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प्रश्न
The feasible region is the set of point which satisfy.
विकल्प
The object functions
All the given constraints
Some of the given constraints
Only one constraint
उत्तर
All the given constraints
संबंधित प्रश्न
Find the feasible solution of the following inequation:
3x + 2y ≤ 18, 2x + y ≤ 10, x ≥ 0, y ≥ 0
A company produces two types of articles A and B which requires silver and gold. Each unit of A requires 3 gm of silver and 1 gm of gold, while each unit of B requires 2 gm of silver and 2 gm of gold. The company has 6 gm of silver and 4 gm of gold. Construct the inequations and find feasible solution graphically.
In a cattle breading firm, it is prescribed that the food ration for one animal must contain 14. 22 and 1 units of nutrients A, B, and C respectively. Two different kinds of fodder are available. Each unit of these two contains the following amounts of these three nutrients:
| Fodder → | Fodder 1 | Fodder 2 |
| Nutrient ↓ | ||
| Nutrients A | 2 | 1 |
| Nutrients B | 2 | 3 |
| Nutrients C | 1 | 1 |
The cost of fodder 1 is ₹ 3 per unit and that of fodder 2 ₹ 2. Formulate the LPP to minimize the cost.
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:
| Fertilizers→ | F1 | F2 | Availability |
| Raw Material ↓ | |||
| A | 2 | 3 | 40 |
| B | 1 | 4 | 70 |
By selling one unit of F1 and one unit of F2, the company gets a profit of ₹ 500 and ₹ 750 respectively. Formulate the problem as LPP to maximize the profit.
A doctor has prescribed two different units of foods A and B to form a weekly diet for a sick person. The minimum requirements of fats, carbohydrates and proteins are 18, 28, 14 units respectively. One unit of food A has 4 units of fat, 14 units of carbohydrates and 8 units of protein. One unit of food B has 6 units of fat, 12 units of carbohydrates and 8 units of protein. The price of food A is ₹ 4.5 per unit and that of food B is ₹ 3.5 per unit. Form the LPP, so that the sick person’s diet meets the requirements at a minimum cost.
The company makes concrete bricks made up of cement and sand. The weight of a concrete brick has to be at least 5 kg. Cement costs ₹ 20 per kg and sand costs of ₹ 6 per kg. Strength consideration dictates that a concrete brick should contain minimum 4 kg of cement and not more than 2 kg of sand. Form the L.P.P. for the cost to be minimum.
Solve the following LPP by graphical method:
Maximize z = 11x + 8y, subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0
Solve the following LPP by graphical method:
Maximize z = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Minimize z = 6x + 2y, subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
The corner points of the feasible solution given by the inequation x + y ≤ 4, 2x + y ≤ 7, x ≥ 0, y ≥ 0 are ______.
The corner points of the feasible solution are (0, 0), (2, 0), `(12/7, 3/7)`, (0, 1). Then z = 7x + y is maximum at ______.
The half-plane represented by 4x + 3y >14 contains the point ______.
Solve the following LPP:
Maximize z = 5x1 + 6x2 subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x1 ≥ 0, x2 ≥ 0.
Solve each of the following inequations graphically using XY-plane:
4x - 18 ≥ 0
Solve each of the following inequations graphically using XY-plane:
5y - 12 ≥ 0
A chemical company produces a chemical containing three basic elements A, B, C, so that it has at least 16 litres of A, 24 litres of B and 18 litres of C. This chemical is made by mixing two compounds I and II. Each unit of compound I has 4 litres of A, 12 litres of B and 2 litres of C. Each unit of compound II has 2 litres of A, 2 litres of B and 6 litres of C. The cost per unit of compound I is ₹ 800 and that of compound II is ₹ 640. Formulate the problems as LPP and solve it 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 printing company prints two types of magazines A and B. The company earns ₹ 10 and ₹ 15 on magazines A and B per copy. These are processed on three machines I, II, III. Magazine A requires 2 hours on Machine I, 5 hours on Machine II and 2 hours on Machine III. Magazine B requires 3 hours on Machine I, 2 hours on Machine II and 6 hours on Machine III. Machines I, II, III are available for 36, 50, 60 hours per week respectively. Formulate the Linear programming problem to maximize the profit.
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 :
Of all the points of the feasible region the optimal value of z is obtained at a point
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 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
If the corner points of the feasible region are (0, 0), (3, 0), (2, 1) and `(0, 7/3)` the maximum value of z = 4x + 5y is ______.
Fill in the blank :
“A gorage employs eight men to work in its shownroom and repair shop. The constraints that there must be at least 3 men in showroom and at least 2 men in repair shop are ______ and _______ respectively.
The point of which the maximum value of z = x + y subject to constraints x + 2y ≤ 70, 2x + y ≤ 90, x ≥ 0, y ≥ 0 is obtained at
Which value of x is in the solution set of inequality − 2X + Y ≥ 17
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.
Minimize z = 2x + 4y is subjected to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0 show that the minimum value of z occurs at more than two points
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?
Solve the following linear programming problems by graphical method.
Maximize Z = 6x1 + 8x2 subject to constraints 30x1 + 20x2 ≤ 300; 5x1 + 10x2 ≤ 110; and x1, x2 ≥ 0.
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.
Maximize: z = 3x1 + 4x2 subject to 2x1 + x2 ≤ 40, 2x1 + 5x2 ≤ 180, x1, x2 ≥ 0. In the LPP, which one of the following is feasible comer point?
In the given graph the coordinates of M1 are
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
Solve the following linear programming problem graphically.
Minimize Z = 200x1 + 500x2 subject to the constraints: x1 + 2x2 ≥ 10; 3x1 + 4x2 ≤ 24 and x1 ≥ 0, x2 ≥ 0.
The maximum value of Z = 3x + 5y, subject to 3x + 2y ≤ 18, x ≤ a, y ≤ 6, x, y ≥ 0 is ______.
Solve the following LP.P.
Maximize z = 13x + 9y,
Subject to 3x + 2y ≤ 12,
x + y ≥ 4,
x ≥ 0,
y ≥ 0.
The optimal value of the objective function is attained at the ______ of feasible region.
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:
Solve the following LPP:
Maximize z = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Food F1 contains 2, 6, 1 units and food F2 contains 1, 1, 3 units of proteins, carbohydrates, fats respectively per kg. 8, 12 and 9 units of proteins, carbohydrates and fats is the weekly minimum requirement for a person. The cost of food F1 is Rs. 85 and food F2 is Rs. 40 per kg. Formulate the L.P.P. to minimize the cost.
