Question
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.
Solution
The least cost of the mixture is Rs 52
Let x kg of food X and y kg of food Y are mixed together to make the mixture. Since one kg of food X contains one unit of vitamin A and one kg of food Y contains 2 units of vitamin A. Therefore, x kg of food X and ykg of food Y will contain x + 2y units of vitamin A. But the mixture should contain at least 10 units of vitamin A. Therefore,
x + 2y ≥ 10
Similarly, x kg of food X and y kg of food Y will produce 2x + 2y units of vitamin B and 3x + y units of vitamin C. But the minimum requirements of vitamins B and C are respectively of 12 and 8 units.
∴ 2x + 2y ≥ 12 and 3x + y ≥ 8
Since the quantity of food X and food Y cannot be negative.
∴ x ≥ 0, y ≥ 0
It is given that one kg of food X costs Rs 6 and one kg of food Y costs Rs 10. So, x kg of food X and y kg of food Y will cost Rs (6x = 10y).
Thus, the given linear programming problem is
Minimize Z = 6x + 10y
Subject to x + 2y ≥ 10
2x + 2y ≥ 12
3x + y ≥ 8
and, x ≥ 0, y ≥ 0
To solve this LPP, we draw the lines
x + 2y = 10, 2x + 2y = 12 and 3x + y = 8.
The feasible region of the LPP is shaded in fig.
The coordinates of the vertices (Corner-points) of shaded feasible regionA1P1P2B3 are A1 (10, 0), P1 (2, 4), P2 (1, 5) and B3 (0, 8). These points have been 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 the objective function Z = 6x + 10y |
|
A1 (10, 0) A2 (2, 4) P2 (1, 5) B3 (0, 8) |
Z = 6 × 10 + 10 × 0 = 60 Z = 6 × 2 + 10 × 4 = 52 Z = 6 × 1 + 10 × 5 = 56 Z = 6 × 0 + 10 × 8 = 80 |
Clearly, Z is minimum at x = 2 and y = 4. The minimum value of Z is 52.
We observe that the open half-plane represented by 6x + 10y < 52 does not have points in common with the feasible region. So, Z has minimum value equal to 52.
Hence, the least cost of the mixture is Rs 52.
SIMILAR QUESTIONS
A company is making two products A and B. The cost of producing one unit of products A and B are Rs 60 and Rs 80 respectively. As per the agreement, the company has to supply at least 200 units of product B to its regular customers. One unit product A requires one machine hour whereas product B has machine hours available abundantly within the company. Total machine hours available for product A are 400 hours. One unit of each product A and B requires one labour hour each and total of 500 labour hours are available. The company wants to minimize the cost of production by satisfying the given requirements. Formulate the problem as a LLP.
A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:
|
|
Product A |
Product B |
Weekly capacity |
|
Department 1 |
3 |
2 |
130 |
|
Department 2 |
4 |
6 |
260 |
|
Selling price per unit |
Rs 25 |
Rs 30 |
|
|
Labour cost per unit |
Rs 16 |
Rs 20 |
|
|
Raw material cost per unit |
Rs 4 |
Rs 4 |
|
The problem is to determine the number of units of produce each product so as to maximize total contribution to profit. Formulate this as a LLP.
Solve the following LPP graphically:
Maximize Z = 5x + 3y
Subject to
3x + 5y ≤ 15
5x + 2y ≤ 10
And, x, y ≥ 0.
Solve the following LPP by graphical method:
Minimize Z = 20x + 10y
Subject to x + 2y ≤ 40
3x + y ≥ 30
4x + 3y ≥ 60
And, x, y ≥ 0
Solve the following LPP graphically:
Minimize and Maximize Z = 5x + 2y
Subject to –2x – 3y ≤ – 6
x – 2y ≤ 2
3x + 2y ≤ 12
–3x + 2y ≤ 3
x, y ≥ 0
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 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.