An LPP in two decision variables is solved graphically, because a region in the plane can be drawn and its corners read off directly. The whole method is four steps: graph the feasible region, find its corner points, evaluate the objective at each corner, and read off the optimum.
Step 1 — draw the feasible region
Treat each constraint inequality as a boundary line by replacing its inequality sign with "$=$", and plot that line (two points fix a line — the $x$- and $y$-intercepts are easiest). The line splits the plane into two half-planes. To decide which side the inequality allows, substitute a test point — usually the origin $(0,0)$, provided the line does not pass through it. If the test point satisfies the inequality, shade its side; otherwise shade the other side.
The feasible region is the overlap of all the shaded half-planes, including the first-quadrant restriction $x \ge 0,\ y \ge 0$. This region is always a convex polygon (possibly unbounded): the line segment joining any two of its points lies entirely inside it. Its vertices are the corner points.
Step 2 — find the corner points
Each corner is where two boundary lines meet. Read the obvious corners straight off the axes (intercepts), and find any interior corner by solving the two relevant boundary equations simultaneously. List every vertex of the feasible region.
Step 3 — the Corner-Point Theorem
This is the engine of the whole method:
$$\text{If an optimum of } Z = ax + by \text{ exists, it occurs at a corner point of the feasible region.}$$
Why corners? The objective function is linear, so its value changes steadily as you move across the region; it can only reach an extreme value at the boundary, and on a polygon the boundary's extremes are its vertices. So you never need to test interior points — evaluating $Z$ at the handful of corners is enough.
| Corner $(x, y)$ | Value of $Z = ax + by$ |
|---|
| list each vertex | substitute and compute |
The largest value in the table is the maximum and the smallest is the minimum. If the largest (or smallest) value is shared by two adjacent corners, then every point on the edge joining them is also optimal — the LPP has infinitely many optimal solutions.
Step 4 — bounded vs unbounded regions
A feasible region is bounded if it can be enclosed in a circle of finite radius, and unbounded if it stretches to infinity. The distinction matters because it decides whether an optimum is guaranteed:
- Bounded region: both a maximum and a minimum are guaranteed, each at a corner point. No extra checking is needed.
- Unbounded region: an optimum may not exist. The corner-point table still tells you the candidate value, but you must confirm it with an open-half-plane test.
The open-half-plane test for unbounded regions
Suppose the corner table gives a smallest value $m$ for a minimisation problem (region unbounded). Draw the open half-plane $ax + by < m$ and ask whether it has any point in common with the feasible region.
- If $ax + by < m$ has no point in common with the feasible region, then $m$ is genuinely the minimum.
- If it does share a point, $Z$ can be pushed below $m$, so no minimum exists.
For a maximisation problem the mirror image applies: with largest corner value $M$, test the open half-plane $ax + by > M$. If it meets the feasible region, $Z$ has no maximum; if not, $M$ is the maximum.
Worked example — full method
Maximise $Z = 4x + y$ subject to $x + y \le 50,\ 3x + y \le 90,\ x \ge 0,\ y \ge 0$.
The boundary lines $x + y = 50$ and $3x + y = 90$ meet where $2x = 40$, i.e. $x = 20,\ y = 30$. The corners of the bounded region are $(0,0),\ (30,0),\ (20,30),\ (0,50)$. Evaluating:
The maximum is $Z = 120$ at $(30, 0)$.
Deeper Insight — why you only check corners, and why unbounded needs a second look. Because both the objective and the constraints are linear, the level lines $ax + by = c$ are a family of parallel straight lines; optimising $Z$ means sliding this line as far as possible in the favourable direction while still touching the feasible region. On a bounded polygon the last point of contact is always a vertex — which is exactly the Corner-Point Theorem. On an unbounded region the line may be able to slide forever, never losing contact, so the objective grows without limit and no optimum exists; the open-half-plane test is simply a clean way of asking "can I still slide further?". Understanding the sliding-line picture turns the whole procedure from a recipe into something you can reconstruct from first principles.
Maximise $Z = 3x + 2y$ over the corner points $(0,0),\ (4,0),\ (0,5),\ (2,3)$.
Solution- $Z(0,0) = 0$.
- $Z(4,0) = 12$.
- $Z(0,5) = 10$.
- $Z(2,3) = 6 + 6 = 12$.
Maximum $Z = 12$, attained at both $(4,0)$ and $(2,3)$ — so every point on the edge joining them is also optimal.
Answer: Maximum $Z = 12$, attained at both $(4,0)$ and $(2,3)$.
