Mathematics Part Ii Chapter 12 Linear Programming

We are to maximise
Z = 3x + 4 y
subject to the constraints
x + y ≤ 4
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + y = 4
For x = 0, y = 4
For y = 0, x = 4
∴ line meets OX in A(4, 0) and OY in L(0, 4)

Since feasible region is the region which satisfies all the constraints.
∴  OAL is the feasible region. The comer points are O(0, 0), A(4, 0), L(0, 4).
At O(0, 0), Z = 0 + 0 = 0
At A(4, 0), Z = 12 + 0 = 12
At L(0, 4), Z = 0 + 16 = 16
∴ maximum value = 16 at (0, 4).

We are to maximise
Z = 4x + y
subject to the constraints
x + y ≤ 50
3x + y ≤ 90
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of the line x + y = 50
For x = 0, y = 50
For y = 0, x = 50
∴ line meets OX in A(50, 0) and OY in L(0, 50)
Let us draw the graph of line 3 x + y = 90
For x = 0, y = 90
For y = 0, 3x = 90 or x = 30
∴ line meets OX in B(30, 0) and OY in M(0, 90).
Since feasible region is the region which satisfies all the constraints.
∴  OBCL is the feasible region, which is bounded.

Tips: -

Note: Coordinates of C can be found by two methods:
Method I: Draw the graph of inequalities on the graph paper. So coordinates of C can be determined.
Method II: Solve the two equation x + y = 50, 3x + y = 90 by any method to find coordinates of C.

We are to maximize
f = x + 2y
subject to the constraints
2x + 3 y ≤ 6
x + 4 y ≤ 4
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of the line 2 x + 3 y = 6.
For x = 0, 3 y = 6, or y = 2
For y = 0, 2 x = 6, or x = 3
∴  line meets OX in A (3, 0) and OY in L (0, 2)
Let us draw the graph of line x + 4 y = 4
For x = 0, 4 y = 4, or y = 1
For y = 0, x = 4
∴  line meets OX in B (4, 0) and OY in M (0, 1)

Since feasible region is the region which satisfies all the constraints
∴    OACM is the feasible region. The comer points are
     
At 

We have to maximize
z = 9x + 3 y
subject to the constraints
2x + 3 y ≤ 13
2x + y ≤ 5
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0,y ≥ 0 lies in the first quadrant.
Let us draw the graph of 2x + 3y = 13
For x = 0,  3y = 13  
For y = 0,  2x = 13    

Again we draw the graph of 2x + y = 5
For x = 0,  y = 5
For y = 0,  2x = 5  


Since feasible region satisfies all the constraints.
    OCEB in the feasibe region. The corner points are O(0, 0),   
       At O(0, 0),  z = 9(0) + 3(0) = 0+ 0 = 0
At   
At   

We are to minimise
Z = - 3x + 4 y
subject to the constraints
x + 2 y ≤ 8
3x + 2 y ≤ 12
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + 2 y = 8
For x = 0, 2 y = 8 or y = 4
For y = 0, x = 8
∴  line meets OX in A(8, 0) and OY in L(0, 4).
Again we draw the graph of 3x + 2 y = 12
For x = 0, 2 y = 12 or y = 6
For y = 0, 3x = 12 or x = 4
∴  line meets OX in B(4, 0) and OY in M(0, 6).
Since feasible region is the region which satisfies all the constraints.

∴ OBCL is the feasible region and O(0, 0), B(4, 0), C(2, 3), L(0, 4) are comer points.
At O(0, 0), Z = 0 + 0 = 0
At B(4, 0), Z = -12 + 0 = -12
At C(2, 3), Z = -6 + 12 = 6
At L(0, 4), Z = 0 + 16 = 16
∴  minimum value = -12 at (4, 0).

We have to maximise
Z = 5x + 3 y
subject to the constraints
3x + 5 y ≤ 15
5x + 2 y ≤ 10
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of 3x + 5 y= 15
For x = 0, 5 y = 15 or y = 3
For y = 0, 3x = 15 or x = 5
∴ line meets OX in A(5, 0) and OY in L(0, 3).
Again we draw the graph of 5x + 2 y = 10
For x = 0, 2 y = 10 or y = 5
For y = 0, 5x = 10 or x = 2
∴ line meets OX in B(2, 0) and OY in M(0, 5).
Since feasible region is the region which satisfies all the constraints.

∴ OBCL is the feasible region and corner points are O(0, 0), B(2, 0),
    
At                                 
At                                  
At  

We are to minimise
Z = 3x + 2y
subject to the constraints
x + 2y ≤ 10
3x + y ≤ 15
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + 2y = 10.
For x = 0, 2y = 10 or y = 5
For y = 0, x = 10
∴  line meets OX in A(10, 0) and OY in L(0, 5).
Again we draw the graph of 3x + y = 15.
For x = 0, y = 15
For y = 0, 3 x = 15 or x = 5
∴  line meets OX in B(5, 0) and OY in M(0, 15).
Since feasible region is the region which satisfies all the constraints.
∴  OBCL is the feasible region.
The comer points O(0, 0), B(5, 0), C(4, 3), L(0, 5).
At O(0, 0), Z = 0 + 0 = 0
At B(5, 0), Z = 15 + 0 = 15
At C(4, 3), Z = 12 + 6 = 18
At L(0, 5), Z = 0 + 10 = 10
∴  maximum value = 18 at (4, 3).

We are to maximize
z = 4x + 7y
subject to the constraints
x + 2y ≤ 20
x + y ≤ 15
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of line
x + 2y = 20.
For x = 0, 2y = 20 or y = 10
For y = 0, x = 20
∴  line meets OX in A (20, 0) and OY in L (0, 10).
Let us draw the graphs of line
x + y = 15.
For x = 0, y = 15
For y = 0, x = 15
∴  line meets OX in B (15, 0) and OY in M (0, 15).

Since feasible region is the region which satisfies all the constraints
∴ OBCL is the feasible region, which is bounded.
The corner points are O (0, 0), B (15, 0), C (10, 5), L (0, 10)
At O (0, 0), z = 0 + 0 = 0
At B (15, 0), z = 4 (15) + 7 (0) = 60 + 0 = 60
At C(10, 5), z = 4 (10) + 7 (5) = 40 + 35 = 75
At L (0, 10), z = 4 (0) + 7 (10) = 0 + 70 = 70
∴ maximum value = 75 at the point (10, 5).

We are to maximize
z = 11x + 5y
subject to the constraints
3x + 2y  ≤ 25
x + y ≤ 10
x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 3x + 2y = 25
For x = 0,   2y = 25   or   
For y = 0,  3x = 25   or   


Again we draw the graph of x + y = 10
For x = 0, y = 10
For y = 0, x = 10
∴  line meets OX in B (10, 0) and OY in M (0, 10).

Since feasible region is the region which satisfies all the constraints
∴   OACM is the feasible region and corner points are

At  
At  

We are to maximize
z = 30x + 19y
subject to the constraints
 x + y ≤ 24,

 x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axis OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 24.
For x = 0, y = 24
For y = 0, x = 24
∴  line meets OX in A (24, 0) and OY in L(0, 24).
Again we draw the graph of 
For  
For  y = 0,  x = 16
∴ line meets OX in B (16, 0) and OY in M (0, 32).

Since feasible region is the region which satisfies all the constraints
∴ OBCL is the feasible region and O (0, 0), B (16, 0), C (8, 16), L (0, 24) are corner points.
At O(0, 0), z = 30(0) + 19 (0) = 0 + 0 = 0
At B(16, 0), z = 30 (16) + 19 (0) = 480 + 0 = 480
At C (8, 16), z = 30 (8) + 19 (16) = 240 + 304 = 544
At L (0, 24), z = 30 (0) + 19 (24) = 0 + 456 = 456
∴ maximum value = 544 at (8, 16).

