NCERT Grade 12-Linear Programming-Answers

NCERT Solutions for Class 12 Maths

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1.    As x ≥ 0, y ≥ 0, therefore we shall shade the other inequalities in the first quadrant only.

Now x + y ≤ 4

Thus the line has 4 and 4 as intercepts along the axes. Now, (0, 0) satisfies the inequation, i.e., 0 + 0 ≤ 4. Therefore, shaded region OAB is the feasible solution.

Its corners are O (0, 0), A (4, 0), B (0, 4)

At O (0, 0) Z = 0

At A (4, 0) Z = 3 x 4 = 12

At B (0, 4) Z = 4 x 4 = 16

Hence, max Z = 16 at x = 0, y = 4.

2.    Consider x + 2y ≤ 8

Let x + 2y = 8

Since, (0, 0) satisfies the inequaitons x + 2y ≤ 8

Therefore, its solution contains (0, 0)

Again 3x + 2y ≤ 12

Let 3x + 2y = 12

Again, (0, 0) satisfies 3x + 2y ≤ 12

Therefore its solution contains (0, 0).

The feasible region is the solution set which is double shaded and is OABCO.

At O (0, 0) Z = 0

At A (4, 0) Z = –3 x 4 = –12

At B (2, 3) Z = –3 x 2 + 4 x 3 = 6

At C (0, 4) Z = 4 x 4 = 16

Hence, minimum Z = –12 at x = 4, y = 0.

3.    We first draw the graph of equation 3x + 5y = 15

For y = 3, x = 0

And for y = 0, x = 0

Similarly, for equation 5x + 2y = 10, the points are (2, 0) and (0, 5).

Similarly, for equation x ≥ 0, y ≥ 0, the points are (2, 0) and (0, 5).

As (0, 0) satisfies both the inequations and also then the feasible require contains the half-plane containing (0, 0).

Therefore, the feasible portion is OABC which is shown as shaded in the graph.

4.    For plotting the graphs of x + 3y = 3 and x + y = 2, we have the following tables:

The feasible portion represented by the inequalities

x + 3y ≥ , x + y ≥ 2 and x, y ≥ 0 is ABC which is shaded

5.    Consider x + 2y ≤ 10

Since, (0, 0) satisfies the inequation, therefore the half plane containing (0, 0) is the required plane.

Again 3x + 2y ≤ 15

Let 3x + y = 15

It also satisfies by (0, 0) and its required half plane contains (0, 0).

Now double shaded region in the first quadrant contains the solution.

Now OABC represents the feasible region.

Z = 3x + 2y

At O (0, 0) Z = 3 x 0 + 2 x 0 = 0

At A (5, 0) Z = 3 x 5 + 2 x 0 = 15

At B (4, 3) Z = 3 x 4 + 2 x 3 = 18

At C (0, 5) Z = 3 x 0 + 2 x 5 = 10

Hence, Z is maximum i.e., 18 at x = 4, y = 3.

6.    Consider 2x + y ≥ 3

Let 2x + y = 3 ⇒ y = 3 – 2x

(0, 0) is not contained in the required half plane as (0, 0) does not satisfy the inequation 2x + y ≥ 3.

Again x + 2y ≥ 6

Let x + 2y = 6

Here also (0, 0) does not contain the required half plane. The double shaded region XABY is the solution set. Its corners are A (6, 0) and B (0, 3).

At A (6, 0) Z = 6 + 0 = 6

At B (0, 3) Z = 0 + 2 x 3 = 6

Therefore, at both points the value of Z = 6 which is minimum. In fact at every point on the line AB makes Z = 6 which is also minimum.

7.     Consider x + 2y ≤ 120

Let x + 2y = 120

The half plane containing(0, 0) is the required half plane as (0, 0) makes x + 2y ≤ 120, true.

Again x + y ≥ 60

Let x + y = 60

Also the half plane containing (0, 0) does not make x + y ≥ 6 true.

Therefore, the require half plane does not contain (0, 0).

Again x – 2y ≥ 0

Let x – 2y = 0 ⇒ x = 2y

Let test point be (30, 0).

⇒ x – 2y ≥ 0 ⇒ 30 – 2 × 0 ≥ 0 It is true.

Therefore, the half plane contains (30, 0).

The region CFEKC represents the feasible region.

