Q.Find the value of the following: Minimise and Maximise $Z = 5x + 10y$ subject to $x + 2y \le 120$, $x + y \ge 60$, $x - 2y \ge 0$, $x, y \ge 0$.

Minimise and maximise $Z=5x+10y$ subject to $x+2y\le120,\ x+y\ge60,\ x-2y\ge0,\ x,y\ge0$ (note $x-2y\ge0$ means $x\ge2y$).

Feasible corner points:

• $x+y=60\ \cap\ y=0$: $(60,0)$
• $x+2y=120\ \cap\ y=0$: $(120,0)$
• $x+2y=120\ \cap\ x-2y=0$: $(60,30)$
• $x-2y=0\ \cap\ x+y=60$: $(40,20)$

Evaluate $Z=5x+10y$: …

Set up

Minimise and maximise $Z = 5x+10y$ subject to

Note $x-2y\ge0$ means $x\ge2y$. The cap $x+2y\le120$ closes the region off, so it is a bounded polygon.

Find the corner points

• $x+y=60$ with $y=0$: $(60,0)$. Check $x-2y=60\ge0$ ✓, $x+2y=60\le120$ ✓.
• $x+2y=120$ with $y=0$: $(120,0)$. Check $x-2y=120\ge0$ ✓, $x+y=120\ge60$ ✓.
• $x+2y=120$ and $x-2y=0$: since $x=2y$, $\;2y+2y=120\Rightarrow y=30,\ x=60$ → $(60,30)$. Check $x+y=90\ge60$ ✓.
• $x-2y=0$ and $x+y=60$: $x=2y$, so $3y=60\Rightarrow y=20,\ x=40$ → $(40,20)$. Check $x+2y=80\le120$ ✓.

So the feasible region is the quadrilateral $(60,0),(120,0),(60,30),(40,20)$.

Evaluate Z at each corner

Method: Finding both minimum and maximum over a bounded polygon

For "minimise and maximise" questions, one corner-point pass delivers both extremes; stay alert for a tie at one end that signals multiple optimal solutions.

Steps

Step 1: Plot every constraint and pin down the feasible polygon.

Translate each inequality (e.g. $x-2y\ge0$ means $x\ge2y$), shade correctly, and enforce $x,y\ge0$. A mix of "$\le$" and "$\ge$" here closes the region into a bounded polygon.

Step 2: Evaluate $Z$ at all valid corners.

$Z=ax+by$ …

Common Mistakes

Mistake 1: Reading $x-2y\ge0$ in the wrong direction.

Why it's wrong: it means $x\ge2y$, not $x\le2y$; shading the wrong half-plane gives a completely different region. Correct approach: rewrite the constraint and test a point before shading.

Mistake 2: Reporting the maximum at one corner when it is tied. …