Learn about linear programming by graphical method. If the objective function Z is a function of two variables only, then the Linear Programming Problem can be solved effectively by the graphical method.
If Z is a function of three variables, then also it can be solved by this method. But in this case the graphical solution becomes complicated enough. The linear programming problems are solved in applied mathematics models.
Method of Linear Programming Graph
• Draw the graph of the constraints.
• Determine the region which satisfies all the constraints and non-negative constraints
(x > 0, y > 0). This region is called the feasible region.
• Determine the co-ordinates of the corners of the feasible region.
• Calculate the values of the objective function at each corner.
• Select the corner point which gives the optimum (maximum or minimum) value of the objective function. The co-ordinates of that point determine the optimal solution.
Below you can see the problems linear programming by graphical method -
Problem 1:-
Use graphical method to solve the following linear programming problem.
Maximize Z = 2x + 10 y
Subject to the constraints 2 x + 5y < 16,
x < 5,
x > 0, y > 0.
Solution:
Since x > 0 and y > 0 the solution set is restricted to the first quadrant.|
i) 2x + 5y < 16 Draw the graph of 2x + 5y = 16
2x + 5y = 16
y =
Determine the region represented by 2x + 5y < 16
ii) x < 5 Draw the graph of x = 5
Determine the region represented by x < 5.
Shade the intersection of the two regions. The shaded region OABC is the feasible region
B(5, 1.2) is the point of intersection of 2x + 5y = 16 and x = 5. The corner points of OABC
are O(0,0), A(5,0), B(5,1.2) and C(0,3.2).
Z is maximum at x = 0, y = 3.2
Maximum value of Z = 32.
Problem 2:
Use graphical method to solve the following linear programming problem.
Maximize Z = 20 x + 15y
Subject to 180x + 120y < 1500,
x + y < 10,
x > 0, y > 0
Solution:
Since x > 0 and y > 0, the solution set is restricted to the first quadrant.
i) 180x + 120 y < 1500
180x + 120y < 1500 => 3x + 2y < 25.
Draw the graph of 3x + 2y = 25
3x + 2y = 25
y =
Determine the region represented by 3x + 2y < 25.
ii) x + y < 10 Draw the graph of x + y = 10
x + y = 10 ⇒ y =10 - x
Determine the region represented by x + y < 10
Shade the intersection of the two regions. The shaded region OABC is the feasible region.
B(5,5) is the point of intersection of 3x + 2y = 25 and x + y = 10. The corner points of OABC
are O(0,0), A(
, 0), B (5,5) and C(0,10).
Z is maximum at x = 5 and y = 5. Maximum value of Z = 175.
• Determine the region which satisfies all the constraints and non-negative constraints
(x > 0, y > 0). This region is called the feasible region.
• Determine the co-ordinates of the corners of the feasible region.
• Calculate the values of the objective function at each corner.
• Select the corner point which gives the optimum (maximum or minimum) value of the objective function. The co-ordinates of that point determine the optimal solution.
How to Graph Linear Programming
Problem 1:-
Use graphical method to solve the following linear programming problem.
Maximize Z = 2x + 10 y
Subject to the constraints 2 x + 5y < 16,
x < 5,
x > 0, y > 0.
Solution:
Since x > 0 and y > 0 the solution set is restricted to the first quadrant.|
i) 2x + 5y < 16 Draw the graph of 2x + 5y = 16
2x + 5y = 16
y =
| x | 8 | 0 | 3 |
| y | 0 | 3.2 | 2 |
ii) x < 5 Draw the graph of x = 5
Determine the region represented by x < 5.
Shade the intersection of the two regions. The shaded region OABC is the feasible region
B(5, 1.2) is the point of intersection of 2x + 5y = 16 and x = 5. The corner points of OABC
are O(0,0), A(5,0), B(5,1.2) and C(0,3.2).
| Corners | O(0,0) | A(5,0) | B(5,1.2) | C(0,3.2) |
| Z = 2x + 10 y | 0 | 10 | 22 | 32 |
Z is maximum at x = 0, y = 3.2
Maximum value of Z = 32.
Problem 2:
Use graphical method to solve the following linear programming problem.
Maximize Z = 20 x + 15y
Subject to 180x + 120y < 1500,
x + y < 10,
x > 0, y > 0
Solution:
Since x > 0 and y > 0, the solution set is restricted to the first quadrant.
i) 180x + 120 y < 1500
180x + 120y < 1500 => 3x + 2y < 25.
Draw the graph of 3x + 2y = 25
3x + 2y = 25
y =
| x | 0 | 5 | |
| y | 0 | 5 |
ii) x + y < 10 Draw the graph of x + y = 10
x + y = 10 ⇒ y =10 - x
| x | 0 | 10 | 5 |
| y | 10 | 0 | 5 |
Determine the region represented by x + y < 10
Shade the intersection of the two regions. The shaded region OABC is the feasible region.
B(5,5) is the point of intersection of 3x + 2y = 25 and x + y = 10. The corner points of OABC
are O(0,0), A(
| Corners | O(0,0) | A( | B(5,5) | C(0,10) |
| Z = 20x + 15 y | 0 | 166.67 | 175 | 150 |
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