Engineering Calculus Notes 347

Engineering Calculus Notes 347 - 335 3.6. EXTREMA while the...

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3.6. EXTREMA 335 while the third says that either λ = 1 or z = 0 . Since only one of the three named λ -values can hold, two of the coordinates must be zero, which means in terms of the constraint that the third is ± 1. Thus we have six relative critical points, with respective f -values f ( ± 1 , 0 , 0) = 2 f (0 , ± 1 , 0) = 1 f (0 , 0 , ± 1) = 1 . Combining this with the critical value 0 at the origin, we have min x 2 + y 2 + z 2 1 (2 x 2 + y 2 z 2 ) = f (0 , 0 , ± 1) = 1 max x 2 + y 2 + z 2 1 (2 x 2 + y 2 z 2 ) = f ( ± 1 , 0 , 0) = 2 . Multiple Constraints The method of Lagrange Multipliers can be extended to problems in which
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