Q.Find the value of the following: Minimise and Maximise $Z = x + 2y$ subject to $x + 2y \ge 100$, $2x - y \le 0$, $2x + y \le 200$; $x, y \ge 0$.

Minimise and maximise $Z=x+2y$ subject to $x+2y\ge100,\ 2x-y\le0,\ 2x+y\le200,\ x,y\ge0$ (note $2x-y\le0$ means $y\ge2x$).

Feasible corner points:

• $x+2y=100\ \cap\ 2x-y=0$: $(20,40)$
• $2x-y=0\ \cap\ 2x+y=200$: $(50,100)$
• $x+2y=100\ \cap\ x=0$: $(0,50)$
• $2x+y=200\ \cap\ x=0$: $(0,200)$

Evaluate $Z=x+2y$: …

Set up

Minimise and maximise $Z=x+2y$ subject to

Here $2x-y\le0$ means $y\ge2x$. The three slanted lines together with the axes enclose a bounded region.

Find the corner points (keep only those satisfying every constraint)

• $x+2y=100$ and $2x-y=0$: $y=2x$, so $x+4x=100\Rightarrow x=20,\ y=40$ → $(20,40)$. Check $2x+y=80\le200$ ✓.
• $2x-y=0$ and $2x+y=200$: adding, $4x=200\Rightarrow x=50,\ y=100$ → $(50,100)$. Check $x+2y=250\ge100$ ✓.
• $x+2y=100$ and $x=0$: $(0,50)$. Check $2x-y=-50\le0$ ✓, $2x+y=50\le200$ ✓.
• $2x+y=200$ and $x=0$: $(0,200)$. Check $x+2y=400\ge100$ ✓, $2x-y=-200\le0$ ✓.

The point $(100,0)$ lies on $x+2y=100$ but fails $2x-y\le0$ (it gives $200\not\le0$), so it is not feasible.

Evaluate Z at each corner

Method: Simultaneous min and max with a mix of $\le$ and $\ge$ constraints

The general graphical method again, with care in reading a constraint's direction and in checking which intercepts actually belong to the feasible region.

Steps

Step 1: Rewrite each constraint in an easy-to-shade form.

For instance $2x-y\le0$ means $y\ge2x$. Shade each half-plane by testing a point, and keep $x,y\ge0$.

Step 2: Compute the corners, discarding infeasible candidates.

Solve boundary lines pairwise; a point such as $(100,0)$ on one line may violate another and must be dropped. Keep only vertices satisfying every constraint. …

Common Mistakes

Mistake 1: Misreading $2x-y\le0$.

Why it's wrong: it means $y\ge2x$; shading it as $y\le2x$ inverts the region. Correct approach: rearrange to $y\ge2x$ and shade accordingly.

Mistake 2: Treating $(100,0)$ as a feasible corner. …