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Class 12 RD Sharma Solutions – Chapter 30 Linear Programming – Exercise 30.2 | Set 2

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Question 11. Minimize Z = 5x + 3y

Subject to

2x + y ≥ 10

x + 3y ≥ 15

x ≤ 10

y ≤ 8

x, y ≥ 0

Solution:

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

2x + y = 10, x + 3y = 15, x = 10, y = 8

Area shown by 2x + y ≥ 10:

The line 2x + y = 10 meets the coordinate axes at A(5, 0) and B(0, 10) respectively. 

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

Thus,

(0,0) does not assure the in equation 2x + y ≥ 10. 

Thus,

The area in xy plane which does not have the origin represents the solution set of the in equation 2x + y ≥ 10.

The area represented by x + 3y ≥ 15:

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

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

Thus,

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

The area in xy plane which does not have the origin represents the solution set of the in equation x + 3y ≥ 15.

The line x = 10 is the line that passes through the point (10, 0) and is parallel to Y axis.x ≤ 10 is the area to the left of the line 

x = 10.

The line y = 8 is the line that passes through the point (0, 8) and is parallel to X axis.y ≤ 8 is the area below the line y = 8.

The area shows by x ≥ 0 and y ≥ 0:

Thus, 

All point in the first quadrant assure these in equations. 

Thus, 

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

The suitable area determined by the system of constraints, 2x + y ≥ 10, x + 3y ≥ 15, x ≤ 10, y ≤ 8, x ≥ 0 and y ≥ 0 are as 

follows.

The corner points of the suitable region are 

E(3, 4),

H(10,\ \frac{5}{3})

F(10, 8) and 

G(1, 8).

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

Corner point

Z = 5x + 3y

E(3, 4)

5 × 3 + 3 × 4 = 27

H(10,53)

5 × 10 + 3× \frac{5}{3}     = 55

F(10, 8)

5 × 10 + 3 × 8 = 74

G(1, 8)

5 × 1 + 3 × 8 = 29

Hence, 

The minimum value of Z is 27 at the point F(3, 4). 

Therefore, 

x = 3 and y =4 is the best solution of the given LPP.

Hence, the best value of Z is 27.

Question 12.  Minimize Z = 30x + 20y

Subject to

x + y ≤ 8

x + 4y ≥ 12

5x + 8y = 20

x, y ≥ 0

Solution:

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

x + y = 8, x + 4y = 12, x = 0 and y = 0

5x + 8y = 20 is already an equation.

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) assures the in equation x + y ≤ 8. 

Thus,

The area in xy plane which have the origin represents the solution set of the in equation x + y ≤ 8.

The area shows 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 does not have the origin shows the solution set of the in equation x + 4y ≥ 12.

The line 5x + 8y = 20 is the line that passes through E(4, 0) and F (0,\ \frac{5}{2})    .

The area shows by x ≥ 0 and y ≥ 0:

Thus, 

All point in the first quadrant assure these in equations. 

Thus, 

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

The suitable area determined by the system of constraints, x + y ≤ 8, x + 4y ≥ 12, 5x + 8y = 20, x ≥ 0 and y ≥ 0 are as follows.

The corner points of the suitable area are B(0,8), D(0,3), G \left(\frac{20}{3},\ \frac{4}{3}\right)    .
The values of Z at these corner points are as follows.

Corner point

Z = 30x + 20y

B(0,8)

160

D(0,3)

60

\left(\frac{20}{3},\ \frac{4}{3}\right)

266.66

Hence, 

The minimum value of Z is 60 at the point D(0,3).  

Therefore, 

x = 0 and y =3 is the best solution of the given LPP.

Therefore, 

The best value of Z is 60.

Question 13.  Maximize Z = 4x + 3y

Subject to

3x + 4y ≤ 24

8x + 6y ≤ 48

           x ≤ 6000

           x ≤ y

       x, y ≥ 0

Solution:

Here, we need 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). 

Connects 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 represents 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) assure the in equation 8x + 6y  ≤ 48. 

Thus, 

The area in xy-plane that have the origin represents 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 points in the first quadrant assure these in equations. 

Thus, 

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

These lines are drawn using a suitable scale.

The corner points of the feasible region are O(0, 0), G(5, 0), F (5,\ \frac{4}{3}), E (\frac{24}{7},\ \frac{24}{7}) 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) G5, 0

4 × 5 + 3 × 0 = 20

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

4 × 5 + 3 × \frac{4}{3}     = 24

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

4 × \frac{24}{7}     + 3 × \frac{24}{7}     = \frac{168}{7}    = 24

B(0, 6)B0, 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 (5,\ \frac{4}{3}) and E (\frac{24}{7},\ \frac{24}{7}).

