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Class 12 RD Sharma Solutions – Chapter 30 Linear Programming – Exercise 30.2 | Set 1

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  • Last Updated : 20 May, 2021
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Question 1. Maximize Z = 5x + 2y, 

Subject to

3x + 5y ≤ 15

5x + 2y ≤ 10

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

3x + 5y = 15, 

5x + 2y = 10, 

x = 0 and 

y = 0

Area represented by 3x + 5y ≤ 15:

The line 3x + 5y = 15 connects the coordinate axes at A(5,0) and B(0,3) respectively. 

By connecting these points we will get the line 3x + 5y = 15.

Thus,

(0,0) assure the in equation 3x + 5y ≤ 15. 

Hence,

The area having the origin shows the solution set of the in equation 3x + 5y ≤ 15.

Area shows by 5x + 2y ≤ 10:

The line 5x + 2y = 10 connects the coordinate axes at C(2,0) and D(0,5) respectively. 

By connecting these points we will get the line 5x + 2y = 10.

Thus, 

(0,0) satisfies the in equation 5x + 2y ≤ 10. 

Hence,

The area having the origin shows the solution set of the in equation 5x + 2y ≤ 10.

Area shows by x ≥ 0 and y ≥ 0:

Here, 

All the point in the first quadrant assures these in equations. 

Thus, 

The first quadrant is the area shows by the in equations x ≥ 0, and y ≥ 0.

The feasible area determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

 The corner points of the feasible area are O(0, 0), C(2, 0), E\left(\frac{20}{19},\ \frac{45}{19}\right)    and B(0, 3).

The values of Z at these corner points are as follows.

Corner point

Z = 5+ 3y

O(0, 0)

5 × 0 + 3 × 0 = 0

C(2, 0)

5 × 2 + 3 × 0 = 10

\left(\frac{20}{19},\ \frac{45}{19}\right)

5\times\frac{20}{19}+3\times\frac{45}{19}=\frac{235}{19}

B(0, 3) 

5 × 0 + 3 × 3 = 9

Hence, 

The maximum value of Z 

\frac{235}{19} at the point \left(\frac{20}{19},\ \frac{45}{19}\right)   

Hence, 

x= \frac{20}{19} and y = \frac{45}{19}   is the best solution of the given LPP.

Therefore, 

The best value of Z is \frac{235}{19}.

Question 2. Maximize Z = 9x + 3y

Subject to

2x + 3y ≤ 13

3x + y ≤ 5

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

2x + 3y = 13, 

3x +y = 5, 

x = 0 and 

y = 0

Area shown by 2x + 3y ≤ 13:

The line 2x + 3y = 13 connects the coordinate axes at A \left(\frac{13}{2},\ 0\right) and B \left(0,\ \frac{13}{3}\right) respectively.

By connecting these points we will get the line 2x + 3y = 13.

Thus,

(0,0) assures the line equation 2x + 3y ≤ 13. 

Thus, the area showing the origin represents the solution set of the in equation 2x + 3y ≤ 13.

The area shows by 3x + y ≤ 5:

The line 5x + 2y = 10 connects the coordinate axes at C \left(\frac{5}{3},\ 0\right)   and D(0, 5) respectively. 

After connecting these points, 

We will get the line 3x + y = 5.

Thus,

(0,0) assures the in equation 3x + y ≤ 5. 

Thus,

The area having the origin represents the solution set of the in equation 3x + y ≤ 5.

Area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the point in the first quadrant assures these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x ≥ 0, and y ≥ 0.
 

The suitable area determined by the system of constraints, 

2x + 3y ≤ 13, 

3x + y ≤ 5, 

x ≥ 0, and 

y ≥ 0, are as follows.

The corner points of the suitable area are O(0, 0), C \left(\frac{5}{3},\ 0\right), E \left(\frac{2}{7},\ \frac{29}{7}\right) and B \left(0,\ \frac{13}{3}\right).

The values of Z at these corner points are as follows.