At C (60, 0) Z = 5 x 60 = 300

At F (120, 0) Z = 5 x 120 = 600

At E (60, 30) Z = 5 x 60 + 10 x 30 = 600

At K (40, 20) Z = 5 x 40 + 10 x 20 = 400

Hence, minimum Z = 300 at x = 60, y = 0 and maximum Z = 600 at x = 120, y = 0 or x = 60, y = 30.

8.    Consider x + 2y ≥ 100

Let 

x + 2y ≥ 100 represents which does not include (0, 0) as it does not made it true.

Again consider 2x – y ≤ 0

Let 2x – y = 0 ⇒ y = 2x

Let the test point be (10, 0).

∴ 2 × 10 – 0 ≤ 0 which is false.

Therefore, the required half does not contain (10, 0).

Again consider 2x + y ≤ 200

Let 2x + y = 200

Now (0, 0) satisfies 2x + y ≤ 200

Therefore, the required half place contains (0, 0).

Now triple shaded region is ABCDA which is the required feasible region.

At A (0, 50)

Z = x + 2y = 0 + 2 x 50 = 100

At B (20, 40) Z = 20 + 2 x 40 = 100

At C (50, 100) Z = 50 + 2 x 100 = 250

At D (0, 200) Z = 0 + 2 x 200 = 400

Hence maximum Z = 400 at x = 0, y = 200 and minimum Z = 100 at x = 0, y = 50 or x = 20, y = 40.

9.    Consider x ≥ 3

Let x = 3 which is a line parallel to y-axis at a positive distance of 3 from it.

Since x ≥ 3, therefore the required half-plane does not contain (0, 0).

Now consider x + y ≥ 5

Let x + y = 5

Now (0, 0) does not satisfy x + y ≥ 5, therefore the required half plane does not contain (0, 0).

Again consider x + 2y ≥ 6

Let x + 2y = 6

Here also (0, 0) does not satisfy x + 2y ≥ 6, therefore the required half plane does not contain (0, 0).

The corners of the feasible region are A (6, 0), B (4, 1) and C (3, 2).

At A (6, 0) Z = –6 + 2 x 0 = –6

At B (4, 1) Z = –4 + 2 x 1 = –2

At C (3, 2) Z = –3 + 2 x 2 = 1

Hence, maximum Z = 1 at x = 3, y = 2.

10.  Consider x – y ≤ -1

Let x – y = -1

⇒ x = y – 1

If (0, 0) is the test point then x – y ≤ -1 ⇒ 0 ≤ -1 which is false and thus the required plane does not include (0, 0).

Again -x + y ≤ 0

Let -x + y = 0

⇒ y = x

For (1, 0) – 1 ≤ 0 which is true, therefore the required half-plane include (1, 0).

It is clear that the two required half planes do not intersect at all, i.e., they do not have a common region.

Hence there is no maximum Z.

11.   Let food P consists of x kg and food Q consists of y kg.

Minimum Z = 60x + 80y

According to question 3x + 4y ≥ 8 and 5x + 2y ≥ 11

x ≥ 0 and y ≥ 0

Consider 3x + 4y ≥ 8

But (0, 0) does not satisfy this inequation, therefore the required half-plane does not contain (0, 0).

Again consider 5x + 2y ≥ 11

Again (0, 0) does not satisfy this inequation, therefore the required half-plane does not contain (0, 0).

The double shaded region is our solution set. The corners of this region are D (0, 5.5), Q (2, 1.5) and 

Now Z = 60x + 80y

At D(0,5.5)

Z = 60 x 0 + 80 x 5.5 = 440

At Q (2, 1.5)

Z = 60 x 2 + 80 x 1.5 = 160

At 

Hence, minimum Z = 160 at x = 2, y = 1.5 means minimum cost Z = Rs. 160 when Reshma mixes food P = 2 kg and food Q = 1.5 kg.

12.  Let number of cakes made of first kind are x and that of second kind is y.

∴ Let to maximize Z = x + y

According to question 200x + 100y ≤ 5000 and 25x + 50y ≤ 1000

x ≥ 0, y ≥ 0

Consider 200x + 100y ≤ 5000

Let 200x + 100y = 5000

Here, (0, 0) satisfies this inequation, therefore the required half plane contains (0, 0).

Again consider 25x + 50y ≤ 1000

Let 25x + 50y = 1000

Here, again (0, 0) satisfies this inequation, therefore the required half plane contains (0, 0).