Minimise $Z = x + y$ over the corner points $(2,0),\ (0,3),\ (1,1)$.
Solution- $Z(2,0) = 2$.
- $Z(0,3) = 3$.
- $Z(1,1) = 2$.
Minimum $Z = 2$ (at $(2,0)$ and $(1,1)$).
Answer: Minimum $Z = 2$ (at $(2,0)$ and $(1,1)$).
Where does the optimum of a bounded LPP occur, and why?
Solution- The objective $Z = ax + by$ is linear, so it has no turning point inside the region.
- Its extreme value on a convex polygon must lie on the boundary, at a vertex.
At a corner point of the feasible region, by the Corner-Point Theorem. On a bounded region both the maximum and the minimum are guaranteed.
Answer: At a corner point of the feasible region (Corner-Point Theorem).
Find the corner point common to $x + y = 4$ and $x = y$.
Solution- Substitute $y = x$ into $x + y = 4$: $2x = 4$.
- So $x = 2$, and then $y = 2$.
Corner point $(2, 2)$.
Answer: Corner point $(2, 2)$.
Bounded maximisation, full method. Maximise $Z = 5x + 3y$ subject to $3x + 5y \le 15,\ 5x + 2y \le 10,\ x \ge 0,\ y \ge 0$.
Solution- Intercepts of $3x + 5y = 15$: $(5,0)$ and $(0,3)$. Of $5x + 2y = 10$: $(2,0)$ and $(0,5)$.
- Their intersection: solve $3x + 5y = 15$ and $5x + 2y = 10$. From these, $x = \frac{20}{19},\ y = \frac{45}{19}$.
- Corners of the feasible region: $(0,0),\ (2,0),\ \left(\frac{20}{19}, \frac{45}{19}\right),\ (0,3)$.
- $Z(0,0) = 0;\ Z(2,0) = 10;\ Z\!\left(\frac{20}{19}, \frac{45}{19}\right) = \frac{100 + 135}{19} = \frac{235}{19} \approx 2.37;\ Z(0,3) = 9$.
Maximum $Z = \dfrac{235}{19}$ at $\left(\dfrac{20}{19}, \dfrac{45}{19}\right)$.
Answer: Maximum $Z = \dfrac{235}{19}$ at $\left(\dfrac{20}{19}, \dfrac{45}{19}\right)$.
Diet problem, solved. Minimise the cost $Z = 5x + 4y$ subject to $3x + y \ge 80,\ x + 4y \ge 100,\ x \ge 0,\ y \ge 0$ (the diet problem formulated earlier).
Solution- The region is unbounded (constraints are "$\ge$"). Find the corners.
- $3x + y = 80$ meets the $x$-axis at $\left(\frac{80}{3}, 0\right) \approx (26.7, 0)$; $x + 4y = 100$ meets the $y$-axis at $(0, 25)$.
- Intersection of the two lines: from $3x + y = 80$, $y = 80 - 3x$; substitute: $x + 4(80 - 3x) = 100 \Rightarrow x + 320 - 12x = 100 \Rightarrow -11x = -220 \Rightarrow x = 20,\ y = 20$.
- Corners: $\left(\frac{80}{3}, 0\right),\ (20, 20),\ (0, 80)$ — using whichever pieces bound the region.
- $Z\!\left(\frac{80}{3}, 0\right) = \frac{400}{3} \approx 33.3;\ Z(20,20) = 100 + 80 = 180;\ Z(0,80) = 320$.
- Smallest corner value is $m = \frac{400}{3}$. Test the open half-plane $5x + 4y < \frac{400}{3}$ against the feasible region.
The half-plane $5x + 4y < \frac{400}{3}$ shares no point with the feasible region, so the minimum cost is ₹$Z = \dfrac{400}{3} \approx \text{}133.3$ at $\left(\dfrac{80}{3}, 0\right)$.
Answer: Minimum cost ₹$Z = \dfrac{400}{3} \approx 33.3$ at $\left(\dfrac{80}{3}, 0\right)$.
Manufacturing problem, solved. Maximise profit $Z = 17.5x + 7y$ subject to $x + 3y \le 12,\ 3x + y \le 12,\ x \ge 0,\ y \ge 0$.
Solution- Intercepts of $x + 3y = 12$: $(12,0),\ (0,4)$. Of $3x + y = 12$: $(4,0),\ (0,12)$.
- Intersection: $x + 3y = 12$ and $3x + y = 12$ give $x = 3,\ y = 3$.
- Corners of the bounded region: $(0,0),\ (4,0),\ (3,3),\ (0,4)$.
