NCERT Solutions›Class 12 Mathematics›Chapter 12

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Chapter 12 · Class 12 Mathematics

Linear Programming

3 exercises31 questions solved

Exercise 12.1 Exercise 12.2 Exercise Misc

Exercise 12.1Linear Programming Problems — Graphical Method

Maximise Z = 3x + 4y subject to constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.

Solution

Corner points of feasible region: O(0,0), A(4,0), B(0,4). Z(O)=0, Z(A)=12, Z(B)=16. Maximum Z=16 at (0,4).

Minimise Z = −3x + 4y subject to: x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Solution

Corner points: O(0,0), A(4,0), B(2,3), C(0,4). Z(O)=0, Z(A)=−12, Z(B)=−6+12=6, Z(C)=16. Minimum Z=−12 at (4,0).

Maximise Z = 5x + 3y subject to: 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

Solution

Find corner points. Intersection: 3x+5y=15 and 5x+2y=10. Solve: y=3(1−x)/5... From eqs: multiply: 6x+10y=30, 25x+10y=50. 19x=20→x=20/19, y=(15−60/19)/5=(285−60)/(95)=225/95=45/19. Z at (20/19, 45/19)=100/19+135/19=235/19≈12.37. Z(0,3)=9, Z(2,0)=10. Maximum Z=235/19 at (20/19, 45/19).

Minimise Z = 3x + 5y such that: x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0.

Solution

Corner points of feasible region (unbounded from above): intersection x+3y=3 and x+y=2: 2y=1→y=1/2, x=3/2. Corner points: A(3,0), B(3/2,1/2), C(0,2). Z(A)=9, Z(B)=9/2+5/2=7, Z(C)=10. Minimum Z=7 at (3/2,1/2). Check: no values < 7 in feasible region → minimum=7.

Maximise Z = 3x + 2y, subject to: x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, y ≥ 0.

Solution

Intersection: x+2y=10 and 3x+y=15: x=4,y=3. Corner points: O(0,0), A(5,0), B(4,3), C(0,5). Z: 0, 15, 18, 10. Maximum Z=18 at (4,3).

Minimise Z = x + 2y subject to: 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0.

Solution

Feasible region is unbounded below. Corner points: A(6,0), B(0,3). Intersection: 2x+y=3 and x+2y=6: 3x=0→x=0 and y=3; or y=0 and x=3 for first eq, x=6 for second. Corner points: (3,0) and (0,3): Z(3,0)=3, Z(0,3)=6. Check if minimum exists: draw Z=x+2y=3; if no feasible point satisfies x+2y<3 then Z_min=3 at (3,0). Need to check: 2(3)+0=6≥3 ✓ and 3+0=3 which is not ≥6 ✗. So (3,0) is not feasible. Redo: Intersection of 2x+y=3 and x+2y=6: multiply first by 2: 4x+2y=6. Subtract: 3x=0→x=0,y=3. Only corner point intersection at (0,3). And individual: (6,0) on x+2y=6. Z(6,0)=6, Z(0,3)=6. Minimum=6 at line segment from (6,0) to (0,3).

Minimise and Maximise Z = 5x + 10y subject to: x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 0, x ≥ 0, y ≥ 0.

Solution

Find feasible region. x=2y from x−2y=0. Corners: A(60,0), B(120,0), C(60,30), D(40,20). Z(A)=300, Z(B)=600, Z(C)=600, Z(D)=400. Min Z=300 at (60,0). Max Z=600 at (120,0) and (60,30) — multiple optimal.

Minimise and Maximise Z = x + 2y subject to: x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200, x ≥ 0, y ≥ 0.

Solution

Find corners: From 2x=y and x+2y=100: x+4x=100→x=20,y=40. From 2x=y and 2x+y=200: 2x+2x=200→x=50,y=100. Corner points: A(0,50), B(20,40), C(50,100), D(0,200). Z: A=100, B=100, C=250, D=400. Min=100 (at A and B, and along AB), Max=400 at (0,200).

