Chapter 7 Linear Programming Miscellaneous Exercise 7 Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 7 Linear Programming Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Miscellaneous Exercise 7

I) Select the appropriate alternatives for each of the following :
Question 1.
The value of objective function is maximum under linear constraints _______.
(A) at the centre of feasible region
(B) at (0, 0)
(C) at a vertex of feasible region
(D) the vertex which is of maximum distance from (0, 0)
Solution:
(C) at a vertex of feasible region

II) Solve the following :
Question 1.
Solve each of the following inequations graphically using X Y plane.
(i) 4x – 18 ≥ 0
Solution:
Consider the line whose equation is 4x – 18 ≥ 0 i.e. x = = 4.5
This represents a line parallel to Y-axis passing3through the point (4.5, 0)
Draw the line x = 4.5
To find the solution set we have to check the position of the origin (0, 0).
When x = 0, 4x – 18 = 4 × 0 – 18 = -18 > 0
∴ the coordinates of the origin does not satisfy thegiven inequality.
∴ the solution set consists of the line x = 4.5 and the non-origin side of the line which is shaded in the graph.

(ii) -11x – 55 ≤ 0
Solution:
Consider the line whose equation is -11x – 55 ≤ 0 i.e. x = -5
This represents a line parallel to Y-axis passing3through the point (-5, 0)
Draw the line x = – 5
To find the solution set we have to check the position of the origin (0, 0).
When x = 0, -11x – 55 = – 11(0) – 55 = -55 > 0
∴ the coordinates of the origin does not satisfy thegiven inequality.
∴ the solution set consists of the line x = -5 and the non-origin side of the line which is shaded in the graph.

(iii) 5y – 12 ≥ 0
Solution:
Consider the line whose equation is 5y – 12 ≥ 0 i.e. y =
This represents a line parallel to X-axis passing through the point (o, )
Draw the line y =
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 5y – 12 = 5(0) – 12 = -12 > 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = and the non-origin side of the line which is shaded in the graph.

(iv) y ≤ -3.5
Solution:
Consider the line whose equation is y ≤ – 3.5 i.e. y = – 3.5
This represents a line parallel to X-axis passing3through the point (0, -3.5)
Draw the line y = – 3.5
To find the solution set, we have to check the position of the origin (0, 0).
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = – 3.5 and the non-origin side of the line which is shaded in the graph.

(ii) x + y≤ 0
Solution:

(iii) 2y – 5x ≥ 0
Solution:

(iv) |x + 5| ≤ y
Solution:
|x + 5| ≤ y
∴ -y ≤ x + 5 ≤ y
∴ -y ≤ x + 5 and x + 5 ≤ y
∴ x + y ≥ -5 and x – y ≤ -5
First we draw the lines AB and AC whose equations are
x + y= -5 and x – y = -5 respectively.

The graph of |x + 5| ≤ y is as below:

(ii) x + 2y ≥ 4, 2x – y ≤ 6
Solution:

(iii) 3x + 4y ≤ 12, x – 2y ≥ 2, y ≥ -1
Solution:
First we draw the lines AB, CD and ED whose equations are 3x + 4y = 12, x – 2y = 2 and y = -1 respectively.


The solution set of given system of inequation is shaded in the graph.

(ii) 3x + 4y ≥ 12, 4x + 7y ≤ 28, x ≥ 0, y ≥ 0
Solution:

The feasible solution is ACDBA.

(ii) Maximize z = 4x + 2y subject to 3x + y ≥ 27, x + y ≥ 21
Question is modified.
Maximize z = 4x + 2y subject to 3x + y ≤ 27, x + y ≤ 21, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + y = 27 and x + y = 21 respectively.


The feasible region is OAPDO which is shaded region in the graph. The vertices of the feasible region are 0(0, 0), A (9, 0), P and D(0, 21). P is the point of intersection of lines
3x + y = 27 … (1)
and x + y = 21 … (2)
On substracting, we get 2x = 6 ∴ x = 3
Substituting x = 3 in equation (1), we get
9 + y = 27 ∴ y = 18
∴ P = (3, 18)
The values of the objective function z = 4x + 2y at these vertices are
z(O) = 4(0) + 2(0) = 0 + 0 = 0
z(a) = 4(9) + 2(0) = 36 + 0 = 36
z(P) = 4(3) + 2(18) = 12 + 36 = 48
z (D) = 4(0) + 2(21) = 0 + 42 = 42
∴ 2 has minimum value 48 when x = 3, y = 18.

