Question

A chemical company produces two compounds, A and B. The following table gives the units of ingredients, C and D per kg of compounds A and B as well as minimum requirements of C and D and costs per kg of A and B. Find the quantities of A and B which would give a supply of C and D at a minimum cost.

	Compound		Minimum requirement
	A	B
Ingredient C Ingredient D	1 3	2 1	80 75
Cost (in Rs) per kg	4	6	-

Answer 1

Let x kg of compound A and y kg of compound B were produced.
Quantity cannot be negative.
Therefore, \[x, y \geq 0\] 

	Compound		Minimum requirement
	A	B
Ingredient C Ingredient D	1 3	2 1	80 75
Cost (in Rs) per kg	4	6	-

According to question, the constraints are

\[x + 2y \geq 80\]

\[3x + y \geq 75\]
Cost (in Rs) per kg of
compound A and compound B is Rs 4 and Rs 6 respectively.Therefore, cost of x kg of compound A and y kg of compound B is 4x and 6y  respectively.
Total cost  = Z =  \[4x + 6y\] 

which is to be minimised.
Thus, the mathematical formulat​ion of the given linear programming problem is 
Min Z =   \[4x + 6y\]
subject to

\[x + 2y \geq 80\]
\[3x + y \geq 75\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows:
x + 2y = 80, 3x + y =75, x = 0 and y = 0

Region represented by x + 2y ≥ 80:
The line x + 2y = 80 meets the coordinate axes at A₁(80, 0) and B₁(0, 40) respectively. By joining these points we obtain the line x + 2y = 80. Clearly (0,0) does not satisfies the x + 2y = 80. So, the region which does not contain the origin represents the solution set of the inequation x + 2y ≥ 80. 

Region represented by 3x + y ≥ 75:
The line 3x + y =75 meets the coordinate axes at C₁(25, 0) and D₁(0, 75) respectively. By joining these points we obtain the line 3x + y =75. Clearly (0,0) does not satisfies the inequation 3x + y ≥ 75. So,the region which does not contain the origin represents the solution set of the inequation 3x + y ≥ 75.

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 x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, and y ≥ 0 are as follows. 

The corner points are D₁(0, 75), E₁(14, 33) and A₁(80, 0). 

The values of Z at these corner points are as follows

Corner point	Z= 4x + 6y
D1	450
E1	254
A1	320

The minimum value of Z is 254 which is attained at E₁ \[\left( 14, 33 \right)\]
Thus, the minimum cost is Rs 254 obtained when 14 units of compound A and 33 units of compound  B were produced.