Maximise Z = −x + 2y, subject to: x ≥ 3, x + y ≤ 5, x − y ≤ −1, y ≥ 0.

Solution

x≥3, x+y≤5, y−x≥1. Corners: (3,4): Z=−3+8=5; (4,1): check y−x=1−4=−3 not ≥1. Feasible: x=3,y≥4,y≤2? Contradiction unless: x+y≤5 and y≥x+1. At x=3: y≥4 and y≤2 → no feasible region. Check: x+y≤5 and x−y≤−1→y≥x+1. At x=3: y≥4 and y≤2 — empty! Feasible region may be empty or at boundary. NCERT answer: Z_max=1 is not achievable → unbounded or no solution.

Q10

Maximise Z = x + y, subject to: x − y ≤ −1, −x + y ≤ 0, x ≥ 0, y ≥ 0.

Solution

x−y≤−1→y≥x+1. −x+y≤0→y≤x. y≥x+1 and y≤x: impossible. Feasible region is empty. No solution.

Exercise 12.2Different Types of Linear Programming Problems

A manufacturer produces two types of steel trunks. A trunk of type I requires 3 machine hours and 1 carpenter hour; trunk II requires 3 machine hours and 2 carpenter hours. Machine hours available: 18, carpenter hours: 8. Type I gives profit ₹170, type II gives ₹300. How many of each should be made to maximise profit?

Solution

Let x=type I, y=type II. Maximise Z=170x+300y. Constraints: 3x+3y≤18 (x+y≤6), x+2y≤8, x≥0, y≥0. Corners: O(0,0), A(6,0), B(4,2), C(0,4). Z: 0, 1020, 1280, 1200. Maximum Z=₹1280 at x=4, y=2.

A factory produces two types of screws A and B. Each type A requires 2 machine minutes and 3 man-minutes; B requires 3 machine minutes and 2 man-minutes. Machine time available: 8 hours = 480 minutes; man time: 8 hours = 480 minutes. Profit: A→₹0.35, B→₹0.25. Formulate and solve.

Solution

Maximise Z=0.35x+0.25y. Constraints: 2x+3y≤480, 3x+2y≤480. Corners: (0,0),(160,0),(0,160),(96,96). Z at (96,96)=0.35(96)+0.25(96)=33.6+24=57.6. Z at (160,0)=56. Z at (0,160)=40. Maximum Z=₹57.60 at x=96 (A=96), y=96 (B=96).

A factory makes tennis rackets and cricket bats. Each tennis racket takes 1.5 hours of machine time and 3 hours of craftsman's time; bat takes 3 hours machine and 1 hour craftsman. Machine hours available: 42, craftsman hours: 24. Profit: racket ₹20, bat ₹10. Find optimal production.

Solution

Let x=rackets, y=bats. Z=20x+10y. Constraints: 1.5x+3y≤42 (x+2y≤28), 3x+y≤24. Intersection: x+2y=28 and 3x+y=24: 5x=20→x=4,y=12. Corners: (0,0),(8,0),(4,12),(0,14). Z: 0,160,200,140. Maximum Z=₹200 at x=4, y=12.

A manufacturer of a furniture company manufactures two types of tables X and Y. A table of type X requires 8 man-hours of carpentry and 2 man-hours of painting; type Y requires 5 man-hours carpentry and 5 man-hours painting. Total hours available: 80 carpentry, 20 painting. Profit: X→₹90, Y→₹72. Find optimal production.

Solution

Let x=X, y=Y. Z=90x+72y. Constraints: 8x+5y≤80, 2x+5y≤20. Corners: (0,0),(10,0),(0,4),(5/... wait: 8x+5y=80 and 2x+5y=20: 6x=60→x=10,y=−8 (not feasible). So intersection outside feasible region. Check corners of feasible region: (0,0),(10,0) from 8x≤80, and (0,4) from 5y≤20. Z: 0,900,288. Maximum Z=₹900 at x=10, y=0.

A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food I costs ₹2 per kg and contains 1 unit vitamin A and 2 units C; Food II costs ₹5/kg with 2 units A and 1 unit C. Minimise cost.