Corner point

Z = 9+ 3y

O(0, 0)

9 × 0 + 3 × 0 = 0

\left(\frac{5}{3},\ 0\right)

9 × \frac{5}{3}   + 3 × 0 = 15

\left(\frac{2}{7},\ \frac{29}{7}\right)

9 × \frac{2}{7}   + 3 × \frac{29}{7}   = 15

\left(0,\ \frac{13}{3}\right)

9 × 0 + 3 ×\frac{13}{3}   = 13

Here,

We can see that the maximum value of the objective function Z is 15 which is at C \left(\frac{5}{3},\ 0\right) and E \left(\frac{2}{7},\ \frac{29}{7}\right).

Thus, 

The best value of Z is 15.

Question 3. Minimize Z = 18x + 10y

Subject to

4x + y ≥ 20

2x + 3y ≥ 30

x, y ≥ 0

Solution:

Convert the given in equations into equations,

We will get the following equations:

4x + y = 20, 

2x +3y = 30, 

x = 0 and 

y = 0

Area shown by 4x + y ≥ 20 :

The line 4x + y = 20 connects the coordinate axes at A(5, 0) and B(0, 20) respectively. 

By connecting these points we will get the line 4x + y = 20.

Thus,

(0,0) does not assure the in equation 4x + y ≥ 20. 

Thus,

The area in xy plane which does not have the origin represents the solution set of the in equation 4x + y ≥ 20.

Area shown by 2x +3y ≥ 30:

The line 2x +3y = 30 connects the coordinate axes at C(15,0) and D(0, 10) respectively. 

By joining these points we will get the line 2x +3y = 30.

Thus, (0,0) does not assure the in equation 2x +3y ≥ 30.

Thus,

The area which does not have the origin shows the solution set of the in equation 2x +3y ≥ 30.

Area shown by x ≥ 0 and y ≥ 0:

Hence, 

Every point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints, 4x + y ≥ 20, 2x +3y ≥ 30, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the suitable area are 

B(0, 20), 

C(15,0), 

E(3,8) and 

C(15,0).

The values of Z at these corner points are as follows.

Corner point

Z = 18+ 10y

B(0, 20)

18 × 0 + 10 × 20 = 200

E(3,8)

18 × 3 + 10 × 8 = 134

C(15,0) 

18 × 15 + 10 ×0 = 270

Hence, 

The minimum value of Z is 134 at the point E(3,8). Hence, x = 3 and y =8 is the best solution of the given LPP.

Therefore, 

The best value of Z is 134.

Question 4.  Maximize Z = 50x + 30y

Subject to

2x + y ≤ 18

3x + 2y ≤ 34

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

2x + y = 18, 3x + 2y = 34

The area shown by 2x + y ≥ 18:

The line 2x + y = 18 connects the coordinate axes at A(9, 0) and B(0, 18) respectively. 

After connecting these points 

We will get the line 2x + y = 18.

Thus,

(0,0) does not assure the in equation 2x + y ≥ 18. 

Thus,

The region in xy plane which does not have the origin represents the solution set of the in equation 2x + y ≥ 18.

The area shown by 3x + 2y ≤ 34:

The line 3x + 2y = 34 connects the coordinate axes at C \left(\frac{34}{3},\ 0\right), 0 and D(0, 17) respectively. 

After joining these points we will get the line 3x + 2y = 34.

Thus,

(0,0) assure the in equation 3x + 2y ≤ 34. 

Thus,

The area having the origin shows the solution set of the in equation 3x + 2y ≤ 34.

The corner points of the suitable area are 

A(9, 0), C \left(\frac{34}{3},\ 0\right)   and E(2, 14).

 

The values of Z at these corner points are as follows.

Corner point

Z = 50+ 30y

A(9, 0)

50 × 9 + 3 × 0 = 450

 C \left(\frac{34}{3},\ 0\right)

50 × \frac{34}{3}   + 30 × 0 = \frac{1700}{3}

E(2, 14)

50 × 2 + 30 × 14 = 520

Hence, 

The maximum value of Z is \frac{1700}{3} at the point \left(\frac{34}{3},\ 0\right).

Hence,

x = \frac{34}{3} and y = 0 is the best solution of the given LPP.