Therefore, 

The best value of Z is 24.

Question 14. Minimize Z = x – 5y + 20

Subject to

    x – y ≥ 0

-x + 2y ≥ 2

          x ≥ 3

          y ≤  4

      x, y ≥ 0

Solution:

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

x − y = 0, − x + 2y = 2, x = 3, y = 4, x = 0 and y = 0. 

The area shown by x − y ≥ 0 or x ≥ y:

The line x − y = 0 or x = y passes through the origin.The area to the right of the line x = y will assure the given in equation.

Now we will check by taking an example like if we take a point (4, 3) to the right of the line x = y .

Here, x ≥ y.

Thus, 

It assure the given in equation.  

Take a point (4, 5) to the left of the line x = y. Here, x ≤ y. That means it does not assure the given in equation.
 
The area shown by − x + 2y ≥ 2:

The line − x + 2y = 2 connects the coordinate axes at A(−2, 0) and B(0, 1) respectively. 

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

Thus, 

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

Thus,

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

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 = 4 is the line that passes through the point (0, 4) and is parallel to X axis. y ≤ 4 is the area below the line y = 4.
 
The area shown by x ≥ 0 and y ≥ 0:

Hence, 

All 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 x − y ≥ 0,− x + 2y ≥ 2, x ≥ 3, y ≤ 4, x ≥ 0 and y ≥ 0 are as follows.

 

The corner points of the suitable area are C (3,\ \frac{5}{2}), D(3, 3), E(4, 4) and F(6, 4).

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

Corner point

Z = x  − 5y + 20

(3,\ \frac{5}{2})

3 − 5 × \frac{5}{2}     + 20 = \frac{21}{2}

D(3, 3)D3, 3

3 − 5 × 3 + 20 = 8

E(4, 4)

4 − 5 × 4 + 20 = 4

F(6, 4)

6 − 5 × 4 + 20 = 6

Hence, 

The minimum value of Z is 4 at the point E(4, 4). 

Therefore, 

x = 4 and y = 4 is the best solution of the given LPP.

Hence, 

The best value of Z is 4.

Question 15.  Maximize Z = 3x + 5y

Subject to

x + 2y ≤ 20

  x + y ≤ 15

         y ≤ 15

     x, y ≥ 0

Solution:

Here we need to maximize Z = 3x + 5y

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

x + 2y = 20, x + y = 15, y = 5, x = 0 and y = 0.

The line x + 2y = 20 connects the coordinate axis at A(20, 0) and B(0,10). 

Connect these points to get the line x + 2y = 20.

Thus, 

(0, 0) assume the in equation x + 2y ≤ 20. 

Thus, 

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

The line x + y = 15 connect the coordinate axis at C(15, 0) and D(0,15). 

Connect these points to get the line x + y = 15.

Thus, 

(0, 0) assume the in equation x + y ≤ 15. 

Thus, 

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

y = 5 is the line passing through (0, 5) and parallel to the X axis. The area below the line y = 5 will assure the given in equation.

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 shown by the in equations.

These lines are drawn using a suitable scale.

The corner points of the suitable area are O(0, 0), C(15, 0), E(10, 5) and F(0, 5)

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

Corner point

Z = 3+ 5y

O(0, 0)

3 × 0 + 5 × 0 = 0

 C(15, 0)

3 × 15 + 5 × 0 = 45

E(10, 5)

3 × 10 + 5 × 5 = 55

F(0, 5)

3 × 0 + 5 × 5 = 25

Here we can see that the maximum value of the objective function Z is 55 which is at E(10, 5).

Therefore, 

The best value of Z is 55.

Question 16. Minimize Z = 3x1 + 5x2

Subject to

x1 + x2 ≥ 3

x1 + x2 ≥ 2

x1, x2 ≥ 0

Solution:

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

x1 + 3x2 = 3, x1 + x2 = 2, x1 = 0 and x2 = 0

The area shown by x1 + 3x2 ≥ 3 :

The line x1 + 3x2 = 3 connects the coordinate axes at A(3, 0) and B(0, 1) respectively. 

After connecting these points we will get the line x1 + 3x2 = 3.

Thus,

(0,0) does not assure the in equation x1 + 3x2 ≥ 3.

Thus,

The area in the plane which does not have the origin represents the solution set of the in equation x1 + 3x2 ≥ 3.

The area shown by x1 + x2 ≥ 2:

The line x1 + x2 = 2 connects the coordinate axes at C(2, 0) and D(0, 2) respectively. 

