NCERT Exemplar Solutions Class 12 Maths Linear Programming — Short Answer and Long Answer Q1–25

NCERT Exemplar Problems, Class 12 Mathematics — Linear Programming, Short Answer questions 1 to 15 and Long Answer questions 16 to 25. All 25 questions solved.

Every question here is solved by the corner-point method: draw the feasible region, list its corner points, evaluate the objective $Z$ at each, and read off the optimum. Questions 1 to 10 are the warm-up, with the region either given as inequalities or drawn. Questions 11 to 15 only ask you to write the problem as a linear programming problem (LPP). Questions 16 to 20 then solve the problems formulated in 11 to 15, and Questions 21 to 25 are fresh problems, two of which have unbounded regions.

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: problems already written as inequalities, solved by the corner-point (graphical) method. It still explains formulation in section 12.2.1, but the word-problem exercises (manufacturing, diet, transport) were removed. On this exemplar unit that means Questions 11 to 20 and 22 to 24 are word problems beyond the current exercises; the method is the same.

Key insight. If a linear objective has an optimum on a feasible region, it occurs at a corner point. For a bounded region the maximum and minimum both exist and sit at corners. For an unbounded region, a corner-point value is only a candidate: confirm it by showing that no feasible point does better (Questions 9, 23, 24), or find a direction along which $Z$ runs off (Question 25).

Question 1

Determine the maximum value of $Z=11x+7y$ subject to the constraints $2x+y\le6$, $x\le2$, $x\ge0$, $y\ge0$.

Solution. The boundary lines are $2x+y=6$ and $x=2$, with the feasible side towards the origin for each, and the axes. The region is the quadrilateral bounded by the axes, $x=2$ and $2x+y=6$. Its corner points are the origin, $(2,0)$, $(2,2)$ (where $x=2$ meets $2x+y=6$) and $(0,6)$.

Corner	$Z=11x+7y$
$(0,0)$	$0$
$(2,0)$	$22$
$(2,2)$	$36$
$(0,6)$	$42$

The region is bounded, so the largest value is at a corner.

Question 2

Maximise $Z=3x+4y$ subject to $x+y\le1$, $x\ge0$, $y\ge0$.

Solution. The region is the triangle with corners $(0,0)$, $(1,0)$ and $(0,1)$, on the origin’s side of $x+y=1$.

Corner	$Z=3x+4y$
$(0,0)$	$0$
$(1,0)$	$3$
$(0,1)$	$4$

Question 3

Maximise $Z=11x+7y$ subject to $x\le3$, $y\le2$, $x\ge0$, $y\ge0$.

Solution. The region is the rectangle $0\le x\le3$, $0\le y\le2$, with corners $(0,0)$, $(3,0)$, $(3,2)$ and $(0,2)$.

Corner	$Z=11x+7y$
$(0,0)$	$0$
$(3,0)$	$33$
$(3,2)$	$47$
$(0,2)$	$14$

Question 4

Minimise $Z=13x-15y$ subject to $x+y\le7$, $2x-3y+6\ge0$, $x\ge0$, $y\ge0$.

Solution. The line $x+y=7$ has the origin on the feasible side. The line $2x-3y+6=0$ crosses the axes at $(-3,0)$ and $(0,2)$; the origin gives $6\ge0$, so the feasible side again contains the origin. The two lines meet where $y=\tfrac{2x+6}{3}$ and $x+y=7$: $3x+2x+6=21$, so $x=3$, $y=4$. The corner points are $(0,0)$, $(7,0)$, $(3,4)$ and $(0,2)$.

Corner	$Z=13x-15y$
$(0,0)$	$0$
$(7,0)$	$91$
$(3,4)$	$-21$
$(0,2)$	$-30$

Question 5

Determine the maximum value of $Z=3x+4y$ if the feasible region (shaded) for an LPP is shown in Fig. 12.7.

Solution. The figure shows two lines, $2x+y=104$ and $x+2y=76$, and the shaded region lies between the axes and these lines, to the left of their intersection. Its labelled corners are $O$, $A$ on the $x$-axis, $E$ where the two lines meet, and $D$ on the $y$-axis. The intercepts give $A=(52,0)$ (from $2x+y=104$) and $D=(0,38)$ (from $x+2y=76$). For $E$, solve the two lines together: doubling the second gives $2x+4y=152$, and subtracting $2x+y=104$ gives $3y=48$, so $y=16$ and $x=44$.

