NCERT Class 12 Mathematics — Linear Programming, Exercise 12.1. All 10 questions solved.

This is the only exercise in Chapter 12, and every question follows the same four steps:

1. Draw each constraint as a straight line, and shade the half-plane that satisfies the inequality. Testing the origin is the quickest way to decide which side, provided the line does not pass through it.
2. Identify the feasible region — the intersection of all the shaded half-planes together with $x \ge 0$, $y \ge 0$.
3. Find every corner point by solving the pairs of boundary lines that meet there.
4. Evaluate $\mathrm{Z}$ at each corner and take the largest or smallest.

Key insight. Step 4 is only valid without qualification when the feasible region is bounded. If it is unbounded, a corner value $M$ is the true maximum only if the open half-plane $\mathrm{Z} > M$ has no point in common with the region — and the true minimum only if $\mathrm{Z} < m$ has none. Questions 4, 6 and 9 all have unbounded regions, and question 9 is precisely the case where the check fails and there is no maximum at all.

The other thing to watch for: when two corner points give the same optimal value, every point on the segment joining them is also optimal. That is questions 6, 7 and 8.

Solve the following Linear Programming Problems graphically.

Question 1

Maximise $\mathrm{Z} = 3x + 4y$ subject to the constraints $x + y \le 4$, $x \ge 0$, $y \ge 0$.

Solution. The line $x + y = 4$ meets the axes at $(4, 0)$ and $(0, 4)$. Since $0 + 0 = 0 \le 4$, the origin satisfies the inequality, so the feasible region is the triangle below the line in the first quadrant — bounded.

The region is bounded, so the largest corner value is the maximum.

Question 2

Minimise $\mathrm{Z} = -3x + 4y$ subject to $x + 2y \le 8$, $3x + 2y \le 12$, $x \ge 0$, $y \ge 0$.

Solution. $x + 2y = 8$ meets the axes at $(8, 0)$ and $(0, 4)$; $3x + 2y = 12$ meets them at $(4, 0)$ and $(0, 6)$. Subtracting the first equation from the second gives $2x = 4$, so the two lines cross at $(2, 3)$.

Note that the negative coefficient on $x$ pushes the minimum to the corner with the largest $x$ and smallest $y$.

Question 3

Maximise $\mathrm{Z} = 5x + 3y$ subject to $3x + 5y \le 15$, $5x + 2y \le 10$, $x \ge 0$, $y \ge 0$.

Solution. $3x + 5y = 15$ meets the axes at $(5, 0)$ and $(0, 3)$; $5x + 2y = 10$ at $(2, 0)$ and $(0, 5)$. Solving the pair simultaneously: from the second, $y = \tfrac{10 – 5x}{2}$, and substituting into the first gives $6x + 50 – 25x = 30$, so $x = \tfrac{20}{19}$ and $y = \tfrac{45}{19}$.

Question 4

Minimise $\mathrm{Z} = 3x + 5y$ such that $x + 3y \ge 3$, $x + y \ge 2$, $x, y \ge 0$.

Solution. Both inequalities are $\ge$, and the origin fails both, so the feasible region lies away from the origin and is unbounded. The boundary lines meet where $x + 3y = 3$ and $x + y = 2$: subtracting gives $2y = 1$, so $y = \tfrac12$ and $x = \tfrac32$.

Because the region is unbounded, the value $7$ is only a candidate until checked. Draw $3x + 5y < 7$: this open half-plane lies strictly on the origin side of the line through $\left(\tfrac32, \tfrac12\right)$, and the feasible region lies entirely on the other side. They have no point in common, so $7$ really is the minimum.

Question 5

Maximise $\mathrm{Z} = 3x + 2y$ subject to $x + 2y \le 10$, $3x + y \le 15$, $x, y \ge 0$.

Solution. $x + 2y = 10$ meets the axes at $(10, 0)$ and $(0, 5)$; $3x + y = 15$ at $(5, 0)$ and $(0, 15)$. Substituting $y = 15 – 3x$ into the first gives $x + 30 – 6x = 10$, so $x = 4$ and $y = 3$.

Question 6

Minimise $\mathrm{Z} = x + 2y$ subject to $2x + y \ge 3$, $x + 2y \ge 6$, $x, y \ge 0$. Show that the minimum of $\mathrm{Z}$ occurs at more than two points.

Solution. $2x + y = 3$ meets the axes at $\left(\tfrac32, 0\right)$ and $(0, 3)$; $x + 2y = 6$ at $(6, 0)$ and $(0, 3)$. Both lines pass through $(0, 3)$, so the corner points of the unbounded region are just $(0, 3)$ and $(6, 0)$.

The two corners give the same value. Since $\mathrm{Z} = x + 2y$ is a constant multiple of the left-hand side of the constraint $x + 2y \ge 6$, the objective function is constant along the whole boundary segment $x + 2y = 6$ joining $(0, 3)$ to $(6, 0)$ — so every point on that segment is a minimiser, which is what the question asks you to show.

