Q.Solve the following problem graphically: Minimise and Maximise $Z = 3x + 9y$ subject to the constraints: $x + 3y \le 60$, $x + y \ge 10$, $x \le y$, $x \ge 0,\ y \ge 0$.

Minimise and maximise $Z=3x+9y$ subject to $x+3y\le60,\ x+y\ge10,\ x\le y,\ x,y\ge0$.

Feasible corner points:

• $x+y=10\ \cap\ x=0$: $(0,10)$
• $x+y=10\ \cap\ x=y$: $(5,5)$
• $x=y\ \cap\ x+3y=60$: $(15,15)$
• $x+3y=60\ \cap\ x=0$: $(0,20)$ …

Set up

Minimise and maximise $Z=3x+9y$ subject to

So we stay below $x+3y=60$, above $x+y=10$, and above the line $y=x$ (since $x\le y$), in the first quadrant.

Find the corner points (each must satisfy all constraints)

• $x+y=10$ and $x=0$: $(0,10)$. Check $x\le y$ ✓, $x+3y=30\le60$ ✓.
• $x+y=10$ and $x=y$: $2x=10\Rightarrow x=y=5$ → $(5,5)$. Check $x+3y=20\le60$ ✓.
• $x=y$ and $x+3y=60$: $x+3x=60\Rightarrow x=15,\ y=15$ → $(15,15)$. Check $x+y=30\ge10$ ✓.
• $x+3y=60$ and $x=0$: $(0,20)$. Check $x\le y$ ✓, $x+y=20\ge10$ ✓.

Points like $(10,0)$ or $(60,0)$ satisfy $x+y\ge10$ but fail $x\le y$, so they are not corners.

Evaluate Z at each corner

Method: Corner-Point Method for Both Minimum and Maximum (and Multiple Optima)

Some problems ask for both the minimum and the maximum of $Z=ax+by$ over one feasible region. The graphical corner-point technique handles both at once, and also flags when the optimum is not unique.

Steps

Step 1: Build the feasible region.

Plot every constraint line, shade the correct half-plane for each (origin test), and intersect them, keeping $x\ge 0,\ y\ge 0$. A constraint like $x\le y$ is the line $y=x$ with the region kept on the side where $y$ is larger.

Step 2: List all corner points.

Solve each pair of adjacent boundary lines and keep only intersections that satisfy every constraint. An intersection that fails even one constraint is not a vertex of this region.

Step 3: Evaluate $Z$ at every corner.

Build a small table of $Z=ax+by$ for each vertex. The smallest entry gives the minimum; the largest gives the maximum — read both off the same table.

Step 4: Watch for a tie — the sign of multiple optimal solutions. …

Common Mistakes

Mistake 1: Reporting the maximum only at $(0,20)$ (or only at $(15,15)$).

Why it's wrong: both corners give $Z=180$, so $Z$ is constant along the whole edge joining them — every point on that segment is optimal. Correct approach: when two adjacent corners tie for the optimum, state that the optimum holds along the entire connecting segment.

Mistake 2: Ignoring the constraint $x\le y$ and admitting points like $(10,0)$.

Why it's wrong: $(10,0)$ satisfies $x+y\ge 10$ but violates $x\le y$ (since $10\not\le 0$), so it is not in the feasible region. Correct approach: every corner must satisfy all constraints, including $x\le y$ (i.e. it lies on or above the line $y=x$).

Mistake 3: Confusing minimum and maximum in the corner table. …