We are to minimize
z = 2x + 3y
subject to the constraints
x + 2y ≤ 10
x + 2y ≥ 1
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + 2y = 10
For x = 0, 2y =10 or y = 5
For y = 0, x = 10
∴  line meets OX in A (10, 0) and OY in L (0, 5).
Again we draw the graph of x + 2y = 1
For x = 0, 2 y = 1 or y = 0.5
For y = 0, x = 1
∴ line meets OX is B (1, 0) and OY in (0, 0.5)

Since feasible region satisfies all the constraints
∴ in this case, BALM is the feasible region and corner points are
B(1, 0), A(10, 0), L(0, 5), M(0, 0.5).
At B(1, 0), z = 2(1) + 3(0) = 2 + 0 = 2
At A(10, 0), z = 2 (10) + 3 (0) = 20 + 0 = 20
At L(0, 5), z = 2(0) + 3(5) = 0 +15 = 15
At M(0, 0.5), z = 2 (0) + 3 (0.5) = 0 + 1.5 = 1.5
∴ minimum value = 1.5 at (1, 0).

We are to minimise
Z = 200x + 500 y
subject to the constraints
x + 2y ≥ 10
3x + 4y ≤ 24
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of the line x + 2y = 10
For  x = 0, 2 y = 10 or y = 5
For y = 0, x = 10
∴ line meets OX in A(10, 0) and OY in L(0, 5).
Let us draw the graph of the line 3x + 4y = 24
For x = 0, 4 y = 24 or y = 6
For y = 0, 3x = 24 or x = 8
∴ line meets OX in B(8, 0) and OY in M(0, 6)
Since feasible region is the region which satisfies all the constraints.

∴  CML is the feasible region, which is bounded.
The comer points are C(4, 3), M(0, 6), L(0, 5).
At C(4, 3), Z = 800 + 1500 = 2300
At M(0, 6), Z = 0 + 3000 = 3000
At L(0, 5), Z = 0 + 2500 = 2500
∴ minimum value = 2300 at (4, 3).

We are minimise and maximise
Z = 3x + 9 y
subject to constraints
x + 3 y ≤ 60
x + y ≥ 10
x ≤ y
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of the line x + 3y = 60
For x = 0, 3 y = 60 or y = 20
For y = 0, x = 60
∴ line meets OX in A(60, 0) and OY in L(0, 20)
Let us draw the graph of
x + y = 10
For  x = 0, y = 10
For y = 0, x = 10

∴ line meets OX in B(10, 0) and OY in M(0, 10).
Again we draw the graph of x = y
This is a straight line passing through O and meeting AL in C(15, 15) and BM in D(5, 5).
Since feasible region is the region which satisfies all the constraints.
∴ DCLM is the feasible region, which is bounded. The corner points are D(5, 5), C(15, 15), L(0, 20), M(0, 10).
At D(5, 5), Z = 15 + 45 = 60
At C(15, 15), Z = 45 + 135 = 180
At L(0, 20), Z = 0 + 180 = 180
At M(0, 10), Z = 0 + 90 = 90
∴ minimum value = 60 at (5, 5)
and maximum value = 180 at (15, 15) or (0, 20).

Tips: -

We are to minimise and maximise Z = 5x + 10y subject to constraints x + 2 y ≤ 120
x + y ≥ 60
x - 2y ≥ 0
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + 2 y = 120
For x = 0, 2 y = 120 or y = 60
For y = 0, x = 120
∴ line meets OX in A( 120, 0) and OY in L(0, 60).
Also we draw the graph of
x + y = 60.
For r = 0, y = 60
For y = 0, x = 60
∴ line meets OX in B(60, 0) and OY in L(0, 60).
Again we draw the graph of
x - 2y = 0
This is a line through the origin and C(40, 0), which is point of intersection of x - 2 y = 0 and x + y = 60

Since feasible region satisfies all the constraints.
∴ BADC is the feasible region.
Comer points are B(60, 0), A(120, 0), D(60, 30), C(40, 20).
At B(60, 0), Z = 300 + 0 = 300
At A(120, 0), Z = 600+ 0 = 600
At D(60, 30), Z = 300 + 300 = 600
At C(40, 20), Z = 200 + 200 = 400
∴ minimum value = 300 at (60, 0) and maximum value = 600 at (120, 0) and (60, 30).

We are to minimise
z = 5x + 7y
subject to the constraints
2x + y ≥ 8
x + 2y ≥ 10
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of
2x + y = 8
For x = 0, y = 8
For y = 0, 2 x = 8 or x = 4
∴ line meets OX in A (4, 0) and OY in L (0, 8).

Again we draw the graph of
x + 2y = 10.
For x = 0, 2y = 10 or y = 5
For y = 0, x = 10
∴ line meets OX in B (10, 0) and OY in M (0, 5)
Since feasible region is the region which satisfies all the constraints
∴ shaded region is the feasible region and comer points are B (10, 0), C (2, 4), L (0, 8).
At B (10, 0), z = 5 (10) + 7 (0) = 50 + 0 = 50
At C (2, 4), z = 5 (2) +7 (4) = 10 + 28 = 38
At L (0, 8), z = 5 (0) + 7 (8) = 0 + 56 = 56
∴ 38 is the smallest value of z at (2, 4)
Since feasible region is unbounded
∴ we are to check whether this value is minimum.
For this we draw the graph of
5x + 7y < 38    ...(1)
Since (1) has no common point with feasible region.
∴ minimum value = 38 at (2, 4).

We are to minimise and maximise
Z = x + 2y subject to constraints x + 2y ≥ 100, 2x - y ≤ 0, 2x + y ≤ 200, x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + 2 y = 100
For x = 0, 2 y = 100 or y = 50
For y = 0, x = 100
∴  line meets OX in A(100, 0) and OY in L(0, 50).
2x - y = 0 is a straight line passing through origin and C(20, 4), which is point of intersection of 2x - y = 0 and x + 2y = 100.
Again we draw the graph of 2x + y = 200.
For x = 0, y = 200
For y = 0, 2x = 200 or x = 100
∴ line meets OX in A(100, 0) and OY in M(0, 200).
Since feasible region satisfies all the constraints.
∴ CDML is the feasible region.
The comer points are C(20, 40), D(50. 100), M(0, 200), L(0, 50).
At C(20, 40), Z = 20 + 80 = 100
At D(50, 100), Z = 50 + 200 = 250
At M(0, 200), Z = 0 + 400 = 400
At L(0, 50), Z = 0 + 100 = 100

∴  maximum value = 400 at (0, 200)
and minimum value = 100 at (20, 40) and (0, 50) i.e. along the segment joining (20, 40) and (0, 50).

We are to minimise
Z = - 50x + 20 y subject to the constraints 2x - y ≥ - 5, 3x + y ≥ 3, 2x - 3y ≤ 12, x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 2x - y = - 5
For x = 0,   - y = -5   or  y = 5
For y = 0,  2x =  -5,  or  x = -
 line meets OX in 
Now we draw the graph of 3x + y = 3
For x = 0, y = 3
For y = 0, 3x = 3 or x = 1
∴ line meets OX in B(1, 0) and OY in M(0, 3)
Again we draw the graph of 2x - 3 y - 12
For x = 0, - 3 y = 12 or y = - 4
For y = 0, 2 x = 12 or x = 6
∴ line meets OX in C(6, 0) and OY in N(0, - 4)
Since feasible region is the region which satisfies alt the constraints.
∴  feasible region (shaded) is unbounded and has corner points B(1, 0), C(6, 0), L(0, 5), M(0,3)
At B(1, 0), Z = - 50 + 0 = - 50
At C(6, 0), Z = - 300 + 0 = - 300
At L(0, 5), Z = 0 + 100 = 100
At M(0, 3), Z = 0 + 60 = 60
∴ - 300 is the smallest value of Z at the comer point (6, 0).
Since the feasible region is unbounded.
∴ we are to check whether this value is minimum.

For this, we draw the graph of
- 50x + 20 y < - 300 i.e. - 5x + 2y < - 30    ...(1)
Consider - 5x + 2 y = - 30
This line passes through C(6, 0), D(8, 5).
Now (1) has common points with feasible region.
∴  Z = - 50x + 20y has no minimum value.

