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: 11


With $x=8, y=16, f(x,y) =3xy^{2}$ is maximized at 6144

Work Step by Step

We are given $f(x,y)=3xy^{2}$ with $x+y=24$ so $y=24-x$ $f(x,y)=3x(24-x)^{2}=3x(576-48x+x^{2})=1728x-144x^{2}+3x^{3}$ Now differentiate f(x) with respect to x and equate it to '0' as shown below $f_{x}(x,y)=1728-288x+9x^{2} = 0$ $x=24 \rightarrow y=0 \rightarrow $ The value of $3xy^{2}$ is 0 or $x=8 \rightarrow y=16 \rightarrow $ The value of $3xy^{2}$ is 6144 Since $6144\gt0$, with $x=8, y=16, f(x,y) =3xy^{2}$ is maximized at 6144
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.