Corner	$Z=3x+4y$
$O(0,0)$	$0$
$A(52,0)$	$156$
$E(44,16)$	$196$
$D(0,38)$	$152$

Question 6

The feasible region (shaded) for an LPP is shown in Fig. 12.8. Maximise $Z=5x+7y$.

Solution. The figure shows a quadrilateral with corners $O(0,0)$, $A(7,0)$ on the $x$-axis, $B(3,4)$ at the top and $(0,2)$ on the $y$-axis. Read the corners straight from the labels.

Corner	$Z=5x+7y$
$(0,0)$	$0$
$(7,0)$	$35$
$(3,4)$	$43$
$(0,2)$	$14$

Question 7

The feasible region for an LPP is shown in Fig. 12.9. Find the minimum value of $Z=11x+7y$.

Solution. The figure shows the lines $x+y=5$ and $x+3y=9$. The shaded region is a small triangle against the $y$-axis, above $x+3y=9$ and below $x+y=5$. Its corners are the $y$-intercepts $(0,3)$ (of $x+3y=9$) and $(0,5)$ (of $x+y=5$), and the point where the lines meet: subtracting gives $2y=4$, so $y=2$, $x=3$, that is $(3,2)$.

Corner	$Z=11x+7y$
$(0,3)$	$21$
$(0,5)$	$35$
$(3,2)$	$47$

Question 8

Refer to Question 7 above. Find the maximum value of $Z$.

Solution. The region is the same bounded triangle, so the table above gives the largest value directly: $47$ at $(3,2)$.

Question 9

The feasible region for an LPP is shown in Fig. 12.10. Evaluate $Z=4x+y$ at each of the corner points of this region. Find the minimum value of $Z$, if it exists.

Solution. The figure shows the lines $x+y=3$ and $x+2y=4$, with arrows showing that the feasible side is away from the origin (above both lines and to the right of the $y$-axis). The region is therefore $x+y\ge3$, $x+2y\ge4$, $x\ge0$, $y\ge0$, and it is unbounded. The corner points are $(0,3)$ on $x+y=3$, $(4,0)$ on $x+2y=4$, and their intersection: subtracting the equations gives $y=1$, $x=2$, so $(2,1)$.

Corner	$Z=4x+y$
$(0,3)$	$3$
$(2,1)$	$9$
$(4,0)$	$16$

The smallest corner value is $3$, but the region is unbounded, so it needs checking. The open half-plane $4x+y<3$ would have to contain a feasible point. It cannot: subtracting $x+y\ge3$ from $4x+y<3$ gives $3x<0$, contradicting $x\ge0$. So no feasible point has $Z<3$.

Question 10

In Fig. 12.11, the feasible region (shaded) for an LPP is shown. Determine the maximum and minimum value of $Z=x+2y$.

Solution. The figure shows a shaded quadrilateral with labelled corners $P\!\left(\tfrac{3}{13},\tfrac{24}{13}\right)$, $Q\!\left(\tfrac32,\tfrac{15}{4}\right)$, $R\!\left(\tfrac72,\tfrac34\right)$ and $S\!\left(\tfrac{18}{7},\tfrac27\right)$. The region is bounded, so the extremes are at the corners.

Corner	$Z=x+2y$
$P\left(\tfrac{3}{13},\tfrac{24}{13}\right)$	$\tfrac{3}{13}+\tfrac{48}{13}=\tfrac{51}{13}$
$Q\left(\tfrac32,\tfrac{15}{4}\right)$	$\tfrac32+\tfrac{15}{2}=9$
$R\left(\tfrac72,\tfrac34\right)$	$\tfrac72+\tfrac32=5$
$S\left(\tfrac{18}{7},\tfrac27\right)$	$\tfrac{18}{7}+\tfrac47=\tfrac{22}{7}$

Since $\tfrac{22}{7}\approx3.14<\tfrac{51}{13}\approx3.92<5<9$, the maximum is at $Q$ and the minimum at $S$.