After connecting these points we will get the line x1 + x2 = 2.

Thus, 

(0,0) does not assure the in equation x1 + x2 ≥ 2. 

Thus,

The area having the origin represents the solution set of the in equation x1 + x2 ≥ 2.

The area shown by x1 ≥ 0 and x2 ≥ 0:

Hence, 

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

Thus, 

The first quadrant is the area shown by the in equations x1 ≥ 0 and x2 ≥ 0.

The suitable area determined by the system of constraints, x1 + 3x2 ≥ 3 , x1 + x2 ≥ 2, x1 ≥ 0, and x2 ≥ 0, are as follows.

The corner points of the suitable area are O(0, 0), B(0, 1), E (\frac{3}{2},\ \frac{1}{2}) and C(2, 0).

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

 

Corner point

Z = 3x1 + 5x2

O(0, 0)

3 × 0 + 5 × 0 = 0

B(0, 1)

3 × 0 + 5 × 1 = 5

(\frac{3}{2},\ \frac{1}{2})

3 × \frac{3}{2}    + 5 × \frac{1}{2}    = 7

C(2, 0)

3 × 2 + 5 × 0 = 6

Hence, 

The minimum value of Z is 0 at the point O(0, 0). 

Hence, 

x1 = 0 and x2 = 0 is the best solution of the given LPP.

Therefore, 

The best value of Z is 0.

Question 17.  Maximize Z = 2x + 3y

Subject to

      x + y ≥ 1

  10x + y ≥ 5

  x + 10y ≥ 1

         x, y ≥ 0

Solution:

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

x + y = 1, 10x +y = 5, x + 10y = 1, x = 0 and y = 0

The area shown by x + y ≥ 1:

The line x + y = 1 connects the coordinate axes at A(1, 0) and B(0,1) respectively. 

After connecting these points we will get the line x + y = 1.

Thus,

(0,0) does not assure the in equation x + y ≥ 1. 

Thus,

The area in xy plane which does not have the origin represents the solution set of the in equation x + y ≥ 1.

The area shown by 10x +y ≥ 5:

The line 10x +y = 5 connect the coordinate axes at C (\frac{1}{2},\ 0)    and D(0, 5) respectively. 

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

Thus,

(0,0) does not assure the in equation 10x +y ≥ 5. 

Thus, the area which does not have the origin represents the solution set of the in equation 10x +y ≥ 5.

The area shown by x + 10y ≥ 1:

The line x + 10y = 1 connect the coordinate axes at A(1, 0) and F (0,\ \frac{1}{10})    respectively. 

After connecting these points we will have the line x + 10y = 1.

Thus, 

(0,0) does not assure the in equation x + 10y ≥ 1. 

Thus, the area which does not have the origin represents the solution set of the in equation x + 10y ≥ 1.

The area shown by x ≥ 0 and y ≥ 0:

Here, 

All the points 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 x + y ≥ 1, 10x +y ≥ 5, x + 10y ≥ 1, x ≥ 0, and y ≥ 0, are as follows.

The suitable area is unbounded.

Hence, 

The maximum value is infinity i.e. the solution is unbounded.

Question 18. Maximize Z = -x1 + 2x2

Subject to

  -x1 + 3x2 ≤ 10

     x1 + x2 ≤ 6

       x1 – x2 ≤ 2

        x1, x2 ≥ 0

Solution:

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

−x1 + 3x2 = 10, x1 + x2 = 6, x1 + x2 = 2, x1 = 0 and x2 = 0

The area shown by −x1 + 3x2 ≤ 10:

The line −x1 + 3x2 = 10 coincide the coordinate axes at A(−10, 0) and B (0,\ \frac{10}{3})    respectively. 

After connecting these points we will get the line −x1 + 3x2 = 10.

Thus, 

(0,0) satisfies the in equation −x1 + 3x2 ≤ 10 .

Thus,

The area region in the plane which have the origin shows the solution set of the in equation

−x1 + 3x2 ≤ 10.

The area shown by x1 + x2 ≤ 6:

The line x1 + x2 = 6 connects the coordinate axes at C(6, 0) and D(0, 6) respectively. 

After connecting these points we will get the line x1 + x2 = 6.

Thus, 

(0,0) assure the in equation x1 + x2 ≤ 6. 

Thus,

The area having the origin represents the solution set of the in equation x1 + x2 ≤ 6.

The area shown by x1− x2 ≤ 2:

The line x1 − x2 = 2 coincide the coordinate axes at E(2, 0) and F(0, −2) respectively. 

After connecting these points we will get the line x1 − x2 = 2.

