The programming problem Max $Z=2x+3y$ subject to the conditions $0 ≤x≤3, 0 ≤y≤4$ is:
Answer & explanation
Correct answer: option 3
objective fn → $Z=2x+3y$
constraints : $0 ≤x≤3$ → rectangular region
$0 ≤y≤4$
corner point $z(x,y)=2x+3y$
(0, 0) → Z(0, 0) = 0 + 0 = 0 (Z-min)
(3, 0) → Z(3, 0) = 6 + 0 = 6
(0, 4) → Z(0, 4) = 0 + 12 = 12
(3, 4) → Z(3, 4) = 6 + 12 = 18 (Z-max)