Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.6 - Linear Programming - Exercise Set - Page 871: 4

Answer

$z(0,0)= 0$, $z(0,9)= 405$, $z(4,4)= 300$, $z(3,0)= 90$ maximum $z(0,9)= 405$ minimum $z(0,0)= 0$

Work Step by Step

Step 1. Given the objective function $z(x,y)=30x+45y$, we can obtain the function values at each corner as $z(0,0)=30(0)+45(0)=0$ $z(0,9)=30(0)+45(9)=405$ $z(4,4)=30(4)+45(4)=300$, $z(3,0)=30(3)+45(0)=90$ Step 2. We can find the maximum of $z$ as $z(0,9)= 405$ Step 3. We can find the minimum of $z$ as $z(0,0)= 0$
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