The double shaded region is the feasible region which is solution set.

The corner points of this region are O (0, 0), A (25, 0), B (20, 10) and C (0, 20).

∴ Z = x + y

At O (0, 0) Z = 0 + 0 = 0

At A (25, 0) Z = 25 + 0 = 25

At B (20, 10) Z = 20 + 10 = 30

AtC (0, 20) Z = 0 + 20 = 20

Hence, maximum number of cakes Z = 30 when x = 20, y = 10.

13.  Let number of rackets = x and number of bats = y

(i) To maximize Z = x + y

According to question 1.5x + 3y ≤ 42 and 3x + y ≤ 24

x ≥ 0, y ≥ 0

Consider 1.5x + 3y ≤ 42

Let 1.5x + 3y = 42 ⇒ x + 2y = 28

⇒ x = 28 – 2y

Here (0, 0) satisfies this inequaiton, therefore the required half-plane contains (0, 0).

Again consider 3x + y ≤ 24

Let 3x + y = 24

Here (0, 0) satisfies this inequaiton, therefore the required half-plane contains (0, 0).

Now the feasible region is the double-shaded region i.e., the solution set. The corner points of this region are O (0, 0), P (8, 0), Q (4, 12) and R (0, 14).

Now Z = x + y

At O (0, 0) Z = 0 + 0 = 0

At P (8, 0) Z = 8 + 0 = 0

At Q (4, 12) Z = 4 + 12 = 16

At R (0, 14) Z = 0 + 14 = 14

Now maximum Z = 16 at x = 4, y = 12.

(ii) The profit on a racket = Rs. 20 and the profit on a bat = Rs. 10

Let maximum profit P = 20x + 10y

Hence maximum profit P = 20 x 4 + 10 x 12 = Rs.200.

14.  Let number of package of nuts = x and number of packages of bolts = y

To maximum profit Z = 17.50x + 7y

According to question x + 3y ≤ 12 and 3x + y ≤ 12 x ≥ 0, y ≥ 0

Therefore, points are A (12, 0), B (0, 4). Also, (0, 0) satisfies this inequation, therefore the required half-plane contains (0, 0).

Again consider 3x + y ≤ 12

Let 3x + y = 12

Therefore, points are C (4, 0), D (0, 12).

Again (0, 0) satisfies this inequation, therefore the required half-plane contains (0, 0).

The double shaded region is OBPCO and its corners are O (0, 0), B (0, 4), D (3, 3), C (4, 0).

Now Z = 17.50x + 7y

At O (0, 0) Z = 17.5 x 0 + 7 x 0 = 0

At B (0, 4) Z = 17.5 x 0 + 7 x 4 = 28

At P (3, 3) Z = 17.5 x 3 + 7 x 3 = 73.50

At C (4, 0) Z = 17.5 x 4 + 7 x 0 = 70

∴ Maximum profit = Rs. 73.50 at x = 3, y = 3.

Hence maximum profit Z = Rs. 73.50 when he produces number of packets of nuts = 3 and number of packets of bolts = 3.

15.  Let the manufacturers produces x packages of screws A and y packages of screws B, then time taken x by packages of screws A and y packages of screws B on automatic machine = (4x + 6y) minutes. And time taken by x packages of screws A and y packages of screws B on hand operated machine = (6x + 3y) minutes.

Since, each machine is available for at the most 4 hours, i.e., 4 x 60 = 240 minutes.

Therefore, we have 4x + 6y ≤ 240 ⇒ 2x + 3y ≤ 120 and 6x + 3y ≤ 240 ⇒ 2x + y ≤ 80.

Profit on selling x packages of screws A and y packages of screws B = Z = 7x + 10y

∴ To find: x and y such that Z = 7x + 10y is maximum subject to 2x + 3y ≤ 120, and 2x + y ≤ 80, x ≥ 0 and y ≥ 0.

Consider 2x + 3y ≤ 120

Let 2x + 3y = 120

A (60, 0) and B (0, 4)

Here, (0, 0) satisfies this inequation, therefore the required hal-plane contains (0, 0).

Again 2x + y ≤ 80

Let 2x + y = 80

∴ C (40, 0) and D (0, 80)

Here also (0, 0) satisfies this inequation, therefore the required hal-plane contains (0, 0).

The feasible portion of the graph satisfying the inequalities 2x + 3y ≤ 120 and 2x + y ≤ 80 is OABC which is shown shaded in the figure.