Solution

Let x=Food I (kg), y=Food II (kg). Minimise Z=2x+5y. Constraints: x+2y≥8, 2x+y≥10, x≥0, y≥0. Corner points: (0,10) from 2y=10... intersection: x+2y=8 and 2x+y=10: 3x=12→x=4,y=2. Corner (0,8): check 2(0)+8=8≥10? No. Correct corners: (0,10) on 2x+y=10, intersection (4,2), (8,0) on x+2y=8. Z: Z(0,10)=50, Z(4,2)=18, Z(8,0)=16. Minimum Z=₹16 at x=8, y=0.

A fruit grower can use two types of fertilizers. Brand P supplies 1.5 kg nitrogen and 3 kg phosphoric acid per bag; Brand Q supplies 3 kg nitrogen and 1.5 kg phosphoric acid per bag. Tests indicate at least 4.5 kg nitrogen and 9 kg phosphoric acid needed. Cost: P=₹100/bag, Q=₹200/bag. How many bags of each brand to minimize cost?

Solution

Let x=bags P, y=bags Q. Minimise Z=100x+200y. Constraints: 1.5x+3y≥4.5 (x+2y≥3), 3x+1.5y≥9 (2x+y≥6). Corners: (0,6) on 2x+y=6, intersection: x+2y=3 and 2x+y=6: 3x=9→x=3,y=0. Corner (3,0): check x+2y=3≥3 ✓, 2(3)=6≥6 ✓. Z(0,6)=1200, Z(3,0)=300. Minimum Z=₹300 at x=3, y=0.

An oil company has two depots A and B with capacities 7000L and 4000L. The company is to supply oil to three petrol pumps D, E, F with requirements 4500L, 3000L, 3500L respectively. Transportation costs per litre: A→D ₹1, A→E ₹2, A→F ₹3; B→D ₹3, B→E ₹2, B→F ₹1. Minimise transportation cost.

Solution

Let x litres from A to D, y litres from A to E. From A to F = 7000−x−y. From B to D=4500−x, B to E=3000−y, B to F=3500−(7000−x−y)=x+y−3500. Z=1(x)+2y+3(7000−x−y)+3(4500−x)+2(3000−y)+1(x+y−3500). Simplify: Z=x+2y+21000−3x−3y+13500−3x+6000−2y+x+y−3500. Z=−4x−2y+37000. Minimise: maximise 4x+2y subject to constraints. x≤4500, y≤3000, x+y≤7000, x+y≥3500, x≥0, y≥0. Max 4x+2y occurs at x=4500,y=3000 (if feasible: 7500>7000, not feasible). Try x=4000,y=3000: 7000 ✓. Z=−16000−6000+37000=15000. Or x=4500,y=2500: Z=−18000−5000+37000=14000. Or try corners systematically. Optimal: x=4500,y=2500 if feasible: 7000 ✓. Constraints: x≤4500 ✓, y≤3000 ✓. B to F=4500+2500−3500=3500 ✓, B to D=0 ✓. Min cost=₹14000.

A company produces two types of products A and B. Product A costs ₹5 and B costs ₹4. Availability: 120 man-hours and 40 machines hours. A requires 2 man-hours and 1 machine-hour; B requires 1 man-hour and 2 machine-hours. Selling prices: A→₹8, B→₹7. Find maximise profit.

Solution

Profit: A=3, B=3. Z=3x+3y=3(x+y). Maximise x+y subject to 2x+y≤120, x+2y≤40... wait: Constraints: 2x+y≤120, x+2y≤40. x+y max: Intersection: 2x+y=120 and x+2y=40: x−y=80, x=40+y; 3y=−40... not feasible in first quadrant with these numbers. Check: if B=2 machine hours means small availability. Corners: (0,20) from x+2y=40; (60,0) from 2x=120... (60,0): check x+2y=60≤40 ✗. So (0,20) and corner of 2x+y=120 and x=0: y=120 but x+2(120)=240≤40 ✗. Feasible region is {2x+y≤120, x+2y≤40, x,y≥0}. Corner (0,20): 2(0)+20=20≤120 ✓. (40,0): 2(40)=80≤120 ✓, 40+0=40≤40 ✓. Intersection: from 2x+y=120 and x+2y=40: subtract: x−y=80. y=x−80; x+2(x−80)=40; 3x=200; x=200/3 not integer. Z=3(x+y) max at... actually profit per unit: need to recalculate. At (40,0): Z=3(40)=120. At (0,20): Z=60. Max=120 at x=40, y=0.

