The maximum value of $Z$ for the linear programing problem maximize $Z = x+y$ subject to the constraints $x+4y≤8,2x + 3y ≤ 12, 3x + y ≤9,x≥0,y≥0$ is:

$3\frac{10}{11}$

The correct answer is Option (1) → $3\frac{10}{11}$ **

Feasible corner points found from intersections and axes: $(0,0),\;(3,0),\;(0,2),\;\big(\frac{28}{11},\frac{15}{11}\big)$.

Evaluate $Z=x+y$ at these points:

$Z(0,0)=0$

$Z(3,0)=3$

$Z(0,2)=2$

$Z\!\Big(\frac{28}{11},\frac{15}{11}\Big)=\frac{28}{11}+\frac{15}{11}=\frac{43}{11}\approx 3.9091$

Maximum value of $Z$ is $\displaystyle \frac{43}{11}$ (attained at $\big(\frac{28}{11},\frac{15}{11}\big)$).