Co-ordinates of O, A, B and C are (0, 0), (0, 40), (30, 20) and (40, 0) respectively.

Now Z = 7x + 10y

At O (0, 0) Z = 7 x 0 + 10 x 0 = 0

At A (0, 40) Z = 7 x 0 + 10 x 40 = 400

At B (30, 20) Z = 7 x 30 + 10 x 20 = 410

At C (40, 0) Z = 7 x 40 + 10 x 0 = 280

maximum profit Z = Rs. 410 at x = 30, y = 20

Hence, if the manufacturer produces 30 screws of type A and 20 screws of type B, he earn a maximum profit of Rs. 410.

16.  Let the manufacturer produces x pedestal lamps and y wooden shades, then the time taken by x pedestal lamps and y wooden shades on grinding/cutting machines = (2x + y) hours and the time taken by x pedestal lamps and y wooden shades on the sprayer = (3x + 2y) hours.

Since grinding/cutting machine is available for at the most 12 hours, i.e., 2x + y ≤ 12 and sprayer is available for at the most 20 hours, i.e., 3x + 2y ≤ 20

Profit from the sale of x lamps and shades Z = 5x + 3y

∴ To find: x and  such that Z = 5x + 3y is maximum subject to constraints , 2x + y ≤ 12, 3x + 2y ≤ 20, x ≥ 0, y ≥ 0.

Consider 3x + 2y  ≤ 20

Let 3x + 2y = 20

Now the area represented by 3x + 2y  ≤ 20 is the half-plane containing (0, 0) as (0, 0) satisfies the inequaiton.

Consider 2x + y ≤ 12

Let 2x+ y = 12

⇒ y = 12 – 2x

The inequation consists of the half-plane containing (0, ) as (0, 0) satisfies this inequation.

The double shaded region OPCAO is our solution where O (0, 0), P (6, 0), C (4, 4), A (0, 10).

Now Z = 5x + 3y

At O (0, 0) Z = 5 x 0 + 3 x 0 = 0

At P (6, 0) Z = 5 x 6 + 3 x 0 = 30

At C (4, 4) Z = 5 x 4 + 3 x 4 = 32

At A (0, 10) Z = 5 x 0 + 3 x 10 = 30

Now maximum Z = 32 at x = 4, y = 4

Hence, maximum profit Z = Rs. 32 when he manufactures 4 pedestal lamps and 4 wooden shades.

17.  Let the company manufactures x souvenirs of type A and y souvenirs of type B, then time taken for cutting x souvenirs of type A and y souvenirs of type B = (5x + 8y) minutes and time taken for assembling x souvenirs of type A and y souvenirs of type B = (10x + 8y) minutes.

Since 3 hours 29 minutes i.e., 200 minutes are available for cutting, therefore we have 5x + 8y ≤ 200

Also since 4 hours i.e., 240 minutes are available for assembling, therefore we have 10x + 8y ≤ 240 ⇒ 5x + 4y ≤ 120

Thus, our L.P.P. is to maximize profit Z = 5x + 6y subject to constraints, 5x + 8y ≤ 200, 5x + 4y ≤ 120, x ≥ 0, y ≥ 0.

Now 5x + 8y ≤ 200

Again 5x + 4y ≤ 120

Leet 5x + 4y = 120

In both the equations origin (0, 0) satisfies them and therefore the required half planes are there which contains (0, 0).

The portion of graph satisfying the inequalities 5x + 8y ≤ 200 and 5x + 4y ≤ 120 is OABC and is shown in shaded in the figure. Coordinates of the points O, A, B and C are (0, 0), (24, 0), (8, 20) and (0, 25) respectively.

Now Z = 5x + 6y

At O (0, 0) Z = 5 x 0 + 6 x 0 = 0

At A (24, 0) Z = 5 x 24 + 6 x 0 = 120

At B (8, 20) Z = 5 x 8 + 6 x 20 = 160

At C (0, 25) Z = 5 x 0 + 6 x 25 = 150

Now maximum Z = 160 at x = 8, y = 20

Hence, maximum profit = Rs. 160 when he manufactures 8 souvenirs of type A and 20 souvenirs of type B.

18.  Let number of desktop model computer = x and number of a portable model computer = y

To maximum profit Z = 4500x + 5000y subject to 25000x + 40000y ≤ 70,00,000, x + y ≤ 250, x ≥ 0, y ≥ 0.

