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: 2

Answer

The answer is At $(3,2)=13$ At $(4,10)=32$ At $(5,12)=39$ At $(7,4)=29$ At $(8,6)=36$ Maximum value $=39$. Minimum value $=13 $.

Work Step by Step

The given objective function is $z=3x+2y $. The value of the function at each corner points of the graph is shown below. At $ (3,2) $ is $z=3(3)+2(2) $ $z=9+4 $ $z=13$. At $ (4,10) $ is $z=3(4)+2(10) $ $z=12+20 $ $z=32$. At $ (5,12) $ is $z=3(5)+2(12) $ $z=15+24 $ $z=39$. At $ (7,4) $ is $z=3(7)+2(4) $ $z=21+8 $ $z=29$. At $ (8,6) $ is $z=3(8)+2(6) $ $z=24+12 $ $z=36$. The maximum value of the objective function is $39$. The minimum value of the objective function is $13$.
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