Question
A company produces soft drinks that has a contract which requires that a minimum of 80 units of the chemical A and 60 units of the chemical B to go into each bottle of the drink. The chemicals are available in a prepared mix from two different suppliers. Supplier S has a mix of 4 units of A and 2 units of B that costs Rs 10, the supplier T has a mix of 1 unit of A and 1 unit of B that costs Rs 4. How many mixes from S and T should the company purchase to honour contract requirement and yet minimize cost?
Solution
10 mixes from supplier S and 40 mixes from supplier T
The given data may be put in the following tabular form:
Supplier Chemical 
S 
T 
Minimum Requirement 
A B 
4 2 
1 1 
80 60 
Cost per unit 
Rs 10 
Rs 4 

Suppose x units of mix are purchased from supplier S and y units per purchased from supplier T.
Total cost Z = 10x + 4y.
Units of chemical A per bottle = 4x + y.
But the minimum requirement of chemical A per bottle = 80
∴ 4x + y ≥ 80
Similarly, 2x + y ≥ 60
Clearly, x ≥ 0, y ≥ 0
Thus, the mathematical formulation of the given LPP is
Minimize Z = 10x + 4y
Subject to
4x + y ≥ 80
2x + y ≥ 60
and, x ≥ 0, y ≥ 0
Now, we find the feasible region which is the set of all points whose coordinates simultaneously satisfy all constraints including nonnegativity restrictions. The shaded region in fig. represents the feasible region of the given LPP. The coordinates of the corner points of the feasible region areA_{2} (30, 0), P (10, 40), B_{1} (0, 80).
These points are obtained by solving the equations of the corresponding intersecting lines, simultaneously.
The values of the objective function at these points are given in the following table:
Point (x, y) 
Value of objective function Z = 10x + 4y 
A_{2} (30, 0) P (10, 40) B_{1} (0, 80) 
Z = 10 × 30 + 4 × 0 = 300 Z = 10 × 10 + 40 × 4 = 260 Z = 10 × 0 + 4 × 80 = 320. 
Clearly, Z is minimum at (10, 40). The feasible region is unbounded and the open half plane represented by 10x + 4y < 260 does not have points in common with the feasible region. So, Z is minimum at x = 10, y = 40. Hence, x = 10, y = 40 is the optimal solution of the given LPP.
Hence, the cost per bottle is minimum when the company purchases 10 mixes from supplier S and 40 mixes from supplier T.
SIMILAR QUESTIONS
Solve the following LPP graphically:
Maximize and Minimize Z = 3x + 5y
Subject to 3x – 4y + 12 ≥ 0
2x – y + 2 ≥ 0
2x + 3y – 12 ≥ 0
0 ≤ x ≤ 4
y ≥ 2.
Solve the following linear programming problem graphically:
Maximize Z = 50x + 15y
Subject to
5x + y ≤ 100
x + y ≤ 60
x, y ≥ 0.
Solve the following LPP graphically:
Maximize Z = 5x + 7y
Subject to
x + y ≤ 4
3x + 8y ≤ 24
10x + 7y ≤ 35
x, y ≥ 0
Solve the following LPP graphically:
Minimize Z = 3x + 5y
Subject to
– 2x + y ≤ 4
x + y ≥ 3
x – 2y ≤ 2
x, y ≥ 0
A house wife wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A,12 units of vitamin B and 8 units of vitamin C.
The vitamin contents of one kg of food is given below:

Vitamin A 
Vitamin B 
Vitamin C 
Food X: 
1 
2 
3 
Food Y: 
2 
2 
1 
One kg of food X costs Rs 6 and one kg of food Y costs Rs 10. Find the least cost of the mixture which will produce the diet.
A dietician wishes to mix two types of food in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C while food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs 5.00 per kg to purchase food ‘I’ and Rs 7.00 per kg to produce food ‘II’. Determine the minimum cost to such a mixture. formulate the above as a LPP and solve it.
Every gram of wheat provides 0.1 gm of proteins and 0.25 gm of carbohydrates. The corresponding values of rice are 0.05 gm and 0.5 gm respectively. Wheat costs Rs. 4 per kg and rice Rs. 6. The minimum daily requirements of proteins and carbohydrates for an average child are 50 gms and 200 gms respectively. In what quantities should wheat and rice be mixed in the daily diet to provide minimum daily requirements of proteins and carbohydrates at minimum cost.
A manufacturer produces nuts and bolts for industrial machinery. It takes 1 hour or work on machine A and 3 hours on machine B to produce a package of nuts while it takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 2.50 per package of nuts and Re 1.00 per package of bolts. How many packages or each should he produce each day so as to maximize hit profit, if he operates his machines for at most 12 hours a day? Formulate this mathematically and then solve it.
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