Consider 25000x + 40000y ≤ 70,00,000 ⇒ 5x + 8y ≤ 1400

Let 5x + 8y = 1400

∴ Points are A (280, 0), B (0, 175)

Now (0, 0) satisfies the inequation, therefore the required half-plane contains (0, 0).

Again consider x + y ≤ 250

Let x + y = 250

∴ Points are C (250, 0), D (0, 250).

Again (0, 0) satisfies the inequation, therefore the required half-plane contains (0, 0).

The double shaded region OCPBO is the solution set. Its corners are O (0, 0), C (280, 0), P (200, 50), B (0, 200).

Now Z = 4500x + 5000y

AtO (0, 0) Z = 4500 x 0 + 5000 x 0 = 0

At C (280, 0) Z = 4500 x 280 + 5000 x 0 = 12,60,000

(Rejected as x ≥ 250)

At P (200, 50) Z = 4500 x 200 + 5000 x 50 = 11,50,000

At B (0, 200) Z = 4500 x 0 + 5000 x 200 = 10,00,000

Now maximum Z = Rs. 11,50,000 at x = 200, y = 50

Hence, maximum profit = Rs. 11,50,000 when he sells 200 desktop models and 50 portable models.

19.  Let the number of units of food F1 = x and number of units of food F2 = y

To minimize Z = 4x + 6y subject to 3x + 6y ≥ 80, 4x + 3y ≥ 100, x ≥ 0, y ≥ 0

Consider 3x + 6y ≥ 80

Clearly, (0, 0) is not included in the half-plane.

Now again consider 4x + 3y ≥ 100

Here, also (0, 0) is not included in the required half-plane.

The double shaded region is our feasible region and its corners are

Therefore, minimum cost is Rs.104 when 24 units of food F1 and mixed with 4/3 units of food F2.

20.  Let the quantity of F1 = kg and the quantity of F2 = kg

We have to minimize Z = 6x + 5y subject to

Consider 

Here, (0, 0) is not included in the required half plane and (0, 0) does not satisfy this inequation.

Again Here, (0, 0) is not included in the required half plane and (0, 0) does not satisfy this inequation.

The shaded region XPCAY is our feasible region. Its corners are , C (100, 80) and A (0, 280).

Now Z = 6x + 5y

At C (100, 80) Z = 6 x 100 + 5 x 80 = 1000

At A (0, 280) Z = 6 x 0 + 5 x 280 = 1400

Now minimum cost Z = Rs. 1000 at x = 100, y = 80.

Therefore, Minimum cost is Rs.1000 when the farmer used 100 kg of fertilizer F1 and 80 kg of fertilizer F2.

21.  To maximize Z = px + qy,    p, q > 0 subject to 2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0, y ≥ 0

The corner points are (0, 0), (5, 0), (3, 4) and (0, 5).

Maximum of Z occurs at both (3, 4) and (0, 5)

Z = px + qy

At (3, 4) Z = 3p + 4q

At (0, 5) Z = 5q

Now maxima occurs at both points,

3p + 4q = 5q

⇒ 3p = q

⇒ q = 3p

Hence option (D) is correct.

22.  Let x and y be the number of packets of food P and Q respectively, x ≥ 0, y ≥ 0.

We have to maximize Z = 6x + 3y (vitamin A) subject to the constraints 12x + 3y ≥ 240 (constraints on Calcium), i.e., 4x + y ≥ 80  ……….(i)

And 4x + 50y ≥ 460 (constraints on Iron), i.e., x + 5y ≥ 115 ……….(ii)

Also 6x + 4y ≤ 300 (constraints on Cholesterol), i.e., 3x + 2y ≤ 150 ……….(iii)

x ≥ 0, y ≥ 0  ……….(iv)

Consider 4x + y ≥ 80

Let 4x + y = 80

⇒ y = 80 – 4x

Here, (0, 0) does not satisfy this inequation, therefore the required half plane does not include the point (0, 0)

Again consider x + 5y ≥ 115

Let x + 5y = 115

⇒ x = 115 – 5y

Here, also (0, 0) does not satisfy this inequation, therefore the required half plane does not include the point (0, 0)

Again consider 3x + 2y ≤ 150

Therefore, G (50, 0) and H (0, 75) satisfy the equation.

