Advertisements
Advertisements
Question
A company manufactures two types of fertilizers F1 and F2. Each type of fertilizer requires two raw materials A and B. The number of units of A and B required to manufacture one unit of fertilizer F1 and F2 and availability of the raw materials A and B per day are given in the table below:
| Raw Material\Fertilizers | F1 | F2 | Availability |
| A | 2 | 3 | 40 |
| B | 1 | 4 | 70 |
By selling one unit of F1 and one unit of F2, company gets a profit of ₹ 500 and ₹ 750 respectively. Formulate the problem as L.P.P. to maximize the profit.
Solution
Let the company manufacture ‘x’ units of fertilizer F1 and ‘y’ units of F2.
The profit on one unit of F1 is ₹ 500 and on unit of F2 is ₹ 750.
∴ Total profit = ₹ (500x + 750y)
From the given table,
The raw material A required for x units of F1 and y units of F2 is (2x + 3y). The raw material B required is (x + 4y).
But maximum availability of raw materials A and B are 40 and 70 units respectively.
∴ The constraints are:
2x + 3y ≤ 40, x + 4y ≤ 70
Since x and y cannot be negative, we have x ≥ 0, y ≥ 0.
∴ Given problem can be formulated as follows:
Maximize Z = 500x + 750y
Subject to 2x + 3y ≤ 40, x + 4y ≤ 70, x ≥ 0, y ≥ 0.
APPEARS IN
RELATED QUESTIONS
Find the feasible solution of the following inequation:
x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B per unit and the number of man-hours available for the firm is as follows :
| Gadgets | Foundry | Machine shop |
| A | 10 | 5 |
| B | 6 | 4 |
| Time available (hour) | 60 | 35 |
Profit on the sale of A is ₹ 30 and B is ₹ 20 per units. Formulate the L.P.P. to have maximum profit.
A company manufactures two types of chemicals Aand B. Each chemical requires two types of raw material P and Q. The table below shows number of units of P and Q required to manufacture one unit of A and one unit of B and the total availability of P and Q.
| Chemical→ | A | B | Availability |
| Raw Material ↓ | |||
| P | 3 | 2 | 120 |
| Q | 2 | 5 | 160 |
The company gets profits of ₹ 350 and ₹ 400 by selling one unit of A and one unit of B respectively. (Assume that the entire production of A and B can be sold). How many units of the chemicals A and B should be manufactured so that the company gets a maximum profit? Formulate the problem as LPP to maximize profit.
Solve the following LPP by graphical method:
Maximize z = 4x + 6y, subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.
Objective function of LPP is ______.
Solve the following LPP:
Maximize z = 4x + 2y subject to 3x + y ≤ 27, x + y ≤ 21, x ≥ 0, y ≥ 0.
Solve the following LPP:
Maximize z = 2x + 3y subject to x - y ≥ 3, x ≥ 0, y ≥ 0.
Find graphical solution for the following system of linear in equation:
3x + 4y ≤ 12, x - 2y ≥ 2, y ≥ - 1
Solve the following LPP:
Maximize z =60x + 50y subject to
x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0.
Solve the following L.P.P. by graphical method:
Maximize: Z = 4x + 6y
Subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.
Choose the correct alternative :
Solution of LPP to minimize z = 2x + 3y st. x ≥ 0, y ≥ 0, 1≤ x + 2y ≤ 10 is
Choose the correct alternative :
The half plane represented by 3x + 2y ≤ 0 constraints the point.
Fill in the blank :
A dish washing machine holds up to 40 pieces of large crockery (x) This constraint is given by_______.
Solve the Linear Programming problem graphically:
Maximize z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0 also find the maximum value of z.
State whether the following statement is True or False:
Objective function of LPP is a relation between the decision variables
A solution which maximizes or minimizes the given LPP is called
The maximum value of the objective function Z = 3x + 5y subject to the constraints x ≥ 0, y ≥ 0 and 2x + 5y ≤ 10 is
The set of feasible solutions of LPP is a ______.
Two kinds of foods A and B are being considered to form a weekly diet. The minimum weekly requirements of fats, Carbohydrates and proteins are 12, 16 and 15 units respectively. One kg of food A has 2, 8 and 5 units respectively of these ingredients and one kg of food B has 6, 2 and 3 units respectively. The price of food A is Rs. 4 per kg and that of food B is Rs. 3 per kg. Formulate the L.P.P. and find the minimum cost.