Therefore,

The best value of Z is \frac{1700}{3}

Question 5. Maximize Z = 4x + 3y

Subject to

3x + 4y ≤ 24

8x + 6y ≤ 48

x ≤ 5

y ≤ 6

x, y ≥ 0

Solution:

Here we have to maximize Z = 4x + 3y

Convert the given in equations into equations, 

We will get the following equations:

3x + 4y = 24, 

8x + 6y = 48, 

x = 5, 

y = 6, 

x = 0 and 

y = 0.

The line 3x + 4y = 24 connects the coordinate axis at A(8, 0) and B(0,6). 

Connect these points to get the line 3x + 4y = 24.

Thus, 

(0, 0) assure the in equation 3x + 4y ≤ 24.

Thus, 

The area in xy-plane that have the origin showing the solution set of the given equation.

The line 8x + 6y = 48 connects the coordinate axis at C(6, 0) and D(0,8). Connect these points to get the line 8x + 6y = 48.

Thus,

(0, 0) assures the in equation 8x + 6y  ≤ 48. 

Thus, 

The area in xy-plane that have the origin shows the solution set of the given equation.

x = 5 is the line passing through x = 5 parallel to the Y axis.

y = 6 is the line passing through y = 6 parallel to the X axis.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shows by the in equations.

These lines are drawn using a suitable scale.

 

The corner points of the suitable area are O(0, 0), G(5, 0), F \left(5,\ \frac{4}{3}\right)  , E \left(\frac{24}{7},\ \frac{24}{7}\right)   

and B(0, 6).

The values of Z at these corner points are as follows.
 

Corner point

Z = 4+ 3y

O(0, 0)

4× 0 + 3 × 0 = 0

 G(5, 0)

4 × 5 + 3 × 0 = 20

\left(5,\ \frac{4}{3}\right)

4 × 5 + 3 × \frac{4}{3}   = 24

\left(\frac{24}{7},\ \frac{24}{7}\right)

4 × \frac{24}{7}   + 3 × \frac{24}{7}   = \frac{196}{7}  = 24

B(0, 6)

4 × 0 + 3 × 6 = 18

Here we can see that the maximum value of the objective function Z is 24 

Which is at F \left(5,\ \frac{4}{3}\right)   and E 

Therefore,

The best value of Z is 24.

Question 6. Maximize Z = 15x + 10y

Subject to

3x + 2y ≤ 80

2x + 3y ≤ 70

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

3x + 2y = 80, 2x + 3y = 70, x = 0 and y = 0

The area shown by 3x + 2y ≤ 80 :

The line 3x + 2y = 80 connects the coordinate axes at A \left(\frac{80}{3},\ 0\right)  , 0 and B(0, 40) respectively. 

By connecting these points we will get the line 3x + 2y = 80.

Thus,

(0,0) assure the in equation 3x + 2y ≤ 80 . 

Thus,

The area having the origin represents the solution set of the in equation 3x + 2y ≤ 80 .

The area shown by 2x + 3y ≤ 70:

The line 2x + 3y = 70 connects the coordinate axes at C(35, 0)C35, 0 and D(0, 703)D0, 703 respectively. 

After connecting these points we will get the line 2x + 3y ≤ 70.

Thus,

(0,0) assure the in equation 2x + 3y ≤ 70. 

Thus,

The area having the origin shows the solution set of the in equation 2x + 3y ≤ 70.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

Every point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x ≥ 0 and y ≥ 0.
 

The suitable area determined by the system of constraints, 3x + 2y ≤ 80, 2x + 3y ≤ 70, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the suitable area are O(0, 0), A \left(\frac{80}{3},\ 0\right)  , 0 ,E(20, 10) and D \left(0,\ \frac{70}{3}\right)  .

The values of Z at these corner points are as follows.

Corner point

Z = 15+ 10y

O(0, 0)

15 × 0 + 10 × 0 = 0

\left(\frac{80}{3},\ 0\right)

15 × \frac{80}{3}   + 10 × 0 = 400

E(20, 10)

15 × 20 + 10 × 10 = 400

\left(0,\ \frac{70}{3}\right)

15 × 0 + 10 ×\frac{70}{3}   = \frac{700}{3}

Thus we can see that the maximum value of the objective function Z is 400 which is at A \left(\frac{80}{3},\ 0\right)

and E(20, 10).