We are to minimise,  Z = 3x + 5y subject to the constraints x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + 3y = 3
For x = 0, 3 y = 3 or y = 1
For y = 0, x = 3
∴ line meets OX in A(3, 0) and OY in L(0, 1).
Again we draw the graph of x + y = 2
For x = 0, y = 2
For y = 0, x = 2
∴ line meets OX in B(2, 0) and OY in M(0, 2).
Since feasible region is the region which satisfies all the constraints.
∴ shaded region is the feasible region which is unbounded and has comer points
   
                  

Since feasible region is unbounded.
∴  we are to check whether this value is minimum.
For this we draw the graph of
3x + 5y < 7    ...(1)
Since (1) has n.o common point with feasible region.

We are to minimise Z = x + 2y subject to the constraints 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of 2x + y = 3
For     x = 0,     y = 3
For    y = 0,      2x = 3     or   x = 

Again we draw the graph of x + 2 y = 6.
For x = 0, 2 y = 6 or y = 3
For y = 0, x = 6
∴ line meets OX in B(6, 0) and OY in L(0, 3).
Since feasible region is the region which satisfies all the constraints.
∴ shaded region is the feasible region and comer points are B(6, 0), L(0, 3).
At B(6, 0), Z = 6 + 0 = 6
At L(0, 3), Z = 0 + 6 = 6
∴ 6 is the greatest value of Z at (6, 0) and (0, 3) and so on the line BL.
Since feasible region is unbounded.
∴ we are to check whether this value is maximum.
    

For this we draw the graph of
x + 2y < 6    ...(1)
Since (1) has no point in common with the feasible region.
∴ minimum value = 6 at all points on the line segment joining the points (6, 0) and (0, 3).
∴  minimum of Z occurs at more than two points.

We are to maximise
Z = - x + 2 y subject to constraints x ≥ 3, x + y ≥ 5, x + 2 y ≥ 6, y ≥ 0.
y = 0 is x-axis.
So y ≥ 0 represents region on and above x-axis.
Now x = 3 is a straight line AL parallel to y-axis at a distance 3.
Now we draw the graph of x + y = 5
For x = 0, y - 5
For y = 0, x = 5
∴ line meets OX in B(5, 0) and OY in M(0, 5).
Again we draw the graph of x + 2y = 6.
For x = 0, 2y = 6 or y = 3
For y = 0, x = 6
∴  line meets OX in C(6, 0) and OY in N(0, 3).
Since feasible region satisfies all the constraints.
             

∴  shaded region is the feasible region, which is unbounded.
The comer points are C(6, 0), D(4, 1), E(3, 2).
At C(6, 0), Z = - 6 + 0 = - 6
At D(4, 1), Z = - 4 + 2 = - 2
At E(3, 2), Z = - 3 + 4 = 1
∴ greatest value of Z is 1 at (3, 2).
Since feasible region is unbounded.
∴ we are to check whether this value is maximum.
For this, we draw the graph of - x + 2y > 1    ...(1)
Since (1) has common points with the feasible region.
∴  Z has no maximum value.

We are to minimise Z = 3x + 2y subject to the constraints x + y ≥ 8 , 3x + 5 y ≤ 15, x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + y = 8
For x = 0, y = 8
For y = 0, x = 8
∴ line meets OX in A(8, 0) and OY in L(0, 8).
Again we draw the graph of 3x + 5y = 15

For x = 0, 5y = 15, or y = 3
For y = 0, 3x = 15 or x = 5
∴ line meets OX in B(5, 0) and OY in M(0, 3).
Since feasible region satisfies all the constraints.
∴ in this case, feasible region is empty and hence no feasible solution.

We are to maximise
Z = x + y subject to the constraints x - y ≤ - 1, - x + y ≤ 0, x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of
x - y = - 1
For x = 0, - y = - 1 ⇒ y = 1
For y = 0, x = - 1
∴  line x - y = 1 meets OX in A (- 1, 0) and OY in B (0, 1)
Again we draw the graph of - x + y = 0
For x = 0, y = 0
For y = 0, x = 0
∴ line - x + y = 0 passes through O (0, 0).

Since feasible region satisfies all the constraints
∴  in this case, feasible region is empty
∴ there exists no solution to the given linear programming problem.

Let x be the number of tables and y be the number of chairs.
Let Z be the profit
∴    We are to maximize
Z = 50x + 15y subject to the constraints
250x + 50y ≤ 5000
x + y ≤ 60
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of 250 x + 50y = 5000
For x = 0, 50 y = 5000 ⇒ y = 100
For y = 0, 250 x = 5000 ⇒ x = 20
∴ line 250 x + 50 y = 5000 meets OX in A (20, 0) and OY in B (0, 100)
Again we draw the graph of the line x + y = 60
For x = 0, y = 60
For y = 0, x = 60
∴ line x + y = 60 meets OX in C (60, 0) and OY in D (0, 60).
Since feasible is the region which satisfies all the constraints
∴ feasible region is the quadrilateral OAED. The comer points are O (0, 0). A (20, 0), E (10, 50), D (0, 60)

At O (0, 0), Z = 0 + 0 = 0
At A (20, 0), Z = 50 (20) + 15 (0) = 1000 + 0 = 1000
At E(10, 50), Z = 50 (10) + 15 (50) = 500 + 750= 1250
At D(0, 60), Z = 50 (0) + 15 (60) = 0 + 900 = 900
∴  maximum value of Z = 1250 at E (10, 50)
∴  maximum profit = Rs. 1250
when x = 10, y = 50 i.e., when number of tables = 10, number of chairs = 50

Let x hectare of land be allocated to crop X and y hectare to crop Y. Clearly, x ≥ 0, y ≥ 0.
Profit per hectare on crop X = Rs. 10500
Profit per hectare on crop Y = Rs. 9000
∴  total profit = Rs. (10500 x + 9000 y)
The mathematical formulation of the problem is as follows:
Maximise    Z = 10500 x + 9000 y
subject to constraints x + y ≤ 50, 20x + 10y ≤ 800
i.e. 2 x + y ≤ 80
and    x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 50
For x = 0, y = 50
For y = 0, x = 50
∴ line meets OX in A(50, 0) and OY in L(0, 50).
Again we draw the graph of 2x + y = 80.
For x = 0, y = 80
For y = 0, 2x = 80 or x = 40

∴  line meets OX in B(40, 0) and OY in M(0, 80).
Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The corner points are O(0, 0), B(40, 0), C(30, 20), L(0, 50).
At O(0, 0), Z = 0 + 0 = 0
At B(40, 0), Z = 420000 + 0 = 420000
At C(30, 20), Z = 315000 + 180000 = 495000
At L(0, 50), Z = 0 + 450000 = 450000
∴ maximum value = 495000 at (30, 20)
∴ society will get the maximum profit of Rs. 495000 by allocating 30 hectares for crop X and 20 hectares for crop Y.

Let x be the number of pieces of Model A and y be the number of pieces of Model B. Then
Total profit = Rs. (8000 x + 12000 y)
Let    Z = 8000 x + 12000 y
The mathematical formulation of the problem is as followings:
Maximise Z = 8000 x + 12000 y subject to constraints 9x + 12y ≤ 180
i.e. 3x + 4 y ≤ 60
x + 3y ≤ 30
x ≥ 0, y ≤ 0
Now we draw the graph of 3x + 4y = 60
For x = 0, 4y = 60 or y = 15
For y = 0, 3x = 60 or x = 20
∴  line meets OX in A (20, 0) and OY in B(0, 15).
Also we draw the graph of x + 3y = 30

Let x and y be the number of cakes of first and second type that can be made. Clearly x ≥ 0, y ≥ 0.
Let Z be the number of cakes.
Table

Mathematical formulation of the problem is as follows:
Maximise Z = x + y subject to the constraints:
200x + 100y ≤ 5000 i.e. 2x + y ≤ 50
25x + 50y ≤ 1000 i.e. x + 2y ≤ 40
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 2x + y = 50
For x = 0, y = 50
For y = 0, 2 x = 50 or x = 25
∴  line meets OX in A(25, 0) and OY in L(0, 50)
Again, we draw the graph of x + 2y = 40
For x = 0, 2y = 40 or y = 20
For y = 0, x = 40
∴ line meets OX in B(40, 0) and OY in M(0, 20).
Since feasible region satisfies all the constraints.
∴ OACM is the feasible region.
The corner points arc O(0, 0), A(25, 0), C(20, 10), M(0, 20).
At O(0, 0), Z = 0 + 0 = 0
At A(25, 0), Z = 25 + 0 = 25
At C(20, 10), Z = 20 + 10 = 30
At M(0, 20), Z = 0 + 20 = 20
∴ maximum value = 30 at (20, 10)
∴ maximum number of cakes is 20 of one kind and 10 of second kind.

