Question: Solve the following Linear Programming Problem graphically : Minimise: Z= 60x + 80y subject to constraints:
3r + 4y ≥ 8
5r + 2y ≥ 11
x, y 20
The correct answer is –
To solve the linear programming problem graphically, we need to plot the feasible region and find the corner points of the region to determine the minimum value of the objective function Z = 60x + 80y.
First, we plot the two constraint equations on a graph:
3x + 4y = 8 (equation 1)
5x + 2y = 11 (equation 2)
To plot each line, we find their x and y intercepts.
-
For equation 1, when x = 0, y = 2, and when y = 0, x = 8/3.
-
For equation 2, when x = 0, y = 11/2, and when y = 0, x = 11/5.
We plot these intercepts and draw the lines passing through them. The feasible region is the area that satisfies the constraints and is bounded by these lines.
|
20 +——–/——–+
| / | |
| / | |
15 +—–/—+——–+
| / | |
| / | |
10 +–/——+——–+
| / | |
|/ | |
5 +———-+——/-
0 2.75 5
x-axis
Next, we need to find the corner points of the feasible region by solving the intersection points of the lines. The corner points are (0, 2), (1.4, 1.5), and (2.75, 0).
We then substitute the coordinates of each corner point into the objective function Z = 60x + 80y to find the minimum value of Z.
- Z at (0, 2): Z = 60(0) + 80(2) = 160
- Z at (1.4, 1.5): Z = 60(1.4) + 80(1.5) = 210
- Z at (2.75, 0): Z = 60(2.75) + 80(0) = 165
Therefore, the minimum value of the objective function Z is 160, which occurs at point (0, 2). Therefore, x = 0 and y = 2, which satisfies the constraints, gives the minimum value of Z.
