Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.4 Lagrange Multipliers - 9.4 Exercises - Page 498: 14



Work Step by Step

Let our three positive numbers be x, y and z. We know that $x + y + z =240 \rightarrow g(x)=x+y+z-240$ Our objective function is $f(x,y,z)=xyz$ $F(x,y,z,λ)=f(x,y,z)-\lambda.g(x,y,z)=xyz-\lambda (x+y+z-240)=xyz -\lambda x -\lambda y -\lambda z +240\lambda$ $F_{x}(x,y,z,\lambda)=yz-\lambda$ $F_{y}(x,y,z,\lambda)=xz-\lambda$ $F_{z}(x,y,z,\lambda)=xy-\lambda$ $F_{\lambda}(x,y,z,\lambda)=-x-y-z+240$ $(1) yz-\lambda=0 \rightarrow \lambda =yz$ $(2) xz-\lambda =0 \rightarrow \lambda =xz$ $(3) xy-\lambda =0 \rightarrow \lambda =xy$ $(4) -x-y-z+240=0$ Set the expressions equal: $yz=xz$ $x=y$ Set the expressions equal: $xz=xy$ $z=y$ Substitute for x=y=z in Equation (4) $-y-y-y+240=0$ $y=80 \rightarrow x=80 \rightarrow z=80$ The value of f is $f(80,80,80)=512000$ Let x=79.9 and y=79.9, z=80.2 then $f(79.9,79.9,80.2)=\approx 511997.602$. Because a nearby point has a value smaller than 512000, the value 512000 is probably not a minimum. Therefore $f_{max}(80,80,80)=512000$
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