Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 4 - Section 4.5 - Linear Programming - Exercise Set - Page 303: 4


The answer is At $(0,0)=0$ At $(0,9)=405$ At $(4,4)=300$ At $(3,0)=90$ Maximum value $405$. Minimum value $0$.

Work Step by Step

The given objective function is $z=30x+45y$. The value of the function at each corner points of the graph is shown below. At $(0,0)$ is $z=30(0)+45(0)$ $z=0+0$ $z=0$. At $(0,9)$ is $z=30(0)+45(9)$ $z=0+405$ $z=405$. At $(4,4)$ is $z=30(4)+45(4)$ $z=120+180$ $z=300$. At $(3,0)$ is $z=30(3)+45(0)$ $z=90+0$ $z=90$. The maximum value of the objective function is $405$. The minimum value of the objective function is $0$.
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.