Linear Programming

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Question
CBSEENMA12033523

Solve the following Linear Programming Problems graphically:
Maximise    Z = 3x + 4y
subject to the constraints:    x + y ≤ 4,  x ≥ 0, y ≥ 0

Solution

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).

Sponsor Area

Question
CBSEENMA12033524

Solve the following linear programming problem graphically:
Maximise    Z = 4x + y
subject to the constraints: x + y ≤ 50,  3x + y ≤ 90,  x ≥ 0, y ≥ 0

Solution

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.

The comer points are
O(0, 0), B(30, 0), C(20, 30), L(0, 50)
At O(0, 0), Z = 0 + 0 = 0
At B(30, 0), Z = 120 + 0 = 120
At C(20, 30), Z = 80 + 30 = 110
At L(0, 50), Z = 0 + 50 = 50
∴ maximum value = 120 at the point (30, 0).

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.

Question
CBSEENMA12033525

Find the maximum value of f = x + 2 y subject to the constraints:
2x + 3 y ≤ 6
x + 4 y ≤ 4
x, y ≥ 0

Solution

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
     straight O left parenthesis 0 comma space 0 right parenthesis comma space space space straight A left parenthesis 3 comma space 0 right parenthesis comma space space space straight C open parentheses 12 over 5 comma space 2 over 5 close parentheses comma space space space straight M left parenthesis 0 comma space 1 right parenthesis
At straight O left parenthesis 0 comma space 0 right parenthesis space straight f space equals 0 plus 0 space equals space 0
At space straight A left parenthesis 3 comma space 0 right parenthesis comma space straight f space space equals space 3 plus 0 space equals space 3
At space straight C open parentheses 12 over 5 comma space 2 over 5 close parentheses comma space space space straight f space equals space 12 over 5 plus 4 over 5 space equals space 16 over 5 space equals space 3.2
At space straight M left parenthesis 0 comma space 1 right parenthesis comma space space straight f space equals space 0 plus 2 space equals space 2
therefore space space space space maximum space value space equals space 3.2 space at space open parentheses 12 over 5 comma space 2 over 5 close parentheses.

Question
CBSEENMA12033526

Maximize z = 9 x + 3 y subject to the constraints
2x + 3y ≤ 13
2x + y ≤ 5
x, y ≥ 0

Solution

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  rightwards double arrow space straight y space equals space 13 over 3
For y = 0,  2x = 13    rightwards double arrow space straight x space equals space 13 over 2
therefore space space space space line space 2 straight x plus 3 straight y space equals space 13 space space meets space OX space in space straight A open parentheses 13 over 2 comma 0 close parentheses space and space OY space in space straight B open parentheses 0 comma space 13 over 3 close parentheses.
Again we draw the graph of 2x + y = 5
For x = 0,  y = 5
For y = 0,  2x = 5  rightwards double arrow space space straight x space equals space 5 over 2
therefore space space space line space 2 straight x plus straight y space equals space 5 space meets space OX space in space straight C open parentheses 5 over 2 comma space 0 close parentheses space and space OY space in space straight D left parenthesis 0 comma space 5 right parenthesis.

Since feasible region satisfies all the constraints.
therefore    OCEB in the feasibe region. The corner points are O(0, 0),  straight C open parentheses 5 over 2 comma space 0 close parentheses comma space straight E left parenthesis 0.5 comma space 4 right parenthesis comma space space straight B open parentheses 0 comma space 13 over 3 close parentheses 
       At O(0, 0),  z = 9(0) + 3(0) = 0+ 0 = 0
At   straight C open parentheses 5 over 2 comma space 0 close parentheses comma space straight z space equals space 9 space open parentheses 5 over 2 close parentheses space plus space 3 space left parenthesis 0 right parenthesis space equals space 45 over 2 plus 0 space space equals space 45 over 2 space equals space 22.5
At   straight E thin space left parenthesis 0.5 comma space 4 right parenthesis comma space space straight z space equals space 9 left parenthesis 0.5 right parenthesis space plus space 3 space left parenthesis 4 right parenthesis space equals space 4.5 space plus space 12 space equals space 16.5
At space straight B space open parentheses 0 comma space 13 over 3 close parentheses comma space space straight z space equals space 9 space left parenthesis 0 right parenthesis space plus space 3 space open parentheses 13 over 3 close parentheses space equals space 0 plus 13 space equals space 13
therefore space space space Maximum space value space of space straight z space is space 22.5 space at space open parentheses 5 over 2 comma 0 close parentheses space space straight i. straight e. comma space space space when space straight x space equals space 5 over 2 comma space space straight y space equals 0