Mathematical formulation of the given problem is as follows:
Maximise Z = x + y, P = 20 x + 10 y
subject to constrains

Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + 2y = 28
For a = 0, 2 y = 28 or y = 14
For y = 0, x = 28
∴ line meets OX in A (28, 0) and OY in L(0, 14)
Again we draw the graph of 3x + y = 24
For x = 0, y = 24
For y = 0, 3x = 24 or x = 8
∴ line meets OX in B(8, 0) and OY in M(0, 24).

Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The comer points are O(0, 0), B(8, 0), C(4, 12), L(0, 14)
(i) For Z
At O(0, 0), Z = 0 + 0 = 0
At B(8, 0), Z = 8 + 0 = 8
At C(4, 12), Z = 4 + 12 = 16
At L(0, 14), Z = 0 + 14 = 14
∴ maximum value of Z is 16 at (4, 12)
∴ 4 tennis rackets and 12 cricket bats are made by the factory to run at full capacity.
(ii) For P
At O(0, 0), P = 0 + 0 = 0
At B(8, 0), P = 160 + 0 = 160
At C(4, 12), P = 80 + 120 = 200
At L(0, 14), P = 0 + 140 = 140
∴ maximum value of P is 200 at (4, 12)
∴ maximum profit = Rs. 200.

Let the manufacturer produce x nuts and y bolts.
Let Z be the profit.
Table:

We are to maximize P = 
subject to constraints
x + 3 y ≤ 12
3 x + y ≤ 12
x, y ≥ 0

Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + 3y = 12
For x = 0, 3 y = 12 or y = 4
For y = 0, x = 12
∴  line meets OX in A(12, 0) and OY in L(0, 4)
Again we draw the graph of
3x + y = 12
For x = 0, y = 12
For y = 0, 3x = 12 or x = 4
∴ line meets OX in B(4, 0) and OY in M(0, 12).
Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The comer points are O(0, 0), B(4, 0), C(3, 3), L(0, 4)


∴ maximum value of P is 73.5 at (3, 3)
∴  3 packages of nuts and 3 packages of bolts are produced for maximum profit of Rs. 73.50.

Let the young man ride x km at the speed of 25 km per hour and y km at the speed of 40 km per hour. Let f be the total distance covered, which is to be maximized.
∴ f = x + y is the objective function.
Cost of travelling per km is Rs. 2 at the speed of 25 km per hour and cost of travelling per km is Rs. 5 at the speed of 40 km per hour.
∴  total cost of travelling = 2x + 5y
Also Rs. 100 are available for petrol
∴  2x + 5y ≤ 100
Time taken to cover x km at the speed of 25 km per hour  = 
Time taken to cover y km at the speed of 40 km per hour  = 
Total time available = 1 hours

∴ we are to maximize f = x + y
subjecl to the constraints
2x + 5y ≤ 100
8x + 5y ≤ 200
x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of the line 2 x + 5 y = 100
For x = 0, 5 y - 100 or y = 20
For y = 0, 2 x = 100 or x = 50
∴  line meets OX in A (50, 0) and OY in L (0, 20)
Again we draw the graph of the line
8x + 5y = 200
For x = 0, 5 y = 200 or y = 40
For y = 0, 8 x = 200 or x = 25
∴ line meets OX in B (25, 0) and OY in M (0, 40)

Since feasible region is the region which satisfies all the constraints,
∴ feasible region is the quadrilateral OBCL. The comer points are

∴ the young man covers the maximum distance of 30 km when he rides km at the speed of 25 km per hour and km at the speed of 40 km per hour. 

Let x and y represent the number of units produced per week of the products A and B respectively. Let Z be the profit corresponding to this rate of production. Then
Z = 40 x + 35 y ...(1)
In order to produce these number of units,
total consumption of raw material = 2x + 3y
and total labour hours needed = 4x + 3y
But total raw material available = 60 kg
and total labour hours available = 96
∴  we have,
                                             ...(2)
∵  it is not possible to produce negative number of units
∴ x ≥ 0, y ≥ 0    ...(3)
∴ firm’s allocation problem can be put in the following mathematical form:
Find two real numbers x and y such that
2x + 3y ≤ 60
4x + 3y ≤ 96
x ≥ 0, y ≥ 0
and for which the objective function
Z = 40x + 35y
is maximum
Consider a set of rectangular axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of
2x + 3y = 60

For x = 0, 3 y = 60 or y = 20
For y = 0, 2 x = 60 or x = 30
∴ line 2x + 3y = 60 meets OX in A (30, 0) and OY in B (0, 20).
Now we draw the graph of 4x + 3y = 96
For x = 0, 3y = 96 or y = 32
For y = 0, 4x = 96 or x = 24
∴ line 4x + 3y = 96 meets OX in C (24, 0) and OY in D (0, 32).
Since feasible region is the region which satisfies all the constraints.
∴  OCEB is the feasible region.
The corner points are O (0, 0), C (24, 0), E (18, 8), B (0, 20).
At O (0, 0), Z = 40 (0) 35 (0) = 0 + 0 = 0
At C (24, 0), Z = 40 (24) + 35 (0) = 960 + 0 = 960
At E (18, 8), Z = 40 (18)+ 35 (8) = 720 + 280 = 1000
At B(0, 20), Z = 40(0) + 35 (20) = 0 + 700 = 700
Here Rs. 1000 is the maximum values of Z and occurs at E (18, 8)
∴  optimal solution is x = 18, y = 8
i.e., 18 units of A and 8 units of B.

Let the manufacturer produce x packages of screw A and y packages of screw B.
Let Z be the profit. 
Table

We are to maximise
P = 7x + 10y
subject to constraints
4x + 6y ≤ 240  or  2x + 3y ≤ 120
6x + 3y ≤ 240 or 2x + y ≤ 80
x ≥ 0. y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
We draw the graph of 2x + 3y = 120
For a = 0, 3y = 120 or y = 40
For y = 0, 2x = 120 or x = 60
∴ line meets OX in A(60, 0) and OY in L(0, 40)
Again we draw the graph of 2x + y = 80
For x = 0, y = 80
For y = 0, 2 x = 80 or x = 40
∴ line meets OX in B(40, 0) and OY in M(0, 80),
Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.

The corner points are O(0, 0), B(40, 0), C(30, 20), L(0, 40)
At O(0, 0), P = 7 × 0 + 10 × 0 = 0 + 0 = 0
At B(40, 0), P = 7 × 40 + 10 × 0 = 280 + 0 = 280
At C(30, 20), P = 7 × 30 + 10 × 20 = 210 + 200 = 410
At L(0, 40), P = 7 × 0 + 10 × 40 = 0 + 400 = 400
∴  maximum value = 410 at (30, 20)
∴ 30 packages of screws A and 20 packages of screws B arc produced for maximum profit of Rs. 410.

Let the manufacturer produce x pedestal lamps and y wooden shades everyday.
Let P be the profits
Table

We are to maximise
P = 5x + 3y
subject to constraints
2x + y ≤ 12
3x + 2y ≤ 20
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
First, we draw the graph of 2x + y = 12
For x = 0, y = 12
For y = 0, 2 x = 12 or x = 6
∴  line meets OX in A(6, 0) and OY in L(0, 12).
              Again we draw the graph of 3x + 2y = 20
               For x = 0,  2y = 20   or  y = 10
               For y = 0,  3x = 20   or 
.
 Since feasible region satisfies all the constraints.

