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Unformatted text preview: g ( x,y,z ) = x 2 + 2 y 2 + 3 z 26. The critical points are at the solutions of the system 5 f = 5 g, g ( x,y,z ) = 0 yz = 2 x, xz = 4 y, xy = 6 z, x 2 + 2 y 2 + 3 z 2 = 6 From the rst three equations we get xyz = 2 x 2 , xyz = 4 y 2 , xyz = 6 z 2 xyz = 2 x 2 = 4 y 2 = 6 z 2 x 2 = 2 y 2 = 3 z 2 So from the fourth equation 3 x 2 = 6, or x = 2. Next, y 2 = x 2 2 = 1 = y = 1. And z = q 2 3 . This gives us 8 critical points. The ones with an odd number of negative components yield 1 the minimum, the rest yield the maximum. Maximum: f ( 2 , 1 , s 2 3 ) = f ( 2 ,1 , s 2 3 ) = f ( 2 , 1 ,s 2 3 ) = f ( 2 ,1 ,s 2 3 ) = 2 3 Minimum: f ( 2 ,1 ,s 2 3 ) = f ( 2 , 1 , s 2 3 ) = f ( 2 , 1 ,s 2 3 ) = f ( 2 , 1 ,s 2 3 ) =2 3 2...
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This note was uploaded on 10/15/2011 for the course MATH 39578 taught by Professor Penner during the Spring '09 term at USC.
 Spring '09
 Penner

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