P14-11-8

# P14-11-8 - x = 2 z , we have x = 2 X 1 6 ~ = 2 6 . At this...

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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 f 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 f 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 .

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This gives us the possible values of z as z = ± 1 6 , and with
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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 . From equation (7), = 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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## This note was uploaded on 10/15/2011 for the course MATH 39578 taught by Professor Penner during the Spring '09 term at USC.

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P14-11-8 - x = 2 z , we have x = 2 X 1 6 ~ = 2 6 . At this...

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