Answer
$z(3,2)= 13$, $z(4,10)= 32$, $z(5,12)= 39$, $z(8,6)= 36$, $z(7,4)= 29$,
maximum $z(5,12)= 39$
minimum $z(3,2)= 13$
Work Step by Step
Step 1. Given the objective function $z(x,y)=3x+2y$, we can obtain the function values at each corner as
$z(3,2)=3(3)+2(2)=13$, $z(4,10)=3(4)+2(10)=32$,
$z(5,12)=3(5)+2(12)=39$, $z(8,6)=3(8)+2(6)=36$, $z(7,4)=3(7)+2(4)=29$,
Step 2. We can find the maximum of $z$ as $z(5,12)= 39$
Step 3. We can find the minimum of $z$ as $z(3,2)= 13$