As (0, 0) satisfies the inequation 3x + 2y = 150, therefore the required half plane contains (0, 0)

The shaded region is the feasible solution and its corners are P (15, 20), Q (40, 15) and R (2, 72).

Now Z = 6x + 3y

At P(15, 20) Z = 6 x 15 + 3 x 20 = 90 + 60 = 150

At Q (40, 15) Z = 6 x 40 + 3 x 15 = 240 + 45 = 285

At R (2, 72) Z = 6 x 2 + 3 x 72 = 12 + 216 = 228

Hence, maximum Z = 285 units of vitamin A at x = 40, y = 15.

23.  Let number of bags of cattle feed brand P = x and number of bags of cattle feed brand Q = y

We have to minimize Z = 250x + 200y subject to the constraints 3x + 1.5 y ≥ 18, 2.5x + 11.25y ≥ 45, 2x + 3y ≥ 24, x ≥ 0, y ≥ 0.

Consider 3x + 1.5y ≥ 18

Let 3x + 1.5y = 18

⇒ 2x + y = 12

Now the points are A (6, 0) and B (0, 12).

Now clearly (0, 0) does not lie in the required half plane as (0, 0) does not satisfy the inequation 3x + 1.5y ≥ 18.

Again consider 2.5x + 11.25y ≥ 45

Let 2.5x + 11.25y = 45

⇒ 2x + 9y = 36

Now again (0, 0) does not lie in the required half plane as (0, 0) does not satisfy the inequation 2.5x + 11.25y ≥ 45.

Again consider 2x + 3y ≥ 24

Here, the points E (12, 0) and F (0, 8) lie on the line.

Again also (0, 0) does not lie on the half plane as (0, 0) does not satisfy this inequation.

The feasible region of XCPQEY and the co-ordinates of corners are C (18, 0), P (9, 2), Q (3, 6) and E (0, 12).

Now Z = 250x + 200y

At C (18, 0) Z = 250 x 18 + 200 x 0 = 4500

At P (9, 2) Z = 250 x 9 + 200 x 2 = 2450

At Q (3, 6) Z = 250 x 3 + 200 x 6 = 1950

At E (0, 12) Z = 250 x 0 + 200 x 12 = 2400

Here, minimum cost Z = Rs. 1950 when x = 3, y = 6

Hence, number of bags of brand P = 3 and number of bags of brand Q = 6 and minimum cost of the mixture per bag = Rs.  = Rs. 216.67 per bag.

24.  Let in the mixture food X weighs = kg x and in the mixture food Y weighs = y kg

We have to minimize Z = 16x + 20y subject to constraints, x + 2y ≥ 10, 2x + 2y ≥ 12, 3x + y ≥ 8, x ≥ 0, y ≥ 0

Consider x + 2y ≥ 10

∴ Points A (10, 0) and B (0, 5) lies on the line.

Here, (0, 0) does not satisfy the inequation x + 2y ≥ 10, therefore the required half plane does not include (0, 0).

Again consider 2x + 2y ≥ 12

Let 2x + 2y = 12

⇒ x + y = 6

∴ Points C (6, 0) and D (0, 6) lies on the line.

Again consider 3x + y ≥ 8

Let 3x + y = 8 ⇒ y = 8 – 3x

Again in the inequation (0, 0) is not included in the required half plane.

The shaded region is our feasible solution A (10, 0), P (2, 4), Q (1, 5), E (0, 8).

The corners of the feasible region are A (10, 0), P (2, 4), Q (1, 5), E (0, 8).

Now Z = 16x + 20y

At A (10, 0) Z = 16 x 10 + 20 x 0 = 160

At P (2, 4) Z = 16 x 2 + 20 x 4 = 112

At Q (1, 5) Z = 16 x 1 + 20 x 5 = 116

At E (0, 8) Z = 16 x 0 + 20 x 8 = 160

Therefore minimum Z = Rs. 112 at x = 2, y = 4

Hence, minimum cost of the mixture = Rs. 112 when he mixes 2 kg of food X and 4 kg of food Y.

25.  Let x units of toys A and y units of toys B are produced by the manufacturer. Time spent on machine I to produce x units of toys A and y units of toys B = (12x + 6y) minutes.

Since each machine is available for a maximum of 6 x 60 = 360 minutes.

