A wholesale dealer deals in two kinds, *A* and *B* (say) of mixture of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew nuts and 30 grams of hazel nuts. Each kg of mixture *B* contains 30 grams of almonds, 60 grams of cashew nuts and 180 grams of hazel nuts. The remainder of both mixtures is per nuts. The dealer is contemplating to use mixtures *A* and *B* to make a bag which will contain at least 240 grams of almonds, 300 grams of cashew nuts and 540 grams of hazel nuts. Mixture *A* costs Rs 8 per kg. and mixture *B* costs Rs 12 per kg. Assuming that mixtures *A* and *B* are uniform, use graphical method to determine the number of kg. of each mixture which he should use to minimise the cost of the bag.

#### Solution

Let *x *kg of kind *A* and *y* kg of kind *B* were used.

Quantity cannot be negative.

Therefore,

The given information can be tabulated as follows:

Nut | Almonds(grams) | Cashewnuts(grams) | Hazel nuts(grams) |

A(x) |
60 | 30 | 30 |

B(y) |
30 | 60 | 180 |

Availability | 240 | 300 | 540 |

Therefore, the constraints are

\[60x + 30y \geq 240\]

\[30x + 60y \geq 300\]

\[30x + 180y \geq 540\]

Mixture *A* costs Rs 8 per kg. and mixture *B* costs Rs 12 per kg.

Total cost = Z = \[8x + 12y\]

which is to be minimised.

Thus, the mathematical formulation of the given linear programmimg problem is

Min Z = \[8x + 12y\] subject to

\[2x + y \geq 8\]

\[ x + 2y \geq 10\]

\[ x + 6y \geq 18\]

First, we will convert the given inequations into equations, we obtain the following equations:

2*x** *+ *y* = 8, *x** *+2*y* = 10, *x** *+6*y* = 18, *x* = 0 and* y* = 0

Region represented by 2*x** *+ *y* ≥ 8:

The line 2*x** *+ *y* = 8 meets the coordinate axes at *A*_{1}(4, 0) and *B*_{1}(0, 8) respectively. By joining these points we obtain the line 2*x** *+ *y* = 8.

Clearly (0,0) does not satisfies the inequation 2*x** *+ *y* ≥ 8. So,the region in *xy* plane which does not contain the origin represents the solution set of the inequation 2*x** *+ *y* ≥ 8.

Region represented by *x** *+2*y* ≥ 10:

The line *x** *+2*y* = 10 meets the coordinate axes at *C*_{1}(10,0) and *D*_{1}(0, 5) respectively. By joining these points we obtain the line*x** *+2*y* = 10. Clearly (0,0) does not satisfies the inequation *x** *+2*y* ≥ 10. So,the region which does not contain the origin represents the solution set of the inequation *x** *+2*y* ≥ 10.

Region represented by *x** *+6*y* ≥ 18:

The line *x** *+6*y* = 18 meets the coordinate axes at *E*_{1}(18,0) and *F*_{1}(0, 3) respectively. By joining these points we obtain the line* x** *+ 6*y* = 18.Clearly (0,0) does not satisfies the inequation *x** *+ 6*y* ≥ 18. So,the region which does not contain the origin represents the solution set of the inequation *x** *+6*y* ≥ 18.

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 2*x** *+ *y* ≥ 8, *x** *+ 2*y* ≥ 10,*x** *+ 6*y* ≥ 18,* x* ≥ 0, and *y* ≥ 0, are as follows.The corner points are *B*_{1}(0, 8), *G*_{1}(2, 4), *H*_{1}(6, 2) and *E*_{1}(18, 0).

The values of Z at these corner points are as follows

Corner point | Z= 8x + 12y |

B_{1} |
96 |

G_{1} |
64 |

H_{1} |
72 |

E_{1} |
144 |

The minimum value of Z is 64 which is attained at *G*_{1}

Thus, the minimum cost is Rs 64 obtained when 2 units of kind *A* and 4 units of kind *B* nuts were used.