A retired person wants to invest ₹20,000. Two types of bonds: A and B. Bond A has 7% interest, Bond B has 10% interest. Condition: amount in A must be at least ₹5000 more than amount in B. How to maximise interest?

Solution

Let x=in A, y=in B. Maximise Z=0.07x+0.1y. Constraints: x+y=20000, x−y≥5000, x≥0, y≥0. From x=20000−y: (20000−y)−y≥5000→2y≤15000→y≤7500. Z=0.07(20000−y)+0.1y=1400+0.03y. Maximise→maximise y→y=7500, x=12500. Max Z=1400+225=₹1625.

Q10

A man has ₹1,500 for purchase of rice and wheat. A bag of rice costs ₹180, bag of wheat costs ₹120. He has storage for at most 10 bags. He earns a profit of ₹11 per bag of rice and ₹8 per bag of wheat. Formulate and solve.

Solution

Let x=bags rice, y=bags wheat. Maximise Z=11x+8y. Constraints: 180x+120y≤1500 (3x+2y≤25), x+y≤10, x≥0, y≥0. Corners: (0,0),(8⅓,0)→(8,0) since integer but for LP: (25/3,0),(0,10),(0,12.5). Intersection: 3x+2y=25 and x+y=10: y=10−x; 3x+20−2x=25; x=5,y=5. Z at (5,5)=55+40=95. Z at (8.33,0)=91.67. Z at (0,10)=80. Maximum Z=₹95 at x=5, y=5.

Q11

An aeroplane can carry maximum 200 passengers. A profit of ₹400 is made on each first class ticket and ₹300 on each economy class ticket. The airline reserves at least 20 seats for first class. However at least 4 times as many passengers prefer economy class to first class. Determine how many tickets of each type must be sold to maximise profit.

Solution

Let x=first class, y=economy. Maximise Z=400x+300y. Constraints: x+y≤200, x≥20, y≥4x. Corners: (20,80) from y=4x and x=20; (20,180) from x=20,x+y=200; (40,160) from y=4x and x+y=200. Z(20,80)=8000+24000=32000. Z(20,180)=8000+54000=62000. Z(40,160)=16000+48000=64000. Maximum Z=₹64000 at x=40, y=160.

Exercise MiscMiscellaneous Exercise

The corner points of the feasible region for an LPP are (0,2), (3,0), (6,0), (6,8), (0,5). Find the minimum and maximum values of Z = 4x + 6y.

Solution

Z(0,2)=12, Z(3,0)=12, Z(6,0)=24, Z(6,8)=72, Z(0,5)=30. Min Z=12 at (0,2) and (3,0). Max Z=72 at (6,8).

For the LPP: Maximise Z = 2x + 3y subject to x + y ≤ 4, x + 3y ≤ 6, x ≥ 0, y ≥ 0. Find the maximum value.

Solution

Corners: (0,0),(4,0),(3,1),(0,2). Z: 0,8,9,6. Max=9 at (3,1).

If the corner points of an LPP are (0,0), (0,8), (4,10), (6,8), (6,0), find the maximum value of Z = 3x + 2y.

Solution

Z(0,0)=0, Z(0,8)=16, Z(4,10)=32, Z(6,8)=34, Z(6,0)=18. Max=34 at (6,8).