(iii) Maximize z = 6x + 10y subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + 5y = 10 and 5x + 3y = 15 respectively.


The feasible region is OCPBD which is shaded in the graph.
The vertices of the feasible region are 0(0, 0), C(3, 0), P and B (0, 2).
P is the point of intersection of the lines
3x + 5y = 10 … (1)
and 5x + 3y = 15 … (2)
Multiplying equation (1) by 5 and equation (2) by 3, we get
15x + 25y = 50
15x + 9y = 45
On subtracting, we get
16y = 5 ∴ y =
Substituting y = in equation (1), we get
3x + = 10 ∴ 3x = 10 –
∴ x = ∴ P ≡
The values of objective function z = 6x + 10y at these vertices are
z(O) = 6(0) + 10(0) = 0 + 0 = 0
z(C) = 6(3) + 10(0) = 18 + 0 = 18
z(P) = 6 + 10 = = 20
z(B) = 6(0) + 10(2) = 0 + 20 = 20
The maximum value of z is 20 at P and B (0, 2) two consecutive vertices.
∴ z has maximum value 20 at each point of line segment PB where B is (0, 2) and P is .
Hence, there are infinite number of optimum solutions.

(iv) Maximize z = 2x + 3y subject to x – y ≥ 3, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB whose equation is x – y = 3.


The feasible region is shaded which is unbounded.
Therefore, the value of objective function can be in- j creased indefinitely. Hence, this LPP has unbounded solution.

(ii) Maximize z = 60x + 50y subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
Solution:
We first draw the lines AB and CD whose equations are x + 2y = 40 and 3x + 2y = 60 respectively.


The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (20, 0), P and B (0, 20).
P is the point of intersection of the lines.
3x + 2y = 60 … (1)
and x + 2y = 40 … (2)
On subtracting, we get
2x = 20 ∴ x = 10
Substituting x = 10 in (2), we get
10 + 2y = 40
∴ 2y = 30 ∴ y = 15 ∴ P is (10, 15)
The values of the objective function z = 60x + 50y at these vertices are
z(O) = 60(0) + 50(0) = 0 + 0 = 0
z(C) = 60(20) + 50(0) = 1200 + 0 = 1200
z(P) = 60(10) + 50(15) = 600 + 750 = 1350
z(B) = 60(0) + 50(20) = 0 + 1000 = 1000.
∴ z has maximum value 1350 at x = 10, y = 15.

(iii) Maximize z = 4x + 2y subject to 3x + y ≥ 27, x + y ≥ 21, x + 2y ≥ 30; x ≥ 0, y ≥ 0
Solution:
We first draw the lines AB, CD and EF whose equations are 3x + y = 27, x + y = 21, x + 2y = 30 respectively.


The feasible region is XEPQBY which is shaded in the graph.
The vertices of the feasible region are E (30,0), P, Q and B (0,27).
P is the point of intersection of the lines
x + 2y = 30 … (1)
and x + y = 21 … (2)
On subtracting, we get
y = 9
Substituting y = 9 in (2), we get
x + 9 = 21 ∴ x = 12
∴ P is (12, 9)
Q is the point of intersection of the lines
x + y = 21 … (2)
and 3x + y = 27 … (3)
On subtracting, we get
2x = 6 ∴ x = 3
Substituting x = 3 in (2), we get
3 + y = 21 ∴ y = 18
∴ Q is (3, 18).
The values of the objective function z = 4x + 2y at these vertices are
z(E) = 4(30) + 2(0) = 120 + 0 = 120
z(P) = 4(12) + 2(9) = 48 + 18 = 66
z(Q) = 4(3) + 2(18) = 12 + 36 = 48
z(B) = 4(0) + 2(27) = 0 + 54 = 54
∴ z has minimum value 48, when x = 3 and y = 18.