Question 11

A manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors and 150 capacitors and is required to produce two types of circuits A and B. Type A requires 20 resistors, 10 transistors and 10 capacitors. Type B requires 10 resistors, 20 transistors and 30 capacitors. If the profit on type A circuit is Rs 50 and that on type B circuit is Rs 60, formulate this problem as a LPP so that the manufacturer can maximise his profit.

Solution. Let the manufacturer make $x$ circuits of Type A and $y$ circuits of Type B. Each component gives one constraint, “used $\le$ stock”:

• resistors: $20x+10y\le200$, that is $2x+y\le20$;
• transistors: $10x+20y\le120$, that is $x+2y\le12$;
• capacitors: $10x+30y\le150$, that is $x+3y\le15$;
• $x,y\ge0$, since a number of circuits cannot be negative.

The profit is $Z=50x+60y$.

Question 12

A firm has to transport 1200 packages using large vans which can carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is Rs 400 and each small van is Rs 200. Not more than Rs 3000 is to be spent on the job and the number of large vans can not exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimise cost.

Solution. Let $x$ be the number of large vans and $y$ the number of small vans. The vans must carry at least 1200 packages: $200x+80y\ge1200$, that is $5x+2y\ge30$. The budget gives $400x+200y\le3000$, that is $2x+y\le15$. Large vans cannot exceed small vans: $x\le y$. And $x,y\ge0$. The cost to minimise is $Z=400x+200y$.

Question 13

A company manufactures two types of screws A and B. All the screws have to pass through a threading machine and a slotting machine. A box of Type A screws requires 2 minutes on the threading machine and 3 minutes on the slotting machine. A box of type B screws requires 8 minutes of threading on the threading machine and 2 minutes on the slotting machine. In a week, each machine is available for 60 hours. On selling these screws, the company gets a profit of Rs 100 per box on type A screws and Rs 170 per box on type B screws. Formulate this problem as a LPP given that the objective is to maximise profit.

Solution. Let $x$ boxes of Type A and $y$ boxes of Type B be made in a week. The machine time is given in minutes per box but the availability in hours, so convert: $60$ hours $=3600$ minutes. Threading: $2x+8y\le3600$, that is $x+4y\le1800$. Slotting: $3x+2y\le3600$. Also $x,y\ge0$. The profit is $Z=100x+170y$.

Question 14

A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100. The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B. Formulate this problem as a LPP to maximise the profit to the company.

Solution. Let $x$ sweaters of type A and $y$ of type B be made in a day. The total count gives $x+y\le300$. The spending gives $360x+120y\le72000$, that is $3x+y\le600$. “Type B cannot exceed type A by more than 100” means $y-x\le100$. Also $x,y\ge0$. The profit is $Z=200x+120y$.

Question 15

A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has atmost Rs 120 to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.

Solution. Let him ride $x$ km at 50 km/hour and $y$ km at 80 km/hour. The distance to maximise is $Z=x+y$. Petrol costs $2x+3y\le120$. The time is $\tfrac{x}{50}+\tfrac{y}{80}\le1$; multiplying by $400$ gives $8x+5y\le400$. Also $x,y\ge0$.

Question 16

Refer to Question 11. How many circuits of Type A and of Type B should be produced by the manufacturer so as to maximise his profit? Determine the maximum profit.

Solution. Maximise $Z=50x+60y$ subject to $2x+y\le20$, $x+2y\le12$, $x+3y\le15$, $x,y\ge0$. The origin satisfies every constraint, so the feasible side of each line is towards the origin. The corner points are:

• $(0,0)$;
• $(10,0)$, where $2x+y=20$ meets the $x$-axis;
• $\left(\tfrac{28}{3},\tfrac43\right)$, where $2x+y=20$ meets $x+2y=12$ (subtract $2\times$ the second from the first: $-3y=-4$);
• $(6,3)$, where $x+2y=12$ meets $x+3y=15$;
• $(0,5)$, where $x+3y=15$ meets the $y$-axis.

