#### 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

#### 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 A_{1}(60, 0) and B_{1}(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 6*x* + 3*y* = 240 meets the coordinate axes at A_{2}(40, 0) and B_{2}(0, 80), respectively. Join these points to obtain the line 6*x* + 3*y* = 240.

Clearly (0,0) satisfies the inequation 6*x* + 3*y* ≤ 240. So, the region containing the origin represents the solution set of the inequation 6*x* + 3*y* ≤ 240.

The line *x* = 10 is the line that passes through A_{3}(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 A_{3}(10, 0), A_{2}(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 |

A_{3}(10, 0) |
Z = 7 × 10 + 10 × 0 = 70 |

A_{2}(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.