∴ OACM is the feasible region.
The corner points are O(0, 0), A(6, 0), C(4, 4), M(0, 10)
At O(0, 0), P = 5 × 0 + 3 × 0 = 0 + 0 = 0
At A(6, 0), P = 5 × 6 + 3 × 0 = 30 + 0 = 30
At C(4, 4), P = 5 × 4 + 3 × 4 = 20 + 12 = 32
At M(0, 10), P = 5 × 0 + 3 × 10 = 0 + 30 = 30
∴  maximum value = 32 at (4, 4)
∴  4 pedestal lamps and 4 wooden shades should be produced for maximum profit of Rs. 32.

Let the company manufacture x Souvenirs of type A and y Souvenirs of type B.
Let P be the profit.
Table

We are to maximise
P = 5x + 6y
subject to constraints
5x + 8y ≤ 200
10x + 8y ≤ 240 or 5x + 4y ≤ 120
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
First we draw the graph of 5x + 8y = 200
For x = 0, 8y = 200 or y = 25
For y = 0, 5 x = 200 or x = 40
∴  line meets OX in A(40, 0) and OY in L(0, 25)
Again we draw the graph of 5x + 4y = 120
For x = 0, 4y = 120 or y = 30
For y = 0, 5x = 120 or x = 24
∴ line meets OX in B(24, 0) and OY in M(0. 30).

Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The comer points are O(0, 0), B(24, 0), C(8, 20), L(0, 25)
At O(0, 0), P = 5 × 0 + 6 × 0 = 0 + 0 = 0
At B(24, 0), P = 5 × 24 + 6 × 0 = 120 + 0 = 120
At C(8, 20), P = 5 × 8 + 6 × 20 = 40 + 120 = 160
At L(0, 25) P = 5 × 0 + 6 × 25 = 0 + 150 = 150
∴ maximum value = 160 at (8, 20)
∴ 8 souvenirs of type A and 20 souvenirs of type B are manufactured for maximum profit of Rs. 160.

Let the merchant stock x desktop computers and y portable computers.
Let P be the profit.
Table

We are to maximise
P = 4500 x + 5000 y
subject to constraints
x + y ≤ 250
25000 x + 40000 y ≤ 7000000
or    5 x + 8 y ≤ 1400
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies a ≥ 0, y ≥  0 lies in the first quadrant.
Now we draw the graph of x + y = 250
For x = 0, y = 250
For y = 0, x = 250
∴  line meets OX in A(250, 0) and OY in L(0, 250)
Again we draw the graph of 5 x + 8 y = 1400
For x = 0, 8 y = 1400 or y = 175
For y = 0, 5 x = 1400 or y = 280
∴ line meets OX in B(280, 0) and OY in M(0, 175).

Since feasible region satisfies all the constraints.
∴  OACM is the feasible region.
The comer points are O(0, 0), A(250, 0), C(200, 50), M(0, 175)
At O(0, 0), P = 4500 × 0 + 5000 × 0 = 0 + 0 = 0
At A(250, 0), P = 4500 × 250 + 5000 × 0 = 1125000
At C(200, 50), P = 4500 × 200 + 5000 × 50 = 900000 + 250000 = 1150000
At M(0, 175), P = 4500 × 0 + 5000 × 175 = 0 + 875000 = 875000
∴ maximum value = 1150000 at (200, 50)
∴ 200 units of desktop models and 50 units of portable models are to be stocked for maximum profit of Rs. 1150000.

Let x units of food A and y units of food B be used where x ≥ 0, y ≥ 0.
Let z be the total cost.
Table

We are to minimise
z = 5x + 4y
subject to the constraints
200x + 100y ≥ 4000 i.e. 2x + y ≥ 40
x + 2y ≥ 50
40x + 40y ≥ 1400  i.e. x + y ≥ 35
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 2 x + y = 40
For x = 0, y = 40
For y = 0, 2 x = 40 or x = 20
∴  line meets OX in A(20, 0) and OY in L(0, 40).
Again we draw the graph of 
x + 2y = 50
For x = 0, 2y = 50 or y = 25
For y = 0, x = 50
∴ line meets OX in B(50, 0) and OY in M(0, 25).
Again we draw the graph of
x + y = 35

For x = 0, y = 35
For y = 0, x = 35
∴ line meets OX in C(35, 0) and OY in N(0, 35).
Since feasible region satisfies all the constraints.
∴ shaded region is the feasible region, which is unbounded , and corner points are B(50, 0), D(20, 15), E(5, 30), L(0, 40).
At B(50, 0), z = 5(50) + 4(0) = 250 + 0 = 250
At D(20, 15), z = 5(20) + 4(15) = 100 + 60 = 160
At E(5, 30), z = 5(5) + 4(30) = 25 + 120 = 145
At L(0, 40), z = 5(0) + 4(40) = 0 + 160 = 160
∴  least cost = Rs. 145 at (5, 30)
Since feasible region is unbounded.
∴ we are to check whether this cost is minimum.
For this we draw the graph of
5x + 4y < 145    ...(1)
Since (1) has no common point with feasible region.
∴  minimum cost = Rs. 145 at (5, 30)
∴ minimum cost is Rs. 145 when 5 units of food A and 30 units of food B are used.

 Let the tailor A work for x days tailor B work for y days where x ≥ 0, y ≥ 0.
Let z be the total cost.

Table

We are to minimise
z = 150 x + 200 y
subject to the constraints
6 x + 10 y ≥ 60
4 x + 4 y ≥ 32
x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 6 x + 10 y = 60
For x = 0, 10 y = 60 or y = 6
For y = 0, 6 x = 60 or x = 10
∴ line meets OX in A(10, 0) and OY in L(0, 6).
Again we draw the graph of
4 x + 4 y = 32
For x = 0, 4 y = 32 or y = 8
For y = 0, 4 x = 32 or x = 8
∴ line meets OX in B(8, 0) and OY in M(0, 8).
Since feasible region satisfies all the constraints.

∴ shaded region is the feasible region, which is unbounded , and comer points are A(10, 0), C(5, 3), M(0, 8).
At A(10, 0), z = 150 (10) + 200 (0) = 1500 + 0 = 1500
At C(5, 3), z = 150 (5) + 200 (3) = 750 + 600 = 1350
At M(0, 8), z = 150 (0) + 200 (8) = 0 + 1600 = 1600
∴ least cost = Rs. 1350 at (5, 3).
Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this we draw the graph of
150 x + 200 y < 1350    ...(1)
Since (1) has no common point with feasible region.
∴ minimum value = Rs. 1350 at (5, 3)
∴ minimum labour cost is Rs. 1350 when tailor A works for 5 days and tailor B works for 3 days.

Let the mixture contain x kg. of Food ‘I’ and y kg. of Food ‘II’.
Clearly x ≥ 0, y ≥ 0. We make the following table from the given data:
Table

Since the mixture must contain at least 8 units of vitamin A and 10 units of vitamin C.
∴  2x + y ≥ 8
and    x + 2y ≥ 10
Total cost Z of purchasing x kg. of food ‘I’ and y kg. of Pood ‘II’ is
Z = 50x + 70y
∴   mathematical formulation of the problem is:
Minimise    Z = 50x + 70y
subject to the constraints:
2x + y ≥ 8
x + 2y ≥ 10
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of 2 x + y = 8.
For x = 0, y = 8
For y = 0, 2 x = 8 or x = 4
∴ line meets OX in A(4, 0) and OY in L(0, 8).
Again, we draw the graph of x + 2y = 10
For x = 0, 2 y = 10 or y = 5
For y = 0, x = 10
∴ line meets OX in B(10, 0) and OY in M(0, 5)
Since feasible region satisfies all the constraints.
∴ shaded region is the feasible region and it is unbounded.