Corner	$Z=50x+60y$
$(0,0)$	$0$
$(10,0)$	$500$
$\left(\tfrac{28}{3},\tfrac43\right)$	$\tfrac{1640}{3}\approx546.67$
$(6,3)$	$480$
$(0,5)$	$300$

The largest value of $Z$ is at $\left(\tfrac{28}{3},\tfrac43\right)$, but circuits come in whole numbers, so $9\tfrac13$ and $1\tfrac13$ circuits cannot be produced. Test the whole-number points near the optimum, which must satisfy the three constraints. $(9,1)$ gives $510$, $(10,0)$ gives $500$, and $(8,2)$ satisfies $18\le20$, $12\le12$, $14\le15$ with $Z=400+120=520$. No whole-number feasible point does better than $520$, so the best plan is $8$ circuits of Type A and $2$ of Type B.

The key prints $6$ of Type A, $3$ of Type B and Rs $480$. That is the corner $(6,3)$, which is feasible but is not the maximum on any reading: the corner-point value $\tfrac{1640}{3}$ is larger, and so is the whole-number plan $(8,2)$ with Rs $520$. The key’s answer is incorrect.

Question 17

Refer to Question 12. What will be the minimum cost?

Solution. Minimise $Z=400x+200y$ subject to $5x+2y\ge30$, $2x+y\le15$, $x\le y$, $x,y\ge0$. The first constraint is feasible away from the origin, the other two towards the origin side. The region is bounded, with corner points $(0,15)$, $\left(\tfrac{30}{7},\tfrac{30}{7}\right)$ (where $5x+2y=30$ meets $x=y$) and $(5,5)$ (where $2x+y=15$ meets $x=y$).

Corner	$Z=400x+200y$
$(0,15)$	$3000$
$\left(\tfrac{30}{7},\tfrac{30}{7}\right)$	$\tfrac{18000}{7}\approx2571.43$
$(5,5)$	$3000$

The corner-point minimum is $\tfrac{18000}{7}\approx2571.43$, which agrees with the key. It needs $\tfrac{30}{7}\approx4.29$ vans of each type, though, and vans are whole. Searching whole-number points on the feasible side: $(4,5)$ carries $800+400=1200$ packages, satisfies $x\le y$ and $2x+y=13\le15$, and costs $1600+1000=2600$. Neighbouring whole points ($(3,8)$ at Rs 2800, $(5,5)$ at Rs 3000) cost more, so with whole vans the least cost is Rs $2600$, using 4 large and 5 small vans.

Question 18

Refer to Question 13. Solve the linear programming problem and determine the maximum profit to the manufacturer.

Solution. Maximise $Z=100x+170y$ subject to $x+4y\le1800$, $3x+2y\le3600$, $x,y\ge0$. Both lines have the origin on the feasible side. The corner points are $(0,0)$, $(1200,0)$ (on $3x+2y=3600$), $(0,450)$ (on $x+4y=1800$), and the intersection: from $x=1800-4y$, $5400-12y+2y=3600$, so $y=180$ and $x=1080$.

Corner	$Z=100x+170y$
$(0,0)$	$0$
$(1200,0)$	$120000$
$(1080,180)$	$138600$
$(0,450)$	$76500$

Question 19

Refer to Question 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?

Solution. Maximise $Z=200x+120y$ subject to $x+y\le300$, $3x+y\le600$, $y-x\le100$, $x,y\ge0$. The feasible region has corners $(0,0)$, $(200,0)$ (on $3x+y=600$), $(150,150)$ (where $x+y=300$ meets $3x+y=600$: subtract to get $2x=300$), $(100,200)$ (where $x+y=300$ meets $y=x+100$) and $(0,100)$ (on $y=x+100$).

Corner	$Z=200x+120y$
$(0,0)$	$0$
$(200,0)$	$40000$
$(150,150)$	$48000$
$(100,200)$	$44000$
$(0,100)$	$12000$

Question 20

Refer to Question 15. Determine the maximum distance that the man can travel.

Solution. Maximise $Z=x+y$ subject to $2x+3y\le120$, $8x+5y\le400$, $x,y\ge0$. The corner points are $(0,0)$, $(50,0)$ (on $8x+5y=400$), $(0,40)$ (on $2x+3y=120$), and the intersection: from $x=\tfrac{120-3y}{2}$, $4(120-3y)+5y=400$, so $480-7y=400$, giving $y=\tfrac{80}{7}$ and $x=\tfrac{300}{7}$.

