Q.Solve the following linear programming problem graphically: Minimise $Z = 200x + 500y$ subject to the constraints: $x + 2y \ge 10$, $3x + 4y \le 24$, $x \ge 0,\ y \ge 0$.

Minimise $Z=200x+500y$ subject to $x+2y\ge10,\ 3x+4y\le24,\ x,y\ge0$.

The feasible region is the triangle with corners:

• $x+2y=10\ \cap\ x=0$: $(0,5)$
• $3x+4y=24\ \cap\ x=0$: $(0,6)$
• $x+2y=10\ \cap\ 3x+4y=24$: $(4,3)$

Evaluate $Z$: $\;Z(0,5)=2500,\ Z(0,6)=3000,\ Z(4,3)=800+1500=2300.$

(The axis points $(10,0)$ and $(8,0)$ are not feasible: e.g. $(8,0)$ fails $x+2y\ge10$ since $8<10$.)

Set up

Minimise $Z=200x+500y$ subject to

The region must lie above $x+2y=10$ and below $3x+4y=24$, in the first quadrant.

Plot the boundary lines

• $x+2y=10$ passes through $(10,0)$ and $(0,5)$.
• $3x+4y=24$ passes through $(8,0)$ and $(0,6)$.

Find the feasible corner points

• On the $y$-axis ($x=0$): the two constraints give $2y\ge10$ (so $y\ge5$) and $4y\le24$ (so $y\le6$). This gives the vertices $(0,5)$ and $(0,6)$.
• Intersection of the two lines: from $x+2y=10$, multiply by $2$: $2x+4y=20$. Subtract from $3x+4y=24$: $x=4$, then $2y=6\Rightarrow y=3$ → $(4,3)$.
• The $x$-axis gives no feasible point: $y=0$ needs $x\ge10$ (first constraint) and $x\le8$ (second) at once, which is impossible. So $(8,0)$ and $(10,0)$ are both outside the region.

Hence the feasible region is the triangle $(0,5),(0,6),(4,3)$.

Evaluate Z at the corners

The smallest value is $2300$ at $(4,3)$. The region is bounded, so this is the true minimum.

Method: Corner-Point (Graphical) Method for a Minimum

Use this to minimise a linear objective $Z=ax+by$ of two variables subject to a mix of $\ge$ and $\le$ linear constraints. The mechanics are identical to the maximisation case — only the final selection changes.

Steps

Step 1: Plot each constraint line.

Turn each inequality into an equation and draw it from its intercepts, keeping $x\ge 0,\ y\ge 0$.

Step 2: Shade each half-plane and take the overlap.

Test the origin in each inequality. Note that a $\ge$ constraint (e.g. $x+2y\ge 10$) typically keeps the side away from the origin, while a $\le$ constraint keeps the side containing the origin. The feasible region is where all kept half-planes overlap.

Step 3: Locate the feasible corner points.

Solve intersecting boundary lines pairwise. Crucially, discard any intersection that violates another constraint — for a mixed $\ge/\le$ system an axis intercept often lies outside the region (e.g. it satisfies one constraint but not the other).

Step 4: Evaluate $Z$ at each valid corner and take the smallest.

Step 5: Confirm the region is bounded.

If the feasible region is a closed polygon, the smallest corner value is the true minimum. (If it were unbounded, you would additionally have to check whether $Z$ can be driven still lower — see the unbounded-region method.) A bounded region guarantees the corner minimum is genuine.

Common Mistakes

Mistake 1: Treating the axis intercepts $(10,0)$ and $(8,0)$ as feasible corners.

Why it's wrong: $(8,0)$ fails $x+2y\ge 10$ since $8<10$, and $(10,0)$ fails $3x+4y\le 24$ since $30>24$. Neither lies in the region. Correct approach: check every candidate corner against all constraints before evaluating $Z$.

Mistake 2: Shading $x+2y\ge 10$ toward the origin.

Why it's wrong: the origin gives $0\ge 10$, which is false, so the $\ge$ constraint keeps the side away from the origin. Shading toward it inverts the whole region. Correct approach: for a $\ge$ constraint that fails the origin test, keep the far side.

Mistake 3: Miscomputing the intersection $(4,3)$.

Why it's wrong: solving $x+2y=10$ and $3x+4y=24$ needs elimination — double the first to $2x+4y=20$, subtract to get $x=4$, then $y=3$. A sign slip here changes the minimum. Correct approach: substitute the found point back into both equations to verify.

Mistake 4: Evaluating $Z=200x+500y$ with the coefficients swapped.

Why it's wrong: writing $500x+200y$ gives the wrong values and can flip which corner is smallest. Correct approach: keep each variable with its own coefficient — $Z(4,3)=200(4)+500(3)=2300$.