Exercise 12.1 — Linear Programming

1. 14 marksNCERT Exercise 12.1

   Solve the following Linear Programming Problem graphically: Maximise $Z=3x+4y$ subject to the constraints $x+y\le 4$, $x\ge 0$, $y\ge 0$.

   Hint. Draw x + y = 4, shade the side towards the origin, and read off the three corner points.

   The feasible region is the triangle cut off by $x+y\le 4$ in the first quadrant, so it is bounded and the Corner Point Method applies directly.

   The corner points are $O(0,0)$, $A(4,0)$ and $B(0,4)$.

   Corner point	$Z=3x+4y$
   $(0,0)$	$0$
   $(4,0)$	$12$
   $(0,4)$	$16$

   Since the region is bounded, the largest of these values is the maximum. The coefficient of $y$ is larger than that of $x$, which is why the optimum sits on the $y$-axis rather than the $x$-axis.

   ✦ Maximum Z = 16 at the point (0, 4)

2. 24 marksNCERT Exercise 12.1

   Solve the following Linear Programming Problem graphically: Minimise $Z=-3x+4y$ subject to $x+2y\le 8$, $3x+2y\le 12$, $x\ge 0$, $y\ge 0$.

   Hint. The two lines meet at (2,3). Because the objective has a negative coefficient on x, expect the minimum where x is largest.

   Both constraints are of $\le$ type with positive coefficients, so the feasible region is bounded and the Corner Point Method gives the answer.

   Solving $x+2y=8$ with $3x+2y=12$ gives $2x=4$, so $x=2$ and $y=3$.

   The corner points are $O(0,0)$, $A(4,0)$, $B(2,3)$ and $C(0,4)$.

   Corner point	$Z=-3x+4y$
   $(0,0)$	$0$
   $(4,0)$	$-12$
   $(2,3)$	$6$
   $(0,4)$	$16$

   The minimum is the most negative value, which occurs where $x$ is as large as possible and $y$ is zero, because the $x$ term carries the negative coefficient.

   ✦ Minimum Z = -12 at the point (4, 0)

3. 34 marksNCERT Exercise 12.1

   Solve the following Linear Programming Problem graphically: Maximise $Z=5x+3y$ subject to $3x+5y\le 15$, $5x+2y\le 10$, $x\ge 0$, $y\ge 0$.

   Hint. Solve 3x + 5y = 15 and 5x + 2y = 10 simultaneously; the intersection has fractional coordinates.

   Here the feasible region is bounded by two slanted lines together with the axes, so it is a bounded quadrilateral.

   To find where the lines cross, solve $3x+5y=15$ and $5x+2y=10$. Multiplying the first by $2$ and the second by $5$ gives $6x+10y=30$ and $25x+10y=50$; subtracting yields $19x=20$, so $x=\frac{20}{19}$ and then $y=\frac{45}{19}$.

   Corner point	$Z=5x+3y$
   $(0,0)$	$0$
   $(2,0)$	$10$
   $\left(\frac{20}{19},\frac{45}{19}\right)$	$\frac{235}{19}$
   $(0,3)$	$9$

   Since $\frac{235}{19}\approx 12.37$ exceeds both $10$ and $9$, the maximum is at the intersection point rather than at either axis intercept.

   ✦ Maximum Z = 235/19 at the point (20/19, 45/19)

4. 44 marksNCERT Exercise 12.1

   Solve the following Linear Programming Problem graphically: Minimise $Z=3x+5y$ such that $x+3y\ge 3$, $x+y\ge 2$, $x,y\ge 0$.

   Hint. Both constraints are of >= type, so the region is unbounded. After finding the smallest corner value, test whether 3x + 5y < that value has any feasible point.

   Both constraints point away from the origin, so the feasible region is unbounded and the smallest corner value must be confirmed before it can be called a minimum.

   Solving $x+3y=3$ with $x+y=2$ gives $2y=1$, so $y=\frac{1}{2}$ and $x=\frac{3}{2}$.

