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: 12


With $x=32, y=16, f(x,y) =5x^{2}y+10$ is maximized at 81930

Work Step by Step

We are given $f(x,y)=5x^{2}y+10$ with $x+y=48$ so $y=48-x$ $f(x,y)=5x^{2}(48-x)+10=240x^{2}-5x^{3}+10$ Now differentiate f(x) with respect to x and equate to '0' as shown below $f_{x}(x,y)=480x-15x^{2} = 0$ $x=32 \rightarrow y=16 \rightarrow $ The value of $5x^{2}y+10$ is 81930 or $x=0 \rightarrow y=48 \rightarrow $ The value of $5x^{2}y+10$ is 10 Since $81930\gt10$, with $x=32, y=16, f(x,y) =5x^{2}y+10$ is maximized at 81930
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