Solve: Maximise Z = 3x + 5y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

Solution

Intersection: 15x+25y=75 and 25x+10y=50. Mult 1st by 2: 30x+50y=150, mult 2nd by 5: 75x+10y=... Better: 3x+5y=15 and 5x+2y=10: from first y=(15−3x)/5; 5x+2(15−3x)/5=10; 25x+30−6x=50; 19x=20; x=20/19, y=45/19. Z=60/19+225/19=285/19=15 (approx). Check: Z=3(20/19)+5(45/19)=(60+225)/19=285/19≈15. Also check (0,3): Z=15, (2,0): Z=6. Max=285/19 at (20/19,45/19) or Z=15 at (0,3) — since (20/19,45/19) gives 285/19≈15 too? Actually 285/19=15. So Z=15 at (0,3) and (20/19,45/19).

Determine the maximum value of Z = 3x + 4y, if the feasible region of an LPP is as shown in the figure (with corners at (0,0),(0,4),(2,3),(3,0)).

Solution

Z(0,0)=0, Z(0,4)=16, Z(2,3)=18, Z(3,0)=9. Maximum Z=18 at (2,3).

If the feasible region for a maximisation LPP is unbounded, does it always have an optimal solution? Explain with an example.

Solution

No. If the feasible region is unbounded and the objective function increases in the direction of the unbounded region, there is no maximum. Example: Maximise Z = x + y subject to x + y ≥ 2, x ≥ 0, y ≥ 0. The feasible region is unbounded and Z can be made arbitrarily large — no maximum exists.

In question 5, if the objective is to Minimise Z = 3x + 4y over the same feasible region, find the answer.

Solution

Using corner points: Z(0,0)=0. Minimum Z=0 at origin (assuming origin is in feasible region).

Minimise Z = 13x − 15y subject to x + y ≤ 7, 2x − 3y + 6 ≥ 0, x ≥ 0, y ≥ 0.

Solution

2x−3y≥−6. Corners: (0,0),(7,0),(0,2) from 2x=3y−6 at x=0: y=2; (3,4): intersection x+y=7 and 2x−3y=−6→2(7−y)−3y=−6→14−5y=−6→y=4,x=3. Z(0,0)=0, Z(7,0)=91, Z(3,4)=39−60=−21, Z(0,2)=0−30=−30. Minimum Z=−30 at (0,2).

Two factories manufacture 3 electrical items — fans, electric irons and mixers. Factory A produces 50 fans, 25 irons and 10 mixers per day; Factory B: 10 fans, 25 irons and 35 mixers per day. To complete an order: 2500 fans, 3000 irons and 1250 mixers needed. Daily cost: A=₹12,500, B=₹15,000. Minimize cost.

Solution

Let x=days of A, y=days of B. Minimise Z=12500x+15000y. Constraints: 50x+10y≥2500 (5x+y≥250), 25x+25y≥3000 (x+y≥120), 10x+35y≥1250 (2x+7y≥250). Corner points: Intersection of 5x+y=250 and x+y=120: 4x=130→x=32.5,y=87.5. Z=12500(32.5)+15000(87.5)=406250+1312500=1718750. Intersection of x+y=120 and 2x+7y=250: 5y=10→y=2,x=118. Z=12500(118)+15000(2)=1475000+30000=1505000. Check all constraints: 5(118)+2=592≥250 ✓, 118+2=120 ✓, 236+14=250 ✓. Min Z=₹15,05,000 at x=118, y=2.

Q10

A manufacturer has three machines I, II and III. Machines I and II can operate 8 hours each, Machine III for 12 hours. Product A requires 1 hour each from I, II and requires 1 hour from III. Product B requires 1 hr from I, II and 2 hrs from III. Profit: A=₹600, B=₹400. Maximise profit.

Solution

Let x=units A, y=units B. Maximise Z=600x+400y. Constraints: x+y≤8 (M-I), x+y≤8 (M-II), x+2y≤12 (M-III). Active constraints: x+y≤8 and x+2y≤12. Corners: (0,0),(8,0),(4,4),(0,6). Z: 0, 4800, 4000, 2400. Maximum Z=₹4800 at x=8, y=0.

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