   Corner point	$Z=3x+5y$
   $(3,0)$	$9$
   $\left(\frac{3}{2},\frac{1}{2}\right)$	$7$
   $(0,2)$	$10$

   The smallest value is $m=7$. For an unbounded region we must check the open half plane $3x+5y<7$: it has no point in common with the feasible region, because every feasible point satisfies both $x+3y\ge 3$ and $x+y\ge 2$, which together force $3x+5y\ge 7$.

   Therefore $7$ is genuinely the minimum and not merely the smallest corner value.

   ✦ Minimum Z = 7 at the point (3/2, 1/2)

5. 54 marksNCERT Exercise 12.1

   Solve the following Linear Programming Problem graphically: Maximise $Z=3x+2y$ subject to $x+2y\le 10$, $3x+y\le 15$, $x,y\ge 0$.

   Hint. The two lines intersect at (4,3). The region is bounded, so simply compare the four corner values.

   With both constraints of $\le$ type the feasible region is a bounded quadrilateral, so no unboundedness check is needed.

   Solving $x+2y=10$ and $3x+y=15$: from the second, $y=15-3x$, and substituting gives $x+30-6x=10$, so $-5x=-20$ and $x=4$, hence $y=3$.

   Corner point	$Z=3x+2y$
   $(0,0)$	$0$
   $(5,0)$	$15$
   $(4,3)$	$18$
   $(0,5)$	$10$

   The largest value occurs at the intersection of the two constraint lines, which is typical when both coefficients of the objective are positive and the region is bounded.

   ✦ Maximum Z = 18 at the point (4, 3)

6. 65 marksNCERT Exercise 12.1

   Solve the following Linear Programming Problem graphically: Minimise $Z=x+2y$ subject to $2x+y\ge 3$, $x+2y\ge 6$, $x,y\ge 0$. Show that the minimum of $Z$ occurs at more than two points.

   Hint. Notice that the objective function x + 2y is identical to the left side of the second constraint.

   The key observation is that the objective function $Z=x+2y$ is exactly the expression appearing in the constraint $x+2y\ge 6$.

   That constraint therefore says $Z\ge 6$ directly, so no feasible point can give a value below $6$.

   The corner points of the unbounded feasible region are $(0,3)$ and $(6,0)$.

   Corner point	$Z=x+2y$
   $(0,3)$	$6$
   $(6,0)$	$6$

   Both corners give the same value $6$. By the result quoted in the textbook, when two corner points produce the same optimal value, every point on the line segment joining them also gives that value.

   So $Z=6$ at every point of the segment joining $(0,3)$ and $(6,0)$ — that is, at infinitely many points, which is what the question asks you to show. The half plane $x+2y<6$ meets the feasible region nowhere, confirming $6$ is the true minimum despite the region being unbounded.

   ✦ Minimum Z = 6, attained at every point on the segment joining (0,3) and (6,0)

7. 75 marksNCERT Exercise 12.1

   Solve the following Linear Programming Problem graphically: Minimise and Maximise $Z=5x+10y$ subject to $x+2y\le 120$, $x+y\ge 60$, $x-2y\ge 0$, $x,y\ge 0$.

   Hint. There are four corner points. Two of them give the same maximum value, so the maximum is attained along a whole edge.

   This problem asks for both extremes, so every corner value must be computed and then read twice.

   The binding intersections are: $x+y=60$ with $x-2y=0$ giving $(40,20)$; $x+2y=120$ with $x-2y=0$ giving $(60,30)$; and the axis points $(60,0)$ and $(120,0)$.

   Corner point	$Z=5x+10y$
   $(60,0)$	$300$
   $(120,0)$	$600$
   $(60,30)$	$600$
   $(40,20)$	$400$

   The region is bounded, so both extremes are attained. The minimum is $300$ at $(60,0)$.