Corner points are B(10, 0), C(2, 4), L(0, 8).
At B(10, 0), Z = 500 + 0 = 500
At C(2, 4), Z = 100 280 = 380
At L(0, 8), Z = 0 + 560 = 560
∴ smallest value of Z is 380 at (2, 4).
Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this, we draw the graph of
50x + 70y < 380 or 5x + 7y < 38    ...(1)
Since (1) has no common point with the feasible region.
∴  minimum value = 380 at (2, 4).
∴ cost is minimum when the dietician mixed 2 kg. of food I with 4 kg. of food II. Minimum cost is Rs. 380.

Let the farmer use x kg of F₁ and y kg of F₂.
Let Z be minimum cost.
Table
 

We are to minimise
Z = 6x + 5y
subject to constraints
                  
   
                       
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 2 x + y = 280
For x = 0, y = 280
For y = 0, 2 x = 280 or x = 140
∴  line meets OX in A(140, 0) and OY in L(0, 280).
Again we draw the graph of 3x + 5y = 700
For x = 0,  5y = 700   or   y = 140
For y = 0,  3x = 700  or  
  line meets OX in 

Since feasible region satisfies all the constraints.
∴ shaded region is the feasible region which is unbounded and has corner points are 

∴ smallest value = 1000 at (100, 80)
Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this we draw the graph of
6x + 5y < 1000    ...(1)
Since (1) has no common point with feasible region.
∴ minimum value = Rs. 1000 at (100, 80)
∴ minimum cost is Rs. 1000 when 100 kg. of fertilizer F₁ and 80 kg. of fertilizer F₂ are used.

Let the mixture contain x kg. of food P and y kg. of food Q.
Clearly x ≥ 0, y ≥ 0.
Let Z be the total cost.
Table

Mathematical formulation of the given problem is as follows:
Minimise Z = 60 x + 80 y
subject to the constraints
3x + 4y ≥ 8
5x + 2y ≥ 11
x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 3x + 4y = 8
For x = 0,   4y = 8   or  y = 2
For y = 0,  3x = 8   or 

and OY in L(0, 2)
              Again we draw the graph of 5x + 2y = 11
              For x = 0,  2y = 11   or   
              For y = 0,   5x = 11   or  


and OY in      

Since feasible region satisfies all the constraints.
∴ shaded region is the feasible region and it is unbounded.
The corner points are 
  

Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this we draw the graph 60x + 80 y < 160 i.e. 3x + 4y < 8    ...(1)
 Since (1) has no common point with feasible region.
         minimum cost  = 160 at  i.e. at points lying on segment joining 

Let the diet contain x units of food F₁ and y units of food F₂.
Let Z be the cost.
Table

We are to minimise
Z = 4x + 6y
subject to constraints
3x + 6y ≥ 80
4x + 3y ≥ 100
x ≥ 0, y ≥ 0
              Now we draw the graph of 3x + 6y = 80
          For x = 0,   6 y = 80  or   y = 
           For y = 0,   3 x = 80   or   
     
Again we draw the graph of 4x + 3 y = 100
For x = 0,  3y = 100   or   y = 
For y = 0,   4x = 100  or  x = 25
   line meets OX in B(25, 0) and OY in M        

Since feasible region satisfies all the constraints.
∴ shaded region is the feasible region which is unbounded and has comer points are 

Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this we draw the graph of
4x + 6y < 104    ...(1)
Since (1) has no common point with feasible region.
                       

D.

The corner points are (0, 0), (5, 0), (3, 4), (0, 5)
At (0, 0), Z = 0 + 0 = 0
At (5, 0), Z = 5 p + 0 = 5 p
At (3, 4), Z = 3 p + 4 q
At (0, 5), Z = 0 + 5 q = 5 q
Since maximum value occurs at both (3, 4) and (0, 5).
∴ 3 p + 4 q = 5 q
∴ q = 3 p
∴ (D) is correct answer.

Let x and y be the number of packets of food P and Q respectively. Clearly x ≥ 0, y ≥ 0.
Let Z be the quantity used of vitamin A.
Mathematical formulation of the given problem is as follows:
Minimise    Z = 6x + 3y
subject to the constraints
12x + 3y ≥ 240 i.e. 4 x + y ≥ 80
4x + 20y ≥ 460 i.e. x + 5 y ≥ 115
6x + 4y ≤ 300 i.e. 3 x + 2 y ≤ 150
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 4x + y = 80
For x = 0, y = 80
For y = 0, 4 x = 80 or x = 20
∴ line meets OX in A(20, 0) and OY in L(0, 80).
Also we draw the graph of x + 5y = 115
For x = 0, 5y = 115 or y = 23
For y = 0, x = 115
∴ line meets OX in E (115, 0) and OY in M(0, 23)
Again consider
3x + 2y = 150
For x = 0, 2 y = 150 or y = 75
For y = 0, 3 x = 150 or x = 50
∴ line meets OX in C(50, 0) and OY in N (0, 75).
Since feasible region satisfies all the constraints.
∴ DEF is the feasible region.

The corner points are D(40, 15), E(15, 20), F(2, 72).
At D(40, 15), Z = 6 × 40 + 3 × 15 = 240 + 45 = 285
At E(15, 20), Z = 6 × 15 + 3 × 20 = 90 + 60 = 150
At F(2, 72), Z = 6 × 2 + 3 × 72 = 12 + 216 = 228
∴ minimum value = 150 at (15, 20).
∴ the amount of vitamin A under the constraints given in the problem will be minimum, if 15 packets of food P and 20 packets of food Q are used in the speciai diet. The minimum amount of vitamin A will be 150 units.
Now instead of minimising Z, we have to maximise Z.
∴ maximum value = 285 at (40, 15)
∴ the amount of vitamin A under the constraints given in the problem will be maximum, if 40 packets of focd P and 15 packets of food Q are used in the special diet. The maximum amount of vitamin A will be 285 units.

Let x and y be the number of items M and N respectively.
Total profit on the production = Rs. (600 x + 400 y)
Let Z be the profit.
Mathematical formulation of the given problem is as follows:
Maximise Z = 600 x + 400 y
subject to the constraints
                

Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + 2y = 12
For x = 0, 2 y = 12 or y = 6
For y = 0, x = 12
∴ line meets OX in A(12, 0) and OY in L(0, 6).
Also we draw the graph of 2x + y = 12
For x = 0, y = 12
For y = 0, 2x = 12 or x = 6
∴ line meets OX in B(6, 0) and OY in M(0, 12).
Again we draw the graph of 
                          
For   x = 0,  
For  y = 0,  x = 5
∴ line meets OX in C(5, 0) and OY in N(0, 4).
Since feasible region satisfies all the constraints.
∴ CBDLN is the feasible region.

The corner points are C(5, 0), B(6, 0),
D(4, 4), L(0, 6), N(0, 4).
At C(5, 0), Z = 600 × 5 + 400 × 0 = 3000 + 0 = 3000
At B(6, 0), Z = 600 × 6 + 400 × 0 = 3600 + 0 = 3600
At D(4, 4), Z = 600 × 4 + 400 × 4 = 2400 + 1600 = 4000
At L(0, 6), Z = 600 × 0 + 400 × 6 = 0 + 2400 = 2400
At N(0, 4), Z = 600 × Ot 400 × 4 = 0 + 1600 = 1600
∴ maximum value = 4000 at (4, 4).
∴ manufacturer has to produce 4 units of each item to get the maximum profit of Rs. 4000.

Table

Again we draw the graph of 2 x + 3 y = 120
For x = 0, 3 y = 120 or y = 40
For y = 0, 2 x = 120 or x = 60
∴ line meets OX in C(60, 0) and OY in N(0, 40).
Since feasible region satisfies all the constraints.
∴ OBDEN is the feasible region.
The corner points are O(0, 0), B(20, 0), D(20, 20), E(15, 30). N(0, 40).
 
∴ maximum value = 262.5 at (15, 30).
∴ maximum profit is Rs. 262.5 when 15 toys of type A and 30 toys of type B are manufactured.