Thus,

(0,0) assure the in equation x1− x2 ≤ 2. 

Thus,

The area having the origin represents the solution set of the in equation x1− x2 ≤ 2.

The area shown by x1 ≥ 0 and x2 ≥ 0:

Hence, 

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

Thus, 

The first quadrant is the region shown by the in equations x1 ≥ 0 and x2 ≥ 0.

The suitable area determined by the system of constraints, −x1 + 3x2 ≤ 10, x1 + x2 ≤ 6, x1− x2 ≤ 2, x1 ≥ 0, and x2 ≥ 0, are as follows.

The corner points of the assure area are O(0, 0), E(2, 0), H(4, 2), G(2, 4) and B (0,\ \frac{10}{3})   .

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

 

Corner point

Z = −x1 + 2x2

O(0, 0)

1 × 0 + 2 × 0 = 0

E(2, 0)

1 × 2 + 2 × 0 = 2

H(4, 2)

1 × 4 + 2 × 2 = 0

G(2, 4)

1 × 2 + 2 × 4 = 6

(0,\ \frac{10}{3})

1 × 0 + 2 × \frac{10}{3}    = \frac{20}{3}

Here we can see that the maximum value of the objective function Z is \frac{20}{3} which is at B (0,\ \frac{10}{3}).

Question 19. Maximize Z = -x + y

Subject to

-2x + y ≤ 1

         x  ≤ 2

    x + y ≤ 3

       x, x ≥ 0

Solution:

Here we need to maximize Z = x + y

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

−2x + y = 1, x = 2, x + y = 3, x = 0 and y = 0.

The line −2x + y = 1 coincide the coordinate axis at A (\frac{-1}{2},\ 0)    and B(0, 1). 

Now connect these points to get the line −2x + y = 1 .

Thus, 

(0, 0) assure the in equation −2x + y ≤ 1. 

Thus, 

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

x = 2 is the line passing through (2, 0) and parallel to the Y axis.

The area below the line x = 2 will assure the given in equation.

The line x + y = 3 coincide the coordinate axis at C(3, 0) and D(0, 3). 

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

Thus, 

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

Thus, 

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

The area shown by x ≥ 0 and y ≥ 0:

Hence, 

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

Thus, 

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

These lines are drawn using a satisfactory scale.

The corner points of the suitable area are O(0, 0),G(2, 0), E(2, 1) and F (\frac{2}{3},\ \frac{7}{3})

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

Corner point

Z = y

O(0, 0)

 0 +  0 = 0

 C(2, 0)

2 + 0 = 2

E(2, 1)

2 +1 = 3

(\frac{2}{3},\ \frac{7}{3})

\frac{2}{3}+\frac{7}{3}=\frac{9}{3}=3

Here we can see that the maximum value of the objective function Z is 3 which is at E(2, 1) and F (\frac{2}{3},\ \frac{7}{3})   .

Therefore, 

The best value of Z is 3.

Question 20.  Maximize Z = -3x1 + 4x2

Subject to

     x1 – x2 ≤ -1

   x1 + x2  ≤ 0

       x1, x2 ≥ 0

Solution:

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

x1 − x2 = −1, −x1 + x2 = 0, x1 = 0 and x2 = 0

The area shown by x1 − x2 ≤ −1:

The line x1 − x2 = −1 coincide the coordinate axes at A(−1, 0) and B(0, 1) respectively. 

After connecting these points we will get the line x1 − x2 = −1.

Thus, 

(0,0) does not assures the in equation x1 − x2 ≤ −1 .

Thus,

The area in the plane which does not have the origin shows the solution set of the in equation x1 − x2 ≤ −1.

The area shown by −x1 + x2 ≤ 0 or x1 ≥ x2:

The line −x1 + x2 = 0 or x1 = x2 is the line passing through (0, 0).

The area to the right of the line x1 = x2 will assure the given in equation −x1 + x2 ≤ 0.

If we mark a point (1, 3) to the left of the line x1 = x2

Here, 1≤3 which is not assuring the in equation x1 ≥ x2

Hence, 

The area to the right of the line x1 = x2 will assure the given in equation −x1 + x2 ≤ 0.

The area shown by x1 ≥ 0 and x2 ≥ 0:

Thus, 

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

Thus, 

The first quadrant is the area shown by the in equations x1 ≥ 0 and x2 ≥ 0.

The suitable area determined by the system of constraints, x1 − x2 ≤ −1, −x1 + x2 ≤ 0, x1 ≥ 0, and x2 ≥ 0, are as follows.

Here we can see that the suitable area of the given LPP does not exist.


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Last Updated : 20 May, 2021
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