Advertisements
Advertisements
प्रश्न
Solve the following LP.P.
Maximize z = 13x + 9y,
Subject to 3x + 2y ≤ 12,
x + y ≥ 4,
x ≥ 0,
y ≥ 0.
उत्तर
| Equation | x | y | Points | Side |
| 3x + 2y = 12 | 0 | 6 | A(0, 6) | Origin |
| 4 | 0 | B(4, 0) | Side | |
| x + y = 4 | 0 | 4 | C(0, 4) | Non-origin |
| 4 | 0 | D(4, 0) | Side |
Shaded region is the feasible region ABCA.
Z = 13x + 9y
Z(A) = 0 + 9 × 6 = 54
Z(B) = 13 × 4 + 0 = 52
Z(C) = 0 + 4 × 9 = 36
∴ Max, value of z is 54 at A(0, 6)
i.e., when x = 0,
y = 6.
APPEARS IN
संबंधित प्रश्न
Which of the following statements is correct?
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.
A furniture dealer deals in tables and chairs. He has ₹ 1,50,000 to invest and a space to store at most 60 pieces. A table costs him ₹ 1500 and a chair ₹ 750. Construct the inequations and find the feasible solution.
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 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.
Solve the following LPP by graphical method:
Maximize z = 11x + 8y, subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0
Minimize z = 6x + 2y, subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
Select the appropriate alternatives for each of the following question:
The value of objective function is maximum under linear constraints
The maximum value of z = 5x + 3y subject to the constraints 3x + 5y ≤ 15, 5x + 2y ≤ 10, x, y ≥ 0 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 ______.
A carpenter makes chairs and tables. Profits are ₹ 140 per chair and ₹ 210 per table. Both products are processed on three machines: Assembling, Finishing and Polishing. The time required for each product in hours and availability of each machine is given by the following table:
| Product → | Chair (x) | Table (y) | Available time (hours) |
| Machine ↓ | |||
| Assembling | 3 | 3 | 36 |
| Finishing | 5 | 2 | 50 |
| Polishing | 2 | 6 | 60 |
Formulate the above problem as LPP. Solve it graphically
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 manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to machine shop for finishing. The number of man hours of labour required in each shop for production of A and B and the number of man hours available for the firm are as follows:
| Gadgets | Foundry | Machine Shop |
| A | 10 | 5 |
| B | 6 | 4 |
| Time available (hours) | 60 | 35 |
Profit on the sale of A is ₹ 30 and B is ₹ 20 per unit. Formulate the L.P.P. to have maximum profit.
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
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 :
The optimal value of the objective function is attained at the _______ points of feasible region.
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.
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
State whether the following statement is True or False:
LPP is related to efficient use of limited resources
A company produces two types of pens A and B. Pen A is of superior quality and pen B is of lower quality. Profits on pens A and B are ₹ 5 and ₹ 3 per pen respectively. Raw materials required for each pen A is twice as that of pen B. The supply of raw material is sufficient only for 1000 pens per day. Pen A requires a special clip and only 400 such clips are available per day. For pen B, only 700 clips are available per day. Formulate this problem as a linear programming problem.
A company manufactures two models of voltage stabilizers viz., ordinary and auto-cut. All components of the stabilizers are purchased from outside sources, assembly and testing is carried out at the company’s own works. The assembly and testing time required for the two models are 0.8 hours each for ordinary and 1.20 hours each for auto-cut. Manufacturing capacity 720 hours at present is available per week. The market for the two models has been surveyed which suggests a maximum weekly sale of 600 units of ordinary and 400 units of auto-cut. Profit per unit for ordinary and auto-cut models has been estimated at ₹ 100 and ₹ 150 respectively. Formulate the linear programming problem.
Solve the following linear programming problems by graphical method.
Maximize Z = 40x1 + 50x2 subject to constraints 3x1 + x2 ≤ 9; x1 + 2x2 ≤ 8 and x1, x2 ≥ 0.
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.
A firm manufactures pills in two sizes A and B. Size A contains 2 mgs of aspirin, 5 mgs of bicarbonate and 1 mg of codeine. Size B contains 1 mg. of aspirin, 8 mgs. of bicarbonate and 6 mgs. of codeine. It is found by users that it requires at least 12 mgs. of aspirin, 74 mgs. of bicarbonate and 24 mgs. of codeine for providing immediate relief. It is required to determine the least number of pills a patient should take to get immediate relief. Formulate the problem as a standard LLP.
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.