Let x quintals of grain be transported from godown A to shop D and y quintals of grain to shop E, then 100 - (x + y) quintals will be transported to shop E.
This means that (60 - x) quintals of grain will be transported from godown B to shop D, (50 - y) quintals of grain to shop E and 40 -{100 - (x + y)} = x + y - 60 quintals will be transported to shop F. For transportation cost, we construct the table
Table

We are to maximise
                               
subject to constraints
                     

Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
x = 60 is a straight line AL parallel to y-axis at a distance of 60.
y = 50 is a straight line BM parallel to .r-axis at a distance of 50.
Now we draw the graph of x + y = 100.
For x = 0, y = 100
For y = 0, x = 100
∴ line meets OX in C(100, 0) and OY in N(0, 100).
Again we draw the graph of x + y = 60.
For x = 0, y = 60
For y = 0, x = 60
∴ line meets OX in A(60, 0) and OY in P(0, 60).

Since feasible region satisfies all the constraints.
∴ AQRS is the feasible region.
The corner points are A(60, 0), Q(60, 40), R(50, 50), S( 10, 50).
 
∴ the minimum cost of transportation is Rs. 510, when from godown A. 10 quintals of grain are sent to shop D, 50 quintals of grain are sent to shop E and 40 quintals ot grain are sent to shop F and from godown B, whole of 50 quintals are sent to shop D.

Since feasible region satisfies all the constraints.
∴ DEF is the feasible region.
The corner points are
D(140, 50), E(20, 140), F(40, 100).

∴ minimum amount of nitrogen is 470 kg. when 40 bags of brand P and 100 bags of brand Q are mixed.
Also maximum value = 595 at (140, 50).
∴ maximum amount of nitrogen is 595 kg. when 140 bags of brand P and 50 bags of brand Q are mixed.

Let the company manufacture x dolls of type A and y dolls of type B.
Let Z be the profit.
We are to maximize
Z = 12x + 16y
subject to the constraints

Now we draw the graph of x + y = 1200
For x = 0, y = 1200
For y = 0, x = 1200
∴ line meets OX in A(1200, 0) and OY in L(0, 1200) x - 2 y = 0 is a straight line passing through (0, 0) and (2. 1).
Again we draw the graph of x - 3 y = 600
For x = 0, - 3 y = 600 or y = - 200
For y = 0, x = 600
∴ line meets OX in B(600, 0) and OY in M(0, - 200)
Since feasible region satisfies all the constraints.

∴ OBCD is the feasible region.
The corner points are O(0, 0), B(600, 0), C(1050, 150), D(800, 400).
At O(0, 0), Z = 12 × 0 + 16 × 0 = 0 + 0 = 0
At B(600, 0), Z = 12 × 600 + 16 × 0 = 7200 + 0 = 7200
At C(1050, 150), Z = 12 × 1050 + 16 × 150 = 12600 + 2400 = 15000
At D(800, 400), Z = 12 × 800 + 16 × 400 = 9600 + 6400 = 16000
∴ maximum profit Rs. 16000 is when 800 dolls of type A and 400 dolls of type B are manufactured and sold.

Let x passengers travel by executive class and y passengers travel by economy class. Let P be the profit.
Table.

We are to maximise
P = 1000 x + 600 y
subject to the constrains
x + y ≤ 200
x ≥ 20
y ≤ 80
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 200
For x = 0, y = 200
For y = 0, x = 200
∴ line meets OX in A(200, 0) and OY in L(0, 200).
x = 20 is a straight line BM parallel to y-axis at a distance of 20.
y = 80 is a straight line CN parallel to x-axis at a distance of 80.
Since feasible region satisfies all the Constraints.
∴ DEF is the feasible region.

∴ x ≥ 0, y ≥ 0 and 8 - x - y ≥ 0
i.e.    x ≥ 0, y ≥ 0 and x + y ≤ 8
The weekly requirement of the depot at A is 5 units of the commodity. Since x units are transported from the factory at P, the remaining (5 - x) units need to be transported from the factory at Q.
Clearly 5 - x ≥ 0, i.e. x ≤ 5.
Similarly, (5 - y) and 6 - (5 - x + 5 - y) = x + y - 4 units are to be transported from the factory at Q to the depots at B and C respectively.
∴ 5 - y ≥ 0, x + y - 4 ≥ 0
i.e.    y ≤ 5, x + y ≥ 4
Total transportation cost Z is given by
Z = 160 x + 100 y + 100(5 - x) + 120 (5 - y) + 100 (x + y - 4) + 150 (8 - x - y)
= 160 x + 100 y + 500 - 100 x + 600 - 120 y + 100 x + 100 y - 400 + 1200 - 150 x - 150 y
= 10 x - 70 y + 1900
= 10(x - 7 y + 190)
∴ the problem reduces to
Minimise Z = 10 (x - 7 y + 190)
subject to the constraints
x ≥ 0, y ≥ 0
x + y ≤ 8
x ≤ 5
y ≤ 5
and    x + y ≥ 4
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 8
For x = 0, y = 8
For y = 0, ,y = 8
∴ line meets OX in A(8, 0) and OY in L(0, 8).
x = 5 is a straight line BM parallel to y-axis.
y = 5 is a straight line CN parallel to x-axis.
Again we draw the graph of x + y = 4
For x = 0, y = 4
For y = 0, x = 4
∴ line meets OX in D(4, 0) and OY in P(0, 4).
Since feasible region satisfies all the constraints.

∴ DBEFCP is the feasible region.
The corner points are D(4, 0), B(5, 0), E(5, 3), F(3, 5), C(0, 5), P(0, 4).
At D(4, 0), Z = 10 (4 - 7 × 0 + 190) = 10 × 194 = 1940
At B(5, 0), Z = 10(5 - 7 × 0 + 190) = 10 × 195 = 1950
At E(5, 3), Z = 10 (5 - 7 × 3 + 190) = 10 × 174 = 1740
At F(3, 5), Z = 10 (3 - 7 × 5 + 190) = 10 × 158 = 1580
At C(0, 5), Z = 10 (0 - 7 × 5 + 190) = 10 × 155 = 1550
At P(0, 4), Z = 10 (0 - 7 × 4 + 190) = 10 × 162 = 1620
∴ minimum value = 1550 at (0, 5).
∴ the optimal transportation strategy will be to deliver 0, 5 and 3 units from the factory at P and 5, 0 and 1 units from the factory at Q to the depots at A. B and C respectively. Corresponding to this strategy, the transportation cost would be minimum, i.e. Rs. 1550.

Suppose that, from depot A,
x litre of oil is transported to pump D
y litre of oil is transported to pump E
and 7000 - (x + y) litre of oil is transported to pump F.
Then, from depot B,
(4500 - x) litres of oil is transported to pump D
(3000 - y) litres of oil is transported to pump E
and (3500 - {7000 - (x + y)}) litres = (x + y - 3500) litres of oil is transported to pump F.
Table

We are to minimise
                                     
subject to the constraints
                              
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.

x = 4500 is a straight line AL parallel to y-axis at a distance of 4500.
y = 3000 is a straight line BM parallel to v-axis at a distance of 3000.
Now we draw the graph of x + y = 3500.
For x = 0, y = 3500
For y = 0, x = 3500
∴ line meets OX in C(3500, 0) and OY in N(0, 3500).
Again we draw the graph of x + y = 7000.
For x = 0, y = 7000
For y = 0, x = 7000
∴ line meets OX in D(7000, 0) and OY in P(0, 7000).

Since feasible region satisfies all the constraints.
∴ CASQR is the feasible region.
The corner points are
C(3500, 0), A(4500, 0), S(4500, 2500), Q(4000, 3000), R(500, 3000).

∴ the minimum cost of transportation is Rs. 4400, when from godown A, 500 litre of oil is transported to pump D, 3000 litre of oil is transported to pump E and 7000 -(500 + 3000) = 3500 litre of oil is transported to pump F.
Hence from godown B whole of 4000 litre of oil is transported to pump D.