Therefore, we have 12x + 6y ≤ 360 ⇒ 2x + y ≤ 60

18x ≤ 360 and 2x + 3y ≤ 120

now, the profit Z earned by the manufacturer to produce x units of type A and y units of type B is 7.50x + 5y

we have to maximize Z = 7.50x + 5y i.e., 4Z = 3x + 2y subject to constraints , 2x + y ≤ 6, x ≤ 20, 2x + 3y ≤ 120 and x ≥ 0, y ≥ 0.

Consider 2x + y ≤ 6

Let 2x + y = 6

∴ Points A (30, 0) and B (0, 60) lies on the line. Also (0, 0) lies in the required half plane.

Again consider x ≤ 20

Let x = 20

It represent the half plane to the left of x = 20.

Again consider 2x+ 3y ≤ 120

∴ Points C (60, 0) and D (0, 40) lies on the line. Therefore, (0, 0) lies in the required half plane.

The shaded portion is our feasible region. Its corners are O (0, 0), P (20, 0), Q (20, 20), R (15, 30), D (0, 40).

Now Z = 7.50x + 5y

At O (0, 0) Z = 7.5 x 0 + 5 x 0 = 0

At P (20, 0) Z = 7.5 x 00 + 5 x 0 = 150

At Q (20, 20) Z = 7.5 x 20 + 5 x 20 = 250

At R (15, 30) Z = 7.5 x 15 + 5 x 30 = 262.50

At D (0, 40) Z = 7.5 x 0 + 5 x 40 = 200

Now maximum profit = Z = Rs. 262.50, when he manufacturers 15 toys of types A and 30 of type B in a day.

26. Let number of tickets of executive class sold = x and number of tickets of economy class sold = y

We have to maximize = Z = 1000x + 600y subject to x + y ≤ 20, x ≤ 20 and y ≥ 4x

x ≥ 0, y ≥ 0

Consider x + y ≤ 200

Let x + y = 200

∴ Points A (200, 0) and B(0, 200) are on the line and therefore (0, 0) is included in the required half plane.

Again consider x ≥ 20

Let x = 20

It is the line parallel to y-axis at a positive distance 20 and the half plane lies towards right of it.

Again consider y ≥ 4x

Let y = 4x

Here, (40, 0) does not satisfy y ≥ 4x, therefore plane does not include (40, 0).

The shaded portion is the feasible region. Its corners are C (20, 80), D (40, 160) and P (20, 180)

Now Z = 1000x + 600y

At C (20, 80) Z = 1000 x 20 + 600 x 80 = 20000 + 48000 = 68,000

At D (40, 60) Z = 1000 x 40 + 600 x 60 = 40000 + 96000 = 1,36,000

At P (20, 180) Z = 1000 x 20 + 600 x 180 = 20000 + 108000 = 1,28,000

Hence Maximum profit Z = Rs. 1,36,000 at x = 40, y = 160.

27.  Let godown A supplies x quintals of grain to the ration shop D and y quintals to ration shop E.

Subject to x + y ≤ 100, x ≤ 6, y ≤ 50 and x + y ≥ 60, x ≥ 0, y ≥ 0.

Consider x + y ≤ 100

Let x + y = 100

∴ Points P (100, 0) and Q (0, 100) lie on the line and it represents the half-plane containing (0, 0).

Again we consider x ≤ 6 and y ≤ 50

We draw x = 60 and y = 50

Again consider x + y ≥ 60

∴ Points A (60, 0) and R (0, 60) lie on the line and it represents the half-plane containing (0, 0)

The shaded region is the feasible solution. Its corners are A (60, 0), B (60, 40), C (50, 50) and D (10, 50).

Hence minimum value is Z = 510 at x = 10, y = 50.

28.  Let x liters of oil is supplied from depot A to petrol pump D and y liters of oil supplied from depot A to petrol pump E then 7000 – (x + y) liters of oil will be supplied from depot A to petrol pump F.

∴ We have x ≥ 0 and y ≥ 0 and 7000 – (x + y) ≥ 0

⇒ x + y ≤ 700

Since requirements of oil at petrol pump, D, E and F are (4500 – x), (5000 – x) and [3500 – (x + y)] liters respectively.

∴ 4500 – x ≥ 0

⇒ x ≤ 4500

And 3000 – y ≥ 0 ⇒ y ≤ 3000

And 3500 – [7000 – (x + y)] ≥ 0 ⇒ x + y ≥ 3500

∵ The cost of transportation per km for 10 liters oil is Re 1

∴ The cost of transportation per km per liter = Rs .

