NCERT Exemplar Solutions Class 12 Maths Linear Programming — Objective, Fill in the Blanks and True or False Q26–45

NCERT Exemplar Problems, Class 12 Mathematics — Linear Programming, Objective Type questions 26 to 34, Fill in the Blanks 35 to 41 and True or False 42 to 45. All 20 questions solved.

Twenty short questions that all rest on one theorem: the maximum and minimum of $Z = ax + by$, when they exist, are found at corner points of the feasible region. Questions 26 to 34 apply it to given corners or figures, Questions 35 to 41 test the vocabulary, and Questions 42 to 45 probe the edge cases: unbounded regions, ties between two corners, and a minimum that is not zero.

One answer in the printed key is wrong, Question 45, and the reason is explained where it occurs.

A note on the syllabus. Linear Programming is in the current NCERT Class 12 textbook (Chapter 12) but is not in the JEE Main 2026 or JEE Advanced 2026 syllabus; it is a board-exam topic. The rationalised Chapter 12 keeps only Exercise 12.1, with problems already written as inequalities and solved by the corner-point (graphical) method. The questions on this page are inside that scope.

Key insight. Evaluate $Z$ at every corner point and compare. If the region is bounded, the largest and smallest values are the answers. If it is unbounded, a maximum or minimum may fail to exist, so test the direction in which the region runs off to infinity.

Question 26

The corner points of the feasible region determined by the system of linear constraints are $(0, 0)$, $(0, 40)$, $(20, 40)$, $(60, 20)$, $(60, 0)$. The objective function is $Z = 4x + 3y$. Compare the quantity in Column A and Column B.

Column A: Maximum of $Z$. Column B: $325$.

Solution. The region is a closed polygon, so the maximum is at a corner. Evaluate $Z = 4x + 3y$:

Corner	$Z = 4x + 3y$
$(0, 0)$	$0$
$(0, 40)$	$120$
$(20, 40)$	$200$
$(60, 20)$	$300$
$(60, 0)$	$240$

The maximum is $300$, which is less than $325$.

Question 27

The feasible solution for a LPP is shown in Fig. 12.12. Let $Z = 3x – 4y$ be the objective function. Minimum of $Z$ occurs at

Solution. Fig. 12.12 is a shaded polygon in the first quadrant, bounded on the left by the $y$-axis and below by the $x$-axis, with labelled corners $(0, 0)$, $(5, 0)$, $(6, 5)$, $(6, 8)$, $(4, 10)$ and $(0, 8)$. Evaluate $Z = 3x – 4y$ at each:

Corner	$Z = 3x – 4y$
$(0, 0)$	$0$
$(5, 0)$	$15$
$(6, 5)$	$-2$
$(6, 8)$	$-14$
$(4, 10)$	$-28$
$(0, 8)$	$-32$

The smallest value is $-32$, at $(0, 8)$. The coefficient $-4$ on $y$ pulls the minimum towards the highest point with small $x$; $(4, 10)$ is higher, but its $3x = 12$ costs more than the $2$ units of height gained are worth.

Question 28

Refer to Exercise 27. Maximum of $Z$ occurs at

Solution. From the table in Question 27, the largest value of $Z = 3x – 4y$ is $15$, at $(5, 0)$. Moving up the right-hand side loses $4$ per unit of $y$ and gains nothing in $x$ beyond the first step.

Question 29

Refer to Exercise 27. (Maximum value of $Z$ + Minimum value of $Z$) is equal to

Solution. Maximum $= 15$ and minimum $= -32$, so the sum is $15 + (-32) = -17$.

Question 30

The feasible region for an LPP is shown in the Fig. 12.13. Let $F = 3x – 4y$ be the objective function. Maximum value of $F$ is

Solution. Fig. 12.13 is the region of Question 25 (the same objective $3x – 4y$) drawn loosely and not to scale. It is bounded by the $y$-axis, a line through the origin and $(12, 6)$, and a line through $(0, 4)$ that runs up to the right. The line through $O$ and $(12, 6)$ is $x – 2y = 0$; the line through $(6, 0)$ and $(12, 6)$ is $x – y = 6$ (drawn nearly vertical in the figure); and the line through $(0, 4)$ is $-3x + y = 4$. The corner points are $(0, 0)$, $(0, 4)$ and $(12, 6)$, and the region is unbounded, running off to the upper right between the lines $y = x – 6$ and $y = 3x + 4$.

Corner	$F = 3x – 4y$
$(0, 0)$	$0$
$(0, 4)$	$-16$
$(12, 6)$	$12$

On an unbounded region the corner value $12$ is the maximum only if $F$ never exceeds it. It does not: for every point of the region $y \ge \frac x2$ and $y \ge x – 6$, so $F = 3x – 4y \le 3x – 2x = x$ and $F \le 3x – 4(x – 6) = 24 – x$. Therefore $F \le \min(x, 24 – x) \le 12$, with equality at $(12, 6)$.

Question 31

Refer to Exercise 30. Minimum value of $F$ is

Solution. The smallest corner value is $-16$ at $(0, 4)$, which is option (B), but the region is unbounded, so the corner value is only a candidate. The direction $(1, 2)$ stays inside the region: the point $(t, 2t)$ satisfies $x – 2y = -3t \le 0$, $-3x + y = -t \le 4$ and $x – y = -t \le 6$ for every $t \ge 0$. Along it $F = 3t – 8t = -5t$, which falls without limit. So $F$ has no minimum.