   For the maximum, two distinct corner points $(120,0)$ and $(60,30)$ both give $600$, which means every point on the segment joining them is also optimal.

   ✦ Minimum Z = 300 at (60, 0); Maximum Z = 600 at every point on the segment joining (120, 0) and (60, 30)

8. 85 marksNCERT Exercise 12.1

   Solve the following Linear Programming Problem graphically: Minimise and Maximise $Z=x+2y$ subject to $x+2y\ge 100$, $2x-y\le 0$, $2x+y\le 200$, $x,y\ge 0$.

   Hint. Here it is the minimum that is attained at two corner points, and hence along the edge joining them.

   Reading both extremes again requires the full table, and this time it is the minimum that repeats.

   The corner points are $(0,50)$ and $(0,200)$ on the $y$-axis, together with $(20,40)$ from $x+2y=100$ meeting $2x-y=0$, and $(50,100)$ from $2x-y=0$ meeting $2x+y=200$.

   Corner point	$Z=x+2y$
   $(0,50)$	$100$
   $(20,40)$	$100$
   $(50,100)$	$250$
   $(0,200)$	$400$

   The maximum is $400$, attained uniquely at $(0,200)$.

   The minimum value $100$ appears at two different corner points, $(0,50)$ and $(20,40)$, so it is attained at every point on the segment joining them. This is the mirror image of the previous question, where the repetition happened at the maximum instead.

   ✦ Minimum Z = 100 at every point on the segment joining (0,50) and (20,40); Maximum Z = 400 at (0, 200)

9. 95 marksNCERT Exercise 12.1

   Solve the following Linear Programming Problem graphically: Maximise $Z=-x+2y$, subject to the constraints $x\ge 3$, $x+y\ge 5$, $x+2y\ge 6$, $y\ge 0$.

   Hint. All constraints are of >= type. Compute the largest corner value, then test the open half plane -x + 2y > that value against the feasible region.

   Every constraint here is of $\ge$ type, so the feasible region extends indefinitely upwards and the unboundedness test is essential rather than optional.

   The corner points are $(6,0)$, $(4,1)$ and $(3,2)$.

   Corner point	$Z=-x+2y$
   $(6,0)$	$-6$
   $(4,1)$	$-2$
   $(3,2)$	$1$

   The largest corner value is $M=1$. Now apply the test for an unbounded region: does the open half plane $-x+2y>1$ contain any feasible point?

   It does. Take $x=3$ and $y=100$, which satisfies $x\ge 3$, $x+y\ge 5$, $x+2y\ge 6$ and $y\ge 0$, yet gives $Z=-3+200=197$, far above $1$.

   Because the half plane has points in common with the feasible region, $Z$ has no maximum value — it can be made arbitrarily large by increasing $y$. Reporting $1$ as the answer is the trap this question is built around.

   ✦ Z has no maximum value; the feasible region is unbounded and Z increases without limit

10. 105 marksNCERT Exercise 12.1

    Solve the following Linear Programming Problem graphically: Maximise $Z=x+y$, subject to $x-y\le -1$, $-x+y\le 0$, $x,y\ge 0$.

    Hint. Rewrite both constraints as inequalities in y and compare them directly.

    Rather than plotting first, rewrite each constraint so that $y$ stands alone, which exposes the difficulty immediately.

    The first constraint $x-y\le -1$ rearranges to $y\ge x+1$.

    The second constraint $-x+y\le 0$ rearranges to $y\le x$.

    Taken together these demand $x+1\le y\le x$, which would require $x+1\le x$, that is $1\le 0$.

    That is impossible, so no point $(x,y)$ satisfies both constraints at once. Graphically the two shaded half planes lie on opposite sides of the line $y=x$ and never overlap.

    Since the feasible region is empty, there are no corner points to evaluate and the problem has no solution at all.

    ✦ There is no feasible region, so Z has no maximum value

Solutions written by the tuition.in editorial team and checked against lemh206.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.