Answer
$44$
Work Step by Step
Step 1. Graph the inequalities
$\begin{cases} x\geq0, y\geq0\\y\leq x\\2x+y\leq12\\2x+3y\geq6 \end{cases}$
Step 2. We can find the vertices of the solution region as
$(\frac{6}{5},\frac{6}{5}), (4,4), (6,0),(3,0)$
Step 3. Evaluate the objective function at the vertices; we have
$z_1=5(\frac{6}{5})+6(\frac{6}{5})=\frac{66}{5}$; $z_2=5(4)+6(4)=44$; $z_3=5(6)+6(0)=30$; $z_4=5(3)+6(0)=15$;
Step 4. Thus the maximum of the objective function values is $z=44$ at $(4,4)$