Question 32

Corner points of the feasible region for an LPP are $(0, 2)$, $(3, 0)$, $(6, 0)$, $(6, 8)$ and $(0, 5)$. Let $F = 4x + 6y$ be the objective function. The minimum value of $F$ occurs at

Solution. The five corners give

Corner	$F = 4x + 6y$
$(0, 2)$	$12$
$(3, 0)$	$12$
$(6, 0)$	$24$
$(6, 8)$	$72$
$(0, 5)$	$30$

The minimum $12$ occurs at two adjacent corners. When $Z$ takes the same optimal value at two corners, it takes that value along the whole segment between them: here $F = 4x + 6y = 12$ is exactly the line $2x + 3y = 6$ through both.

Question 33

Refer to Exercise 32, Maximum of $F$ $-$ Minimum of $F$ $=$

Solution. From the table in Question 32, the maximum is $72$ at $(6, 8)$ and the minimum is $12$, so the difference is $72 – 12 = 60$.

Question 34

Corner points of the feasible region determined by the system of linear constraints are $(0, 3)$, $(1, 1)$ and $(3, 0)$. Let $Z = px + qy$, where $p, q > 0$. Condition on $p$ and $q$ so that the minimum of $Z$ occurs at $(3, 0)$ and $(1, 1)$ is

Solution. The values are $Z(3, 0) = 3p$, $Z(1, 1) = p + q$ and $Z(0, 3) = 3q$. For the minimum to occur at both $(3, 0)$ and $(1, 1)$ the two values must be equal: $3p = p + q$, so $q = 2p$, that is $p = \frac q2$. Then both equal $\frac{3q}{2}$, which is below $Z(0, 3) = 3q$, so $(0, 3)$ is indeed not the minimum.

Question 35

In a LPP, the linear inequalities or restrictions on the variables are called ______.

Solution. The restrictions that define the feasible region are the constraints. Non-negativity ($x, y \ge 0$) is one of them.

Question 36

In a LPP, the objective function is always ______

Solution. “Linear programming” means exactly this: the quantity to be maximised or minimised, $Z = ax + by$, is a linear function of the variables, as are the constraints.

Question 37

If the feasible region for a LPP is ______, then the optimal value of the objective function $Z = ax + by$ may or may not exist.

Solution. A closed polygon always gives a maximum and a minimum. Only when the region is unbounded can the objective run off to infinity, as in Question 31.

Question 38

In a LPP if the objective function $Z = ax + by$ has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same ______ value.

Solution. If $Z$ has the same value at two points, it has that value along the segment joining them, since $Z$ is linear. As in Question 32, the tied value is the maximum (optimal) value.

Question 39

A feasible region of a system of linear inequalities is said to be ______ if it can be enclosed within a circle.

Solution. This is the definition of a bounded region. A region that cannot be enclosed in any circle is unbounded.

Question 40

A corner point of a feasible region is a point in the region which is the ______ of two boundary lines.

Solution. A corner is where two boundary lines meet, and it must lie in the region (an intersection outside the region is not a corner point).

Question 41

The feasible region for an LPP is always a ______ polygon.

Solution. The region is the common part of half-planes, and an intersection of convex sets is convex: the segment joining any two points of the region lies wholly inside it. This convexity is why a linear objective is optimised at a corner.

Question 42

State True or False: If the feasible region for a LPP is unbounded, maximum or minimum of the objective function $Z = ax + by$ may or may not exist.

Solution. It may exist: in Question 30 the region is unbounded and the maximum $12$ still exists. It may not: Question 31 has no minimum. Both cases occur, so the statement is correct.

Question 43

State True or False: Maximum value of the objective function $Z = ax + by$ in a LPP always occurs at only one corner point of the feasible region.

Solution. Question 32 and Question 38 show the counterexample: when the objective line is parallel to a boundary edge, the optimum occurs at both ends of that edge and at every point between them.

Question 44

State True or False: In a LPP, the minimum value of the objective function $Z = ax + by$ is always $0$ if origin is one of the corner point of the feasible region.

Solution. $Z(0, 0) = 0$ is only a candidate. If $a$ or $b$ is negative, another corner may give a smaller value. For example, with the region $x + y \le 1$, $x, y \ge 0$ and $Z = -x – y$, the origin is a corner but the minimum is $-1$, at $(1, 0)$ and $(0, 1)$. Question 27 is a case in point: the origin is a corner and the minimum is $-32$.

Question 45

State True or False: In a LPP, the maximum value of the objective function $Z = ax + by$ is always finite.

Solution. On a bounded region it is finite. On an unbounded region it may not be: for $Z = x + y$ on the region $x, y \ge 0$ the value grows without limit, and there is no maximum. This exercise already contains such cases: Question 25 and Question 31 have an unbounded region and no minimum of $3x – 4y$, and Questions 37 and 42 state that the optimum “may or may not exist”. The word “always” makes the statement false.

The printed key says True, and it is wrong. It treats the bounded case as the only case, which contradicts Questions 37 and 42 of the same exercise.

Common mistakes

• Trusting a corner value on an unbounded region in Questions 30 and 31. The smallest corner value is $-16$, but $F = -5t$ along $(t, 2t)$ is unbounded below. Always check the direction in which the region escapes.
• Missing the tie in Question 32. Two corners with the same optimal value mean a whole segment of optimal points; “(0, 2) only” and “(3, 0) only” are both too narrow.
• Assuming the origin gives the minimum in Question 44. It gives $0$, which is the minimum only when $a, b \ge 0$.
• Reading the figure in Question 30 literally. It is not to scale, and the line through $(6, 0)$ and $(12, 6)$ is drawn almost vertical. Use the corner labels, not the slopes.
• Equating the wrong pair of values in Question 34. The condition is $Z(3, 0) = Z(1, 1)$, not $Z(0, 3) = Z(3, 0)$.

Practise next

• Exemplar Short Answer and Long Answer, Questions 1 to 25, including Question 25, the unbounded region behind Questions 30 and 31.
• NCERT Class 12 Maths Exercise 12.1: linear programming by the corner-point method.