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: 3

Answer

The answer is At $(0,0)=0$ At $(0,8)=400$ At $(4,9)=610$ At $(8,0)=320$ Maximum value $=610$. Minimum value $=0$.

Work Step by Step

The given objective function is $z=40x+50y$. The value of the function at each corner points of the graph is shown below. At $(0,0)$ is $z=40(0)+50(0)$ $z=0+0$ $z=0$. At $(0,8)$ is $z=40(0)+50(8)$ $z=0+400$ $z=400$. At $(4,9)$ is $z=40(4)+50(9)$ $z=160+450$ $z=610$. At $(8,0)$ is $z=40(8)+50(0)$ $z=320+0$ $z=320$. The maximum value of the objective function is $610$. The minimum value of the objective function is $0$.
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