Corner	$Z=x+y$
$(0,0)$	$0$
$(50,0)$	$50$
$\left(\tfrac{300}{7},\tfrac{80}{7}\right)$	$\tfrac{380}{7}$
$(0,40)$	$40$

Since $\tfrac{380}{7}=54\tfrac27>50$, the maximum is at the intersection: $\tfrac{300}{7}$ km at 50 km/hour and $\tfrac{80}{7}$ km at 80 km/hour. Distances need not be whole, so no rounding is needed here.

Question 21

Maximise $Z=x+y$ subject to $x+4y\le8$, $2x+3y\le12$, $3x+y\le9$, $x\ge0$, $y\ge0$.

Solution. All three lines have the origin on the feasible side. The corner points are $(0,0)$, $(3,0)$ (on $3x+y=9$), $(0,2)$ (on $x+4y=8$), and the intersection of $x+4y=8$ with $3x+y=9$: from $x=8-4y$, $24-12y+y=9$, so $y=\tfrac{15}{11}$ and $x=\tfrac{28}{11}$. At this point $2x+3y=\tfrac{101}{11}<12$, and at $(3,0)$ and $(0,2)$ the value is $6<12$, so the line $2x+3y=12$ never touches the region; it is redundant.

Corner	$Z=x+y$
$(0,0)$	$0$
$(3,0)$	$3$
$\left(\tfrac{28}{11},\tfrac{15}{11}\right)$	$\tfrac{43}{11}$
$(0,2)$	$2$

Question 22

A manufacturer produces two Models of bikes, Model X and Model Y. Model X takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.

Solution. Let $x$ bikes of Model X and $y$ of Model Y be made. Man-hours: $6x+10y\le450$, that is $3x+5y\le225$. Funds: $2000x+1000y\le80000$, that is $2x+y\le80$. Also $x,y\ge0$. Maximise $Z=1000x+500y$. The corners are $(0,0)$, $(40,0)$ (on $2x+y=80$), $(0,45)$ (on $3x+5y=225$) and the intersection: from $y=80-2x$, $3x+400-10x=225$, so $x=25$ and $y=30$.

Corner	$Z=1000x+500y$
$(0,0)$	$0$
$(40,0)$	$40000$
$(25,30)$	$40000$
$(0,45)$	$22500$

Two corners tie. That is because $Z=1000x+500y=500(2x+y)$ is parallel to the funds line $2x+y=80$, so $Z=40000$ at every point of the segment joining $(25,30)$ and $(40,0)$. The key gives $25$ of Model X and $30$ of Model Y; $40$ of Model X and none of Model Y is equally good.

Question 23

In order to supplement daily diet, a person wishes to take some X and some Y tablets. The contents of iron, calcium and vitamins in X and Y (in milligrams per tablet) are: X has 6 of iron, 3 of calcium and 2 of vitamin; Y has 2 of iron, 3 of calcium and 4 of vitamin. The person needs at least 18 milligrams of iron, 21 milligrams of calcium and 16 milligrams of vitamins. The price of each tablet of X and Y is Rs 2 and Re 1 respectively. How many tablets of each should the person take in order to satisfy the above requirement at the minimum cost?

Solution. Let $x$ tablets of X and $y$ of Y be taken. Each nutrient needs “at least”: iron $6x+2y\ge18$, that is $3x+y\ge9$; calcium $3x+3y\ge21$, that is $x+y\ge7$; vitamins $2x+4y\ge16$, that is $x+2y\ge8$; and $x,y\ge0$. Minimise $Z=2x+y$. All three constraints are feasible away from the origin, so the region is unbounded. Its corner points are $(0,9)$ (where $3x+y=9$ meets the $y$-axis), $(1,6)$ (where $3x+y=9$ meets $x+y=7$), $(6,1)$ (where $x+y=7$ meets $x+2y=8$) and $(8,0)$ (where $x+2y=8$ meets the $x$-axis).

Corner	$Z=2x+y$
$(0,9)$	$9$
$(1,6)$	$8$
$(6,1)$	$13$
$(8,0)$	$16$

For an unbounded region, the corner minimum $8$ is the true minimum only if $Z<8$ has no feasible point. Here $2x+y=\tfrac12(3x+y)+\tfrac12(x+y)\ge\tfrac92+\tfrac72=8$ at every feasible point, so no point has $Z<8$.

