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P14-11-8

# P14-11-8 - x = − 2 z we have x = − 2 ± X 1 √ 6 ~ =...

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Department of Mathematics University of Southern California March 4, 2011 Class Problem; Section 11-8#14 Find the minimum and the maximum of the function f ( x, y, z ) = 3 x y 3 z, (1) subject to two constraints g ( x, y, z ) = x + y z = 0 , (2) and h ( x, y, z ) = x 2 + 2 z 2 = 1 . (3) Let us fi rst obtain the gradients, f = < 3 , 1 , 3 > g = < 1 , 1 , 1 > h = < 2 x, 0 , 4 z > At extremum values, f = λ g + μ h, i.e., < 3 , 1 , 3 > = λ < 1 , 1 , 1 > + μ < 2 x, 0 , 4 z > . Equivalently, this represents the three equations, 3 = λ + 2 μ x (4) 1 = λ (5) 3 = λ + 4 μ z (6) From equation (5), we determine λ = 1. Using this information in equations (4) and (6) yields, 3 = 1 + 2 μ x, 3 = 1 + 4 μ z, which simplify to 4 = 2 μ x, (7) 4 = 4 μ z. (8) Now, if we divide equation (7) by (8), provided μ W = 0, we fi nd 1 = x 2 z , i.e., x = 2 z. The use of this result in the second constraint (3) leads to h ( x, y, z ) = x 2 + 2 z 2 = 4 z 2 + 2 z 2 = 6 z 2 = 1 . 1

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This gives us the possible values of z as z = ± 1 6 , and with x =
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Unformatted text preview: x = − 2 z , we have x = − 2 ± X 1 √ 6 ~ = ∓ 2 √ 6 . At this point, we can con f rm the value of μ . From equation (7), μ = 2 x = ∓ √ 6 W = 0 . With the f rst constraint (2), the value of y can be found as y = − x + z = − X ∓ 2 √ 6 ~ ± 1 √ 6 = ± 2 √ 6 ± 1 √ 6 = ± 3 √ 6 . Now with ( x, y, z ) determined as ( x, y, z ) = X ∓ 2 √ 6 , ± 3 √ 6 , ± 1 √ 6 ~ , at the extrema, f = 3 x − y − 3 z = 3 X ∓ 2 √ 6 ~ − X ± 3 √ 6 ~ − 3 X ± 1 √ 6 ~ = ∓ 6 ∓ 3 ∓ 3 √ 6 = ∓ 12 √ 6 = ∓ 2 √ 6 From here, we may infer that the maximum for f ( x, y, z ), under the constraints (2) and (3) is 2 √ 6 and the minimum is − 2 √ 6. 2...
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P14-11-8 - x = − 2 z we have x = − 2 ± X 1 √ 6 ~ =...

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