Let the farmer mix x bags of brand P and y bags of brand Q.
Let Z be the cost.
Table,

We are to minimise
                         
subject to the constraints
                                        
                                
Consider a set of rectangular cartesian axes OXY in the plane. 
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 2 x + y = 12
For x = 0, y = 12
For y = 0, 2 x = 12 or x = 6
∴ line meets OX in A(6, 0) and OY in L(0, 12).
Also we draw the graph of 2x + 9y = 36
For x = 0, 9 y = 36 or y = 4
For y = 0, 2 x = 36 or x = 18
∴ line meets OX in B(18, 0) and OY in M(0, 4).
Again we draw the graph of 2x + 3y = 24
For x = 0, 3y = 24 or y = 8
For y = 0, 2x = 24 or x = 12
∴ line meets OX in C(12, 0) and OY in N(0, 8).
Since feasible region satisfies all the constraints.

∴ shaded region is the feasible region which is unbounded and comer points are B(18, 0), D(9, 2), E(3, 6), L(0, 12).
At B(18, 0), Z = 250 × 18 + 200 × 0 = 4500 + 0 = 4500
At D(9, 2), Z = 250 × 9 + 200 × 2 = 2250 + 400 = 2650
At E(3, 6), Z = 250 × 3 + 200 × 6 = 750 + 1200 = 1950
At L(0, 12), Z = 250 × 0 + 200 × 12 = 0 + 2400 = 2400
∴ smallest value = 1950 at (3, 6)
Since feasible region is unbounded.
∴ we are to check w hether this value is minimum.
For this we draw the graph of
250x + 200y < 1950 or 5x + 4y < 39    ...(1)
Since (1) has no common point with feasible region.
∴ minimum value = 1950 at (3, 6)
∴ minimum cost is Rs. 1950 when 3 bags of brand P and 6 bags of brand Q are mixed.

Let the dietician mix x kg of food X and y kg of food Y.
Let Z be the cost.
Table

We are to minimise
Z = 16x + 20y
subject to the constraints
x + 2y ≥ 10
2x + 2y ≥ 12 or x + y ≥ 6
3x + y ≥ 8
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + 2y = 10
For x = 0, 2 y = 10 or y = 5
For y = 0, x = 10
∴ line meets OX in A(10, 0) and OY in L(0, 5).
Also we draw the graph of x + y = 6
For x = 0, y = 6
For y = 0, x = 6
∴ line meets OX in B(6, 0) and OY in M(0, 6).
Again we draw the graph of 3x + y = 8
For x = 0,  y = 8
For y = 0,  3x = 8  or x = 

Since feasible region satisfies all the constraints.
 

∴ shaded region is the feasible region which is unbounded and corner points are A(10, 0), D(2, 4), E(1, 5), N(0, 8).
At A(10, 0), Z = 16 × 10 + 20 × 0 = 160 + 0 = 160
At D(2, 4), Z = 16 × 2 + 20 × 4 = 32 + 80 = 112
At E(1, 5), Z = 16 × 1 + 20 × 5 = 16 + 100 = 116
At N(0, 8), Z = 16 × 0 + 20 × 8 = 0 + 160 = 160
∴ smallest value = 112 at (2, 4)
Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this we draw the graph of
16 x + 20 y < 112 or 4 x + 5 y < 28    ...(1)
Since (1) has no common point with feasible region.
∴ minimum value = 112 at (2, 4)
∴ minimum cost is Rs. 112 when 2 kg of food X and 4 kg. of food Y are mixed.

Let the land allocated for crop A be x hectares and crop B be y hectares. 
Maximum area of the land available for two crops is 50 hectares. 
 
Liquid herbicide to be used for crops A and B are at the rate of 20 litres and 10 litres per hectare respectively. Maximum amount of herbicide to be used is 800 litres. 

The profits from crops A and B per hectare are Rs 10,500 and Rs 9,000 respectively.
Thus, total profit = Rs (10,500x + 9,000y) = Rs 1500 (7x + 6y)
Thus, the linear programming problem is:
Maximize Z = 1500 (7x + 6y) subject to the constraints

The feasible region determined by constraints is represented by the shaded region in the following graph:

The corner points of the feasible region are O (0, 0), A (40, 0), B (30, 20) and C (0, 50). The value of Z at these corner points are

Let the factory manufactures x screws of type A and y screws of type B on each day.

∴ x ≥ 0, y ≥ 0

Given that

The constraints are

4x + 6y ≤ 240

6 x + 3y ≤ 240

Total profit

z = 0.70 x + 1y

∴l.P.P is

maximise z = 0.7 x + y

subject to,

2x +3y ≤ 120

2x + y ≤ 80

x ≥0, y ≥0

∴ common feasible region is OCBAO

The maximum value of 'Z' is 41 at (30,20). Thus the factory showed produce 30 packages at screw A and 20 packages of screw B to ge the maximum profit of Rs.41

Let x and y respectively be the number of machines A and B, which the factory owner should buy.

Now according to the given information, the linear programming problem is:

maximise  Z = 60x + 40y

Subject to the constraints

1000x + 1200 y $\le$ 9000

$$\begin{gathered} \Rightarrow 5\mathrm{x} + 6\mathrm{y} \le 45 ............\left(1\right) \\ 12\mathrm{x} + 8\mathrm{y} \le 72 \\ \Rightarrow 3\mathrm{x} + 2\mathrm{y} \le 18 .............\left(2\right) \\ \mathrm{x}\ge 0, \mathrm{y}\ge 0 .............\left(3\right) \end{gathered}$$

The inequalities (1), (2), and (3) can be graphed as:

The shaded portion OABC is the feasible region.

The value of Z at the corner points are given in the following table.

The maximum value of Z is 360 units, which is attained at             B$\left(\frac{9}{4}, \frac{45}{8}\right)$, and c( 6, 0 ).

Now the number of machines cannot be in fraction.

Thus, to maximise the daily output, 6 machines of type A and no machine of type B need to be bought.                  

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

LPP is,

Minimize  Z = 4x + 6y  such that,

3x + 6y $\ge$ 80

4x + 3y $\ge$ 100

x, y $\ge$ 0

Representing the LPP graphically

Corner points are  $\left[ 0, \frac{100}{3}\right], \left[ 24, \frac{4}{3}\right], \left[ \frac{80}{3}, 0\right]$

From the table it is clear that, minimum cost is 104 and occurs at the point $\left( 24, \frac{4}{3}\right)$.

Let  x  be number of gold rings and  y  be number of chains manufactured 

L.P.P. is

Max Z = 3000 x + 190 y

Substitute in  x + y $\le$ 24

$$\begin{gathered} \mathrm{x} + \frac{\mathrm{y}}{2} \le 16 \mathrm{or} 2 \mathrm{x} + \mathrm{y} \le 32 \\ \mathrm{x} \ge \mathrm{o}, \mathrm{y} \ge 0 \end{gathered}$$

Feasible region          

Hence to make the maximum profit, 8 gold rings and 16 chains must be manufactured.

Let the number of rackets and the number of bats to be made be  x  and  y 

respectively.

The given information can be tabulated as below:

In a day, the machine time is not available for more than  42  hours.

$\therefore 1.5 \mathrm{x} + 3 \mathrm{y} ⩽ 42$

In a day, the craftsman's time can not be  more than  24  hours.

$\therefore 3 \mathrm{x} + \mathrm{y} ⩽ 24$

Let the total profit be Rs. Z.

The profit on a racket is Rs. 20  and on a bat is Rs. 10.

$\therefore \mathrm{Z} = 20 \mathrm{x} + 10 \mathrm{y}$

Thus, the given linear programming problem can be stated as follows:

Maximise   Z = 20 x + 10 y           ...........( i )

Subject to 

1.5 x + 3 y $⩽$ 42                         ...........( ii )

3 x + y $⩽$ 24                               ..........( iii )

x, y $⩾$ 0                                      ..........( iv )

The feasible region can be shaded in the graph as below:

The corner points are  A ( 8, 0 ),  B ( 4, 12 ),  C ( 0, 14 )   and   O ( 0, 0 ).

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

The maximum value of Z is 200,  which occurs at  x = 4  and  y = 12.

Thus, the factory must produce  4  tennis rackets  and 12 cricket bats to earn the maximum profit of Rs. 200.

B.

41

A.

x = 2