Question

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.

Answer 1

Let x kg of kind A and y kg of kind B were used.
Quantity cannot be negative.
Therefore, $x,y\ge 0$

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

$$\begin{gathered} 60x+30y\ge 240 \\ 30x+60y\ge 300 \\ 30x+180y\ge 540 \end{gathered}$$

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 formulat​ion of the given linear programmimg problem is

Min Z = $8x+12y$

subject to

$$\begin{gathered} 2x+y\ge 8 \\ x+2y\ge 10 \\ x+6y\ge 18 \end{gathered}$$
$x,y\ge 0$

First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 8, x +2y = 10, x +6y = 18, x = 0 and y = 0

Region represented by 2x + y ≥ 8:
The line 2x + y = 8 meets the coordinate axes at A₁(4, 0) and B₁(0, 8) respectively. By joining these points we obtain the line 2x + y = 8.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 8. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 8.

Region represented by x +2y ≥ 10:
The line x +2y = 10 meets the coordinate axes at C₁(10,0) and D₁(0, 5) respectively. By joining these points we obtain the line
x +2y = 10. Clearly (0,0) does not satisfies the inequation x +2y ≥ 10. So,the region which does not contain the origin represents the solution set of the inequation x +2y ≥ 10.

Region represented by x +6y ≥ 18:
The line x +6y = 18 meets the coordinate axes at E₁(18,0) and F₁(0, 3) respectively. By joining these points we obtain the line x + 6y = 18.Clearly (0,0) does not satisfies the inequation x + 6y ≥ 18. So,the region which does not contain the origin represents the solution set of the inequation x +6y ≥ 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 2x + y ≥ 8, x + 2y ≥ 10,x + 6y ≥ 18, x ≥ 0, and y ≥ 0, are as follows.


The corner points are B₁(0, 8), G₁(2, 4), H₁(6, 2) and E₁(18, 0).

The values of Z at these corner points are as follows

Corner point	Z= 8x + 12y
B1	96
G1	64
H1	72
E1	144

The minimum value of Z is 64 which is attained at G₁$\left(2,4\right)$.

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