Answer
See below.
Work Step by Step
We evaluate $z$ at the corner points:
$(0,6):z=0+6=6$
$(5,6):z=5+6=11$
$(5,2):z=5+2=7$
$(4,0):z=4+0=0$
$(0,3):z=0+3=3$
Thus the maximum value is $11$, the minimum value is $3$.
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 8 - Section 8.8 - Linear Programming - 8.8 Assess Your Understanding - Page 630: 3
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