find the critical points of the function f ( x, y ) = xy subject to the constraint 2 x 3 y = 6. Solution. The critical points are solutions of the system y = (2) x = ( 3) 2 x 3 y = 6 The first two equations give 2 x = 3 y . Substituting into the third equation, we obtain x = 3 / 2 and y = 1, so (3 / 2 , 1) is the only critical point. 8. Use Lagrange multipliers to find the critical points of the function f ( x, y, z ) = x + y + z subject to the constraints y 2 x 2 = 1 and x + 2 z = 1. Solution. The critical points are solutions of the system 1 = ( 2 x ) + (1) 1 = (2 y ) + (0) 1 = (0) + (2) y 2 x 2 = 1 x + 2 z = 1 The third equation gives = 1 / 2 and then the first two equations simplify to x = 1 / 4 and y = 1 / 2. These equations imply in particular that x and y are nonzero so we can solve them for to get = 1 / 4 x = 1 / 2 y and hence y = 2 x . Plugging this into the fourth equation gives x = radicalbig 1 / 3. Using the fifth equation to obtain the z values, we thus have two critical points ( x, y, z ) = ( radicalbig 1 / 3 , 2 radicalbig 1 / 3 , (1 radicalbig 1 / 3) / 2) and ( radicalbig 1 / 3 , 2 radicalbig 1 / 3 , (1 + radicalbig 1 / 3) / 2) 18. Find the maximum and minimum values of f ( x, y, z ) = x + y z on the sphere x 2 + y 2 + z 2 = 81. Explain how you know there must be both a maximum and a minimum attained. Solution. The critical points are solutions of the system 1 = (2 x ) 1 = (2 y ) 1 = (2 z ) x 2 + y 2 + z 2 = 81 1 Math 2220 Problem Set 4 Solutions Spring 2010 The first three equations give that x = y = z . Substituting into the last equation we get 3 x 2 = 81, i.e., x = 3 3. Thus we have 2 critical points (3 3 , 3 3 , 3 3) and ( 3 3 , 3 3 , 3 3). The values of the function f at these two points are 9 3 and 9 3....
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 Spring '08
 PARKINSON
 Critical Point, Multivariable Calculus

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