Advertisements
Advertisements
प्रश्न
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.
उत्तर
Let the company manufactures x units of chemical A and y units of chemical B. Then the total profit to the company is p = ₹(350x + 400y).
This is a linear function that is to be maximized. Hence, it is an objective function.
The constraints are as per the following table:
Chemical→ | A (x) |
B (y) |
Availability |
Raw Material ↓ | |||
P | 3 | 2 | 120 |
Q | 2 | 5 | 160 |
The raw material P required for x units of chemical A and y units of chemical B is 3x + 2y. Since the maximum availability of P is 120, we have the first constraint as 3x + 2y ≤ 120.
Similarly, considering the raw material Q, we have 2x + 5y ≤ 160.
Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0.
Hence, the given LPP can be formulated as:
Maximize p = 350x + 400y, subject to
3x + 2y ≤ 120, 2x + 5y ≤ 160, x ≥ 0, y ≥ 0
APPEARS IN
संबंधित प्रश्न
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.
Solve the following LPP by graphical method:
Maximize z = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Which of the following is correct?
Of all the points of the feasible region, the optimal value of z obtained at the point lies ______.
Solution of LPP to minimize z = 2x + 3y, such that x ≥ 0, y ≥ 0, 1 ≤ x + 2y ≤ 10 is ______.
The corner points of the feasible solution given by the inequation x + y ≤ 4, 2x + y ≤ 7, x ≥ 0, y ≥ 0 are ______.
If the corner points of the feasible solution are (0, 10), (2, 2) and (4, 0), then the point of minimum z = 3x + 2y is ______.
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 = 4x1 + 3x2 subject to
3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0.
Solve the following LPP:
Minimize z = 4x + 2y
Subject to 3x + y ≥ 27, x + y ≥ 21, x + 2y ≥ 30, x ≥ 0, y ≥ 0
In a cattle breeding firm, it is prescribed that the food ration for one animal must contain 14, 22, and 1 unit of nutrients A, B, and C respectively. Two different kinds of fodder are available. Each unit weight of these two contains the following amounts of these three nutrients:
Nutrient\Fodder | Fodder 1 | Fodder2 |
Nutrient A | 2 | 1 |
Nutrient B | 2 | 3 |
Nutrient C | 1 | 1 |
The cost of fodder 1 is ₹ 3 per unit and that of fodder ₹ 2 per unit. Formulate the L.P.P. to minimize the cost.
A company manufactures two types of chemicals A and 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.
Raw Material \Chemical | A | B | Availability |
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. Formulate the problem as L.P.P. to maximize the profit.
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.
Choose the correct alternative :
Feasible region; the set of points which satify.
Choose the correct alternative :
The corner points of the feasible region given by the inequations x + y ≤ 4, 2x + y ≤ 7, x ≥ 0, y ≥ 0, are
Choose the correct alternative :
The corner points of the feasible region are (0, 0), (2, 0), `(12/7, 3/7)` and (0,1) then the point of maximum z = 7x + y
Choose the correct alternative :
The half plane represented by 3x + 2y ≤ 0 constraints the point.
Fill in the blank :
The optimal value of the objective function is attained at the _______ points of feasible region.
The feasible region is the set of point which satisfy.
Which value of x is in the solution set of inequality − 2X + Y ≥ 17
Minimize z = 7x + y subjected to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
x − y ≤ 1, x − y ≥ 0, x ≥ 0, y ≥ 0 are the constant for the objective function z = x + y. It is solvable for finding optimum value of z? Justify?
Choose the correct alternative:
The feasible region is
Choose the correct alternative:
Z = 9x + 13y subjected to constraints 2x + 3y ≤ 18, 2x + y ≤ 10, 0 ≤ x, y was found to be maximum at the point
State whether the following statement is True or False:
LPP is related to efficient use of limited resources
Constraints are always in the form of ______ or ______.
The constraint that in a particular XII class, number of boys (y) are less than number of girls (x) is given by ______
Solve the following linear programming problems by graphical method.
Maximize Z = 22x1 + 18x2 subject to constraints 960x1 + 640x2 ≤ 15360; x1 + x2 ≤ 20 and x1, x2 ≥ 0.
Solve the following linear programming problems by graphical method.
Maximize Z = 20x1 + 30x2 subject to constraints 3x1 + 3x2 ≤ 36; 5x1 + 2x2 ≤ 50; 2x1 + 6x2 ≤ 60 and x1, x2 ≥ 0.
Solve the following linear programming problems by graphical method.
Minimize Z = 20x1 + 40x2 subject to the constraints 36x1 + 6x2 ≥ 108; 3x1 + 12x2 ≥ 36; 20x1 + 10x2 ≥ 100 and x1, x2 ≥ 0.
The maximum value of the objective function Z = 3x + 5y subject to the constraints x ≥ 0, y ≥ 0 and 2x + 5y ≤ 10 is
The minimum value of the objective function Z = x + 3y subject to the constraints 2x + y ≤ 20, x + 2y ≤ 20, x > 0 and y > 0 is
A firm manufactures two products A and B on which the profits earned per unit are ₹ 3 and ₹ 4 respectively. Each product is processed on two machines M1 and M2. Product A requires one minute of processing time on M1 and two minutes on M2, While B requires one minute on M1 and one minute on M2. Machine M1 is available for not more than 7 hrs 30 minutes while M2 is available for 10 hrs during any working day. Formulate this problem as a linear programming problem to maximize the profit.
Solve the following linear programming problem graphically.
Maximize Z = 3x1 + 5x2 subject to the constraints: x1 + x2 ≤ 6, x1 ≤ 4; x2 ≤ 5, and x1, x2 ≥ 0.
The maximum value of Z = 9x + 13y subject to constraints 2x + 3y ≤ 18, 2x + y ≤ 10, x ≥ 0, y ≥ 0 is ______.
For the following shaded region, the linear constraint are:
Sketch the graph of the following inequation in XOY co-ordinate system.
2y - 5x ≥ 0
Find graphical solution for the following system of linear in equation:
x + 2y ≥ 4, 2x - y ≤ 6