Question 24

A company makes 3 models of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculators of model B and 4800 calculators of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made every day; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made every day. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.

Solution. Let factory I run for $x$ days and factory II for $y$ days. The orders give: model A $50x+40y\ge6400$, that is $5x+4y\ge640$; model B $50x+20y\ge4000$, that is $5x+2y\ge400$; model C $30x+40y\ge4800$, that is $3x+4y\ge480$; and $x,y\ge0$. Minimise $Z=12000x+15000y$. The region lies away from the origin and is unbounded, with corner points $(0,200)$ (where $5x+2y=400$ meets the $y$-axis), $(32,120)$ (where $5x+2y=400$ meets $5x+4y=640$), $(80,60)$ (where $5x+4y=640$ meets $3x+4y=480$) and $(160,0)$ (where $3x+4y=480$ meets the $x$-axis).

Corner	$Z=12000x+15000y$
$(0,200)$	$3000000$
$(32,120)$	$2184000$
$(80,60)$	$1860000$
$(160,0)$	$1920000$

The smallest corner value is $1860000$ at $(80,60)$. To confirm it on this unbounded region, write $12000x+15000y=375(5x+4y)+3375(3x+4y)\ge375\cdot640+3375\cdot480=1860000$ at every feasible point, so $Z$ cannot go lower.

Question 25

Maximise and minimise $Z=3x-4y$ subject to $x-2y\le0$, $-3x+y\le4$, $x-y\le6$, $x,y\ge0$.

Solution. The three lines are $x-2y=0$ (through the origin and $(12,6)$), $-3x+y=4$ (through $(0,4)$) and $x-y=6$ (through $(6,0)$ and $(12,6)$). The feasible side of $x-2y\le0$ is the side with $y\ge\tfrac x2$ (test $(0,1)$); of $-3x+y\le4$ the side with $y\le4+3x$ (test the origin); and of $x-y\le6$ the side with $y\ge x-6$ (test the origin). The corner points are $(0,0)$, $(0,4)$ and $(12,6)$, where $x-2y=0$ meets $x-y=6$. The region extends to infinity between the rays above $y=\tfrac x2$ and below $y=4+3x$, so it is unbounded.

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

Maximum: in the region $y\ge\tfrac x2$ and $y\ge x-6$, so $Z=3x-4y\le3x-2x=x$ and $Z\le3x-4(x-6)=24-x$. Therefore $Z\le\min(x,24-x)\le12$, and $12$ is attained at $(12,6)$. Minimum: the points $(t,2t)$ with $t\ge0$ all lie in the region (check $t-4t\le4$ and $t-2t\le6$), and on them $Z=3t-8t=-5t$, which falls without limit as $t$ grows. So $Z$ has no minimum.

Common mistakes

• Taking the first corner that looks right. In Question 16 the corner $(6,3)$ is feasible and easy to find, but the maximum is at the other corner $\left(\tfrac{28}{3},\tfrac43\right)$. Always evaluate $Z$ at every corner, and re-check each corner against all the constraints.
• Treating the corner-point answer as final for whole-number quantities. Questions 16 and 17 give fractional circuits and vans at the exact LP optimum. If the quantity must be whole, say so and test the nearby whole points that stay feasible, as done above.
• Believing a corner minimum on an unbounded region. For Questions 9, 23 and 24 the corner value is a candidate only; show that $Z$ cannot go lower, by a combination of the constraints or by the half-plane $Z<m$ missing the region. In Question 25 the minimum does not exist at all.
• Forgetting to convert units in the formulation. In Question 13 the machine time is in minutes and the availability in hours; $60$ hours is $3600$ minutes, not $60$.
• Misreading “cannot exceed by more than”. In Question 14, “type B cannot exceed type A by more than 100” is $y-x\le100$, not $y\le100$.
• Stopping at the first tied corner. In Question 22 two corners give $40000$ because $Z$ is parallel to a constraint line; every point on the segment between them is optimal.

Practise next

• Objective, Fill in the Blanks and True or False, Questions 26 to 45
• NCERT Class 12 Maths Exercise 12.1: linear programming by the corner-point method