Q.Determine graphically the minimum value of the objective function $Z = -50x + 20y$ subject to the constraints: $2x - y \ge -5$, $3x + y \ge 3$, $2x - 3y \le 12$, $x \ge 0,\ y \ge 0$.

Minimise $Z=-50x+20y$ subject to $2x-y\ge-5,\ 3x+y\ge3,\ 2x-3y\le12,\ x,y\ge0$.

The corner points of the (unbounded) feasible region are:

• $3x+y=3\ \cap\ x=0$: $(0,3)$, $\;Z=60$
• $2x-y=-5\ \cap\ x=0$: $(0,5)$, $\;Z=100$
• $3x+y=3\ \cap\ y=0$: $(1,0)$, $\;Z=-50$
• $2x-3y=12\ \cap\ y=0$: $(6,0)$, $\;Z=-300$ …

Set up

We want the minimum of $Z=-50x+20y$ subject to

Corner points

Solving the boundary lines pairwise and keeping the first-quadrant points that satisfy every constraint gives four vertices:

The region is unbounded — it opens out to the right and downward between the lines $2x-3y=12$ and $2x-y=-5$ as $x$ grows.

Evaluate Z at the corners

The smallest corner value is $-300$ at $(6,0)$.

Is $-300$ really the minimum?

For an unbounded region, a corner value is the minimum only if the open half-plane $Z<-300$ has no point in common with the feasible region. Here

$-50x+20y<-300 \iff 5x-2y>30,$ …

Method: Graphical Method on an Unbounded Feasible Region

When a feasible region stretches off to infinity, a corner value is not automatically the optimum. This method shows the extra check the syllabus requires for unbounded regions.

Steps

Step 1: Draw the region and find its corners as usual.

Plot each constraint line, shade with the origin test, intersect, and keep $x\ge 0,\ y\ge 0$. Note whether the shaded region closes into a polygon or opens outward without bound.

Step 2: Evaluate $Z=ax+by$ at every corner and pick the best corner value.

Call it $M$ (the smallest corner value when minimising, the largest when maximising).

Step 3: Since the region is unbounded, TEST whether $M$ can be beaten.

Form the open half-plane that would improve on $M$:

Check whether this half-plane has any point in common with the feasible region.

Step 4: Conclude. …

Common Mistakes

Mistake 1: Quoting the smallest corner value $-300$ as the minimum.

Why it's wrong: the region is unbounded, so a corner value is optimal only if no feasible point beats it. Here the half-plane $-50x+20y<-300$ does meet the region, so $Z$ goes lower — the minimum does not exist. Correct approach: for an unbounded region, always run the "can $M$ be beaten?" half-plane check before concluding.

Mistake 2: Not noticing the region is unbounded at all.

Why it's wrong: treating it as a closed polygon and stopping at corners gives a confident but wrong finite answer. Correct approach: after shading, check whether the region opens outward without a bounding edge in the direction that improves $Z$.

Mistake 3: Shading the $\ge$ and $\le$ constraints in the wrong directions. …