- $Z(0,0) = 0;\ Z(4,0) = 70;\ Z(3,3) = 52.5 + 21 = 73.5;\ Z(0,4) = 28$.
Maximum profit $Z = 73.5$ at $(3, 3)$.
Answer: Maximum profit $Z = 73.5$ at $(3, 3)$.
Unbounded with a genuine minimum. Minimise $Z = 3x + 5y$ subject to $x + 3y \ge 3,\ x + y \ge 2,\ x \ge 0,\ y \ge 0$.
Solution- The region is unbounded ("$\ge$" constraints).
- Intersection of $x + 3y = 3$ and $x + y = 2$: subtract to get $2y = 1 \Rightarrow y = \frac{1}{2},\ x = \frac{3}{2}$.
- Corners: $(3, 0),\ \left(\frac{3}{2}, \frac{1}{2}\right),\ (0, 2)$.
- $Z(3,0) = 9;\ Z\!\left(\frac{3}{2}, \frac{1}{2}\right) = \frac{9}{2} + \frac{5}{2} = 7;\ Z(0,2) = 10$.
- Smallest is $m = 7$. Test $3x + 5y < 7$ against the region.
The open half-plane $3x + 5y < 7$ has no point in common with the feasible region, so the minimum is $Z = 7$ at $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$.
Answer: Minimum $Z = 7$ at $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$.
Unbounded with NO maximum. Show that maximising $Z = x + y$ subject to $x - y \ge -1,\ x \ge 0,\ y \ge 0$ has no maximum.
Solution- The constraint $x - y \ge -1$ is $y \le x + 1$; with $x,y \ge 0$ the region is unbounded upward and to the right.
- Pick points moving outward: $(10, 11)$ gives $Z = 21$; $(100, 101)$ gives $Z = 201$.
- $Z$ keeps growing — for any value $M$ the half-plane $x + y > M$ still meets the region.
Since $x + y > M$ always shares points with the feasible region, $Z$ can be made arbitrarily large: the LPP has no maximum.
Answer: The LPP has no maximum ($Z$ is unbounded above).
Transportation-style, solved. A transport cost is $Z = 10x + 12y$ (hundreds of rupees), to be minimised subject to $x + y \ge 8,\ x \le 5,\ y \le 6,\ x \ge 0,\ y \ge 0$. Find the optimum.
Solution- The region is bounded by $x \le 5,\ y \le 6$ from above and $x + y \ge 8$ from below.
- Corners: $x + y = 8$ meets $x = 5$ at $(5, 3)$; meets $y = 6$ at $(2, 6)$. The upper corner is $(5, 6)$.
- $Z(5,3) = 50 + 36 = 86;\ Z(2,6) = 20 + 72 = 92;\ Z(5,6) = 50 + 72 = 122$.
Minimum $Z = 86$ (i.e. ₹$8600$) at $(5, 3)$. Bounded region, so the minimum is guaranteed at a corner.
Answer: Minimum $Z = 86$ (i.e. ₹$8600$) at $(5, 3)$.
Find all the corner points of the feasible region $x \ge 0,\ y \ge 0,\ x + 2y \le 8,\ 3x + 2y \le 12$.
Solution- $x$-axis ($y=0$): $3x \le 12 \Rightarrow x \le 4$, giving $(4, 0)$.
- $y$-axis ($x=0$): $2y \le 8 \Rightarrow y \le 4$, giving $(0, 4)$.
- Intersection of $x + 2y = 8$ and $3x + 2y = 12$: subtract to get $2x = 4 \Rightarrow x = 2,\ y = 3$.
- Include the origin.
The corner points are $(0,0),\ (4,0),\ (2,3),\ (0,4)$.
Answer: The corner points are $(0,0),\ (4,0),\ (2,3),\ (0,4)$.
Maximise $Z = 22x + 18y$ subject to $x + y \le 20,\ 3x + 2y \le 48,\ x \ge 0,\ y \ge 0$.
Solution- Intersection of $x + y = 20$ and $3x + 2y = 48$: from the first $y = 20 - x$; substitute: $3x + 40 - 2x = 48 \Rightarrow x = 8,\ y = 12$.
- Corners: $(0,0),\ (16,0),\ (8,12),\ (0,20)$. (Here $3x = 48$ gives $x = 16$ on the $x$-axis.)
- $Z(0,0) = 0;\ Z(16,0) = 352;\ Z(8,12) = 176 + 216 = 392;\ Z(0,20) = 360$.
Maximum $Z = 392$ at $(8, 12)$.
Answer: Maximum $Z = 392$ at $(8, 12)$.