Therefore, 

The best value of Z is 400.

Question 7.  Maximize Z = 10x + 6y

Subject to

3x + y ≤ 12

2x + 5y ≤ 34

x, y ≥ 0

Solution:

Convert the given in equations into equations, 

We will get the following equations:

3x + y = 12, 2x + 5y = 34, x = 0 and y = 0

The area shown by 3x + y ≤ 12:

The line 3x + y = 12 connects the coordinate axes at A(4, 0) and B(0, 12) respectively. 

After connecting these points we will get the line 3x + y = 12.

Thus,

(0,0) assure the in equation 3x + y ≤ 12. 

Thus,

The area having the origin represents the solution set of the in equation 3x + y ≤ 12.

The area shown by 2x + 5y  ≤ 34:

The line 2x + 5y = 34 connects the coordinate axes at C(17, 0) and D (0,\ \frac{34}{5})   respectively. 

After connecting these points we will get the line 2x + 5y ≤ 34.

Thus,

(0,0) assure the in equation 2x + 5y ≤ 34. 

Thus,

The area having the origin represents the solution set of the in equation 2x + 5y ≤ 34.

Area having by x ≥ 0 and y ≥ 0:

Hence, 

All the point in the first quadrant assure these in equations. 

Thus, 

The first quadrant is the area shown by the in equations x ≥ 0 and y ≥ 0.
 

The suitable area determined by the system of constraints, 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the suitable area are O(0, 0), A(4, 0) ,E(2, 6) and D(0,\ \frac{34}{5})  .

The values of Z at these corner points are as follows:

Corner point

Z = 10+ 6y

O(0, 0)

10 × 0 + 6 × 0 = 0

A(4, 0)A4, 0

10× 4 + 6 × 0 = 40

E(2, 6)E2, 6

10 × 2 + 6 × 6 = 56

D  (0,\ \frac{34}{5})

10 × 0 + 6 ×\frac{34}{5}   = \frac{204}{3}

Here we can see that the maximum value of the objective function Z is 56 which is at E(2, 6) that means at x = 2 and y = 6.

Therefore, 

The best value of Z is 56.

Question 8.  Maximize Z = 3x + 4y

Subject to

2x + 2y ≤ 80

2x + 4y ≤ 120

Solution:

Here we have to maximize Z = 3x + 4y

Convert the given in equations into equations, we will get the following equations:

2x + 2y = 80, 2x + 4y = 120

The area shown by 2x + 2y ≤ 80:

The line 2x + 2y = 80 connects the coordinate axes at A(40, 0) and B(0, 40) respectively. 

After connecting these points we will get the line 2x + 2y = 80.

Thus, 

(0,0) assure the in equation 2x + 2y ≤ 80. 

Thus,

The area having the origin represents the solution set of the in equation 2x + 2y ≤ 80.

The area shown by 2x + 4y ≤ 120:

The line 2x + 4y = 120 connects the coordinate axes at C(60, 0) and D(0, 30) respectively. 

After connecting these points we will get the line 2x + 4y ≤ 120.

Thus,

(0,0) assure the in equation 2x + 4y ≤ 120. 

Thus,

The area having the origin represents the solution set of the in equation 2x + 4y ≤ 120.

The suitable area determined by the system of constraints, 2x + 2y ≤ 80, 2x + 4y ≤ 120 are as follows:

The corner points of the suitable area are O(0, 0), A(40, 0), E(20, 20) and D(0, 30).

The values of Z at these corner points are as follows:

Corner point

Z = 3+ 4y

O(0, 0)

3 × 0 + 4 × 0 = 0

A(40, 0)

3× 40 + 4 × 0 = 120

E(20, 20)

3 × 20 + 4 × 20 = 140

D(0, 30) 

10 × 0 + 4 ×30 = 120

Here we can see that the maximum value of the objective function Z is 140 which is at E(20, 20) that means at

x = 20 and y = 20.

