Maximize $Z=30x-18y$ subject to $3x+4y\le 60$ and $5x-3y\le 20$ ($x,y\ge 0$). In the feasible region, the maximum value occurs at:

Let's find the corner points:$3x+4y=60$, $5x-3y=20$ Multiply eq 1 by 3: $9x+12y=180$ Multiply eq 2 by 4: $20x-12y=80$ Add: $29x=260⟹x=260/29$.$y=(5(260/29)-20)/3=(1300/29-580/29)/3=720/87=240/29$. Point $(260/29,240/29)$.$x$ -intercept of $5x-3y=20$ is $(4,0)$.Origin $(0,0)$ and $y$ -intercept $(0,15)$.Evaluate $Z=30x-18y$:$Z(4,0)=120$$Z(260/29,240/29)=30(260/29)-18(240/29)=(7800-4320)/29=3480/29=120$.Since the maximum value $120$ is achieved at two corner points, it is achieved at infinitely many points on the line segment joining them.