Unboundedness check: $x + 2y < 6$ is exactly the region excluded by the second constraint, so it has no point in common with the feasible region and $6$ is genuinely the minimum.

Question 7

Minimise and Maximise $\mathrm{Z} = 5x + 10y$ subject to $x + 2y \le 120$, $x + y \ge 60$, $x – 2y \ge 0$, $x, y \ge 0$.

Solution. Three constraints, so find each pairwise intersection and keep the ones that satisfy all of them:

• $x – 2y = 0$ with $x + y = 60$: substituting $x = 2y$ gives $3y = 60$, so $(40, 20)$.
• $x – 2y = 0$ with $x + 2y = 120$: gives $4y = 120$, so $(60, 30)$.
• $x + y = 60$ with $y = 0$: $(60, 0)$.
• $x + 2y = 120$ with $y = 0$: $(120, 0)$.

The region is bounded.

Two corners tie at $600$, so — as in question 6 — the maximum is attained along the whole segment joining them.

Question 8

Minimise and Maximise $\mathrm{Z} = x + 2y$ subject to $x + 2y \ge 100$, $2x – y \le 0$, $2x + y \le 200$; $x, y \ge 0$.

Solution. The constraint $2x – y \le 0$ is $y \ge 2x$, a half-plane through the origin. The corners are:

• $x + 2y = 100$ with $x = 0$: $(0, 50)$.
• $x + 2y = 100$ with $y = 2x$: $x + 4x = 100$, so $(20, 40)$.
• $y = 2x$ with $2x + y = 200$: $4x = 200$, so $(50, 100)$.
• $2x + y = 200$ with $x = 0$: $(0, 200)$.

Again the objective function is a multiple of a constraint’s left-hand side, so it is constant along the boundary $x + 2y = 100$ and the minimum is attained on the whole of that segment.

Question 9

Maximise $\mathrm{Z} = -x + 2y$, subject to the constraints $x \ge 3$, $x + y \ge 5$, $x + 2y \ge 6$, $y \ge 0$.

Solution. All three constraints are $\ge$, so the region is unbounded — it extends upwards and to the right without limit. Its corners:

• $x = 3$ with $x + y = 5$: $(3, 2)$.
• $x + y = 5$ with $x + 2y = 6$: subtracting gives $y = 1$, so $(4, 1)$.
• $x + 2y = 6$ with $y = 0$: $(6, 0)$.

The largest corner value is $1$, but the region is unbounded and the check fails: the half-plane $-x + 2y > 1$ contains, for instance, the point $(3, 100)$, which satisfies every constraint. So $1$ is not a maximum, and no maximum exists — moving up the line $x = 3$ increases $\mathrm{Z}$ without bound.

This is the question that shows why the unboundedness check is not a formality.

Question 10

Maximise $\mathrm{Z} = x + y$, subject to $x – y \le -1$, $-x + y \le 0$, $x, y \ge 0$.

Solution. Rewrite both constraints in terms of $y$:

$$x – y \le -1 \iff y \ge x + 1, \qquad -x + y \le 0 \iff y \le x$$

Together these require $x + 1 \le y \le x$, which forces $x + 1 \le x$ — an impossibility. Graphically, $y = x + 1$ and $y = x$ are parallel lines, and the two half-planes lie on opposite sides of the gap between them.

There is no feasible region, so there is nothing to optimise over.

Common mistakes

• Skipping the unboundedness check. In question 9 the corner-point table gives a perfectly plausible answer of $1$, and it is wrong. Whenever the region is unbounded, test the open half-plane before writing anything down.
• Reporting only one optimal point when two corners tie. Questions 6, 7 and 8 each have an entire optimal segment. NCERT expects the segment, not one of its endpoints.
• Shading the wrong half-plane. Test a point — the origin, if the line misses it. For $2x – y \le 0$ in question 8 the line does pass through the origin, so test $(1, 0)$ instead: $2 > 0$, so $(1, 0)$ is excluded and the region lies above the line.
• Missing a corner point. Every pair of boundary lines that meets inside the region contributes one, and so do the intersections with the axes. Question 8 has four corners and it is easy to find only three.
• Forgetting $x \ge 0$ and $y \ge 0$. They are constraints like any other and contribute boundary lines and corners of their own.
• Reading the region as bounded because the picture is cropped. A drawing has to stop somewhere. Decide boundedness from the inequalities — all $\le$ with positive coefficients gives a bounded region; a mixture usually does not.

Practise next

• Exercise 8.1 — a different use of the same first-quadrant geometry, this time for area.
• Class 11 Linear Inequalities, Exercise 5.1 — solving linear inequalities and showing the solution on a number line. The current Class 11 book no longer teaches graphing inequalities in two variables, so the half-plane shading this chapter relies on is set out in the method at the top of this page.