Therefore, 

The best value of Z is 140.

Question 9. Maximize Z = 7x + 10y

Subject to

x + y ≤ 30000

       y ≤ 12000

       x ≥ 6000

       x ≥ y

   x, y ≥ 0

Solution:

Here we have to maximize Z = 7x + 10y

Convert the given in equations into equations, 

We will get the following equations:

x + y = 30000,y = 12000, x = 6000, x = y, x = 0 and y = 0.

Region represented by x + y ≤ 30000:

The line x + y = 30000 connects the coordinate axes at A(30000, 0) and B(0, 30000) respectively. 

After connecting these points we will get the line x + y = 30000.

Thus, (0,0) assure the in equation x + y ≤ 30000. 

Thus,

the area having the origin represents the solution set of the in equation x + y ≤ 30000.

The line y = 12000 is the line that passes through C(0,12000) and parallel to x axis.

The line x = 6000 is the line that passes through (6000, 0) and parallel to y axis.

The area shown by x ≥ y

The line x = y is the line that passes through origin.The points to the right of the line x = y assure the inequation x ≥ y.

Like by taking the point (−12000, 6000).

Here, 6000 > −12000 which implies y > x. 

Hence, 

the points to the left of the line x = y will not assure the given inequation x ≥ y.

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All the point in the first quadrant assure these inequations. 

Thus, 

the first quadrant is the area shown by the inequations x ≥ 0 and y ≥ 0.

The suitable area determined by the system of constraints, x + y ≤ 30000, y ≤ 12000, x ≥ 6000, x ≥  y , x ≥ 0 and y ≥ 0 are as follows:

The corner points of the feasible region are 

D(6000, 0), 

A(3000, 

F(18000, 12000) and 

E(12000, 12000).

The values of Z at these corner points are as follows:

Corner point

Z = 7+ 10y

D(6000, 0)

7 × 6000 + 10 × 0 = 42000

A(3000, 0)A3000, 0

7× 3000 + 10 × 0 = 21000

F(18000, 12000)

7 × 18000 + 10 × 12000 = 246000

E(12000, 12000) 

7 × 12000 + 10 ×12000 = 204000

Here we can see that the maximum value of the objective function Z is 246000 which is at F(18000, 12000) that means at

x = 18000 and y = 12000.

Therefore, 

The best value of Z is 246000.

Question 10. Minimize Z = 2x + 4y

Subject to

x + y ≥ 8

x + 4y ≥ 12

x ≥ 3, y ≥ 2

Solution:

Convert the given in equations into equations, we will get the following equations:

x + y = 8, x + 4y = 12, x = 3, y = 2

The area shows by x + y ≥ 8:

The line x + y = 8 connects the coordinate axes at A(8, 0) and B(0, 8) respectively. 

After connecting these points we will get the line x + y = 8.

Thus,

(0,0) does not assure the in equation x + y ≥ 8. 

Thus,

The region in xy plane which does not contain the origin represents the solution set of the in equation x + y ≥ 8.

The area shown by x + 4y ≥ 12:

The line x + 4y = 12 connects the coordinate axes at C(12, 0) and D(0, 3) respectively. 

After connecting these points we will get the line x + 4y = 12.

Thus,

(0,0) assure the in equation x + 4y ≥ 12. 

Thus,

the area in xy plane which having the origin represents the solution set of the in equation x + 4y ≥ 12.

The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis.x ≥ 3 is the area to the right of the line

x = 3.

The line y = 2 is the line that passes through the point (0, 12) and is parallel to X axis.y ≥ 2 is the area above the line y = 2. 

The corner points of the suitable region are E(3, 5) and F(6, 2).
 

The values of Z at these corner points are as follows.

Corner point

Z = 2x + 4y

E(3, 5)

2 × 3 + 4 × 5 = 26

F(6, 2)

2 × 6 + 4 × 2 = 20

Therefore, 

The minimum value of Z is 20 at the point F(6, 2). 

Hence, x = 6 and y =2 is the best solution of the given LPP.

Therefore, 

The best value of Z is 20.


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