∴ The cost of transportation

Z = 0.7x + 0.6y + 0.3[700 – (x + y)] + 0.3 (4500 – x) + 0.4(3000 – y) + [(x + y) – 3500]

Z = 0.3x + 0.1y + 3950

Therefore, the feasible region is ABECD.

Its corners are A (500, 3000), B (35, 0), E (4500, 0), C (4500, 2500), D (4000, 3000).

Now Z = 0.3x + 0.1y + 3950

At A (500, 3000) Z = 0.3 x 500 + 0.1 x 3000 + 3950 = 4400

At B (3500, 0) Z = 0.3 x 3500 + 0.1 x 0 + 3950 = 5000

At E (4500, 0) Z = 0.3 x 4500 + 0.1 x 0 + 3950 = 5300

At C (4500, 2500) Z = 0.3 x 4500 + 0.1 x 2500 + 3950 = 5550

At D (4000, 3000) Z = 0.3 x 4000 + 0.1 x 3000 + 3950 = 5450

Minimum transportation charges are Rs. 4400 at x = 500, y = 3000

Hence, 500 liters, 3000 liters and 3500 liters of oil should be transported from depot A to petrol pumps D, E, F and 4000 liters, 0 liter and 0 liter of oil be transported from depot B to petrol pumps D, E and F with minimum cost of transportation of Rs. 4400.

29.  Let amount of Brand P of fertilizer = x bags and amount of Brand Q of fertilizer = y bags

We have to minimize Z = 3x + 3.5y subject to x + 2y ≥ 240, 3x + 1.5y ≥ 270, 1.5x + 2y ≤ 310, x ≥ 0, y ≥ 0

Consider x + 2y ≥ 240

Let x + 2y = 240

∴ Points A (240, 0) and B (0, 120) lie on the line.

And (0, 0) does not lie on the required half-plane of this in equation.

Again consider 3x + 1.5y ≥ 270

Let 3x + 1.5y = 270

⇒ 2x + y = 180

∴ Points C (90, 0) and D (0, 180) lie on the line.

Here also (0, 0) does not lie on the required half-plane of this inequation.

Again consider 1.5x + 2y ≤ 310

Here also (0, 0) does not lie on the required half-plane of this inequation.

The shaded portion is the feasible region. Its corners are P (140, 50), Q (20, 140) and R (40, 100).

Now Z = 3x + 3.5y

At P (140, 50) Z = 3 x 140 + 3.5 x 50 = 420 + 175 = 595

At Q (20, 140) Z = 3 x 20 + 3.5 x 140 = 60 + 490 = 550

At R (40, 100) Z = 3 x 40 + 3.5 x 100 = 120 + 350 = 470

Hence minimum Z = 470 at x = 40, y = 100.

Therefore, minimum amount of nitrogen = 470 kg when 40 bags of brand P and 100 bags of brand Q are used.

30.  We have Z = 3x + 3.5y and Z is maximum at P (140, 50).

To maximize the amount of nitrogen, 140 bags of brand P and 50 bags of brand Q are required.

Therefore maximum amount of nitrogen required = 595 kg

31.  Let the number of dolls of type A = x and number of dolls of type B = y

We have to maximize Z = 12x + 16y subject to 

Consider, x + y ≤ 1200

Let x + y = 1200

⇒ y = 1200 – x

Here, (0, 0) is included in the required half plane and satisfies this inequation.

Here, (100, 0) satisfies the inequation , therefore the required half plane includes (100, 0).

Again consider x – 3y ≤ 600

Let x – 3y = 600

⇒ x = 600 – 3y

Here, also (0, 0) is included in the required half-plane.

The shaded region DRSOD is the feasible region whose corners are D(600,0), R(1050,150),

S(800,400) and O(0,0).

Now Z = 12x + 16y

At D (600, 0) Z = 12 x 600 + 16 x 0 = 7200

At R (1050, 150) Z = 12 x 105 + 16 x 150 = 12600 + 2400 = 15,000

At S (800, 400) Z = 12 x 800 + 16 x 400 = 9600 + 6400 = 16,000

At O (0, 0) Z = 12 x 0 + 16 x 0 = 0

Hence maximum profit Z = Rs. 16,000 at x = 800, y = 400.

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