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Question
Solve the following linear programming problem graphically :
Maximise Z = 7x + 10y subject to the constraints
4x + 6y ≤ 240
6x + 3y ≤ 240
x ≥ 10
x ≥ 0, y ≥ 0
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Solution
The given constraints are 4x + 6y ≤ 240, 6x + 3y ≤ 240, x ≥ 10, x ≥ 0, y ≥ 0.
Firstly, convert the given inequations into equations, we obtain the following equations:
4x + 6y = 240, 6x + 3y = 240, x = 10, x = 0 and y = 0.
The line 4x + 6y = 240 meets the coordinate axes at A1(60, 0) and B1(0, 40), respectively. Join these points to obtain the line 4x + 6y = 240
Clearly (0,0) satisfies the inequation 4x + 6y ≤ 240. So, the region containing the origin represents the solution set of the inequation 4x + 6y ≤ 240.
The line 6x + 3y = 240 meets the coordinate axes at A2(40, 0) and B2(0, 80), respectively. Join these points to obtain the line 6x + 3y = 240.
Clearly (0,0) satisfies the inequation 6x + 3y ≤ 240. So, the region containing the origin represents the solution set of the inequation 6x + 3y ≤ 240.
The line x = 10 is the line that passes through A3(10, 0) and parallel to y-axis.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints is shown below:
The corner points of the feasible region are A3(10, 0), A2(40, 0), Q(30, 20) and R (20, `80/3`)
The values of Z at these corner points are as follows:
| Corner point | Value of the objective function Z = 7x + 10y |
| A3(10, 0) | Z = 7 × 10 + 10 × 0 = 70 |
| A2(40, 0) | Z = 7 × 40 + 10 × 0 = 280 |
| Q(30, 20) | Z = 7 × 30 + 10 × 20 = 410 (Maximum) |
| R(20, 80/3) | `Z = 7 xx 20 + xx 80/3 = 1220/3` |
Thus, the maximum value of the objective function Z is 410 which is obtained at x = 30 and y = 20.
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