Advertisements
Advertisements
प्रश्न
Solve the following problem :
A firm manufacturing two types of electrical items A and B, can make a profit of ₹ 20 per unit of A and ₹ 30 per unit of B. Both A and B make use of two essential components, a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should be manufacture per month to maximize profit? How much is the maximum profit?
Advertisements
उत्तर
Let the firm manufacture x units of A and y units of B.
The profit is ₹ 20 per unit of A and ₹ 30 per unit of B.
∴ Total profit = ₹ (20 x + 30 y).
We construct a table with the constraints of number of motors and transformers needed.
| Electrical item\Essential component | A (x) |
B (y) |
Maximum Supply |
| Motors | 3 | 2 | 210 |
| Transformers | 2 | 4 | 300 |
From the table, the total motors required is (3x + 2y) and total motor required is (2x + 4y).
But total supply of components per month is restricted to 210 motors and 300 transformers.
∴ The constraints are 3x + 2y ≤ 210 and 2x + 4y ≤ 300.
As x, y cannot be negative, we have x ≤ 0 and y ≥ 0.
Hence the given LPP can be formulated as follows:
Maximize Z = 20x + 30y
Subject to
3x + 2y ≤ 210,
2x + 4y ≤ 300,
x ≤ 0, y ≥ 0.
For graphical solutions of the inequalities, consider lines L1 : 3x + 2y = 210 and 2x + 4y = 300
For L1 :
| x | y | (x, y) |
| 0 | 105 | (0, 105) |
| 70 | 0 | (70, 0) |
For L2 :
| x | y | (x, y) |
| 0 | 75 | (0, 75) |
| 150 | 0 | (150, 0) |
L1 passes through A (0, 105) and B (70, 0)
L2 passes through P (0, 75) and Q (150, 0)
Solving both lines, we get x = 30, y = 60
The coordinates of origin O (0, 0) satisfies both the inequalities.
∴ The required region is on origin side of both the lines L1 and L2.
As x ≥ 0, y ≥ 0; the feasible region lies in the first quadrant.
OBRP is the required feasible region.
At O (0, 0), Z = 0 + 0 = 0
At B (70, 0), Z = 20 (70) + 0 = 1400
At R (30, 60), Z = 20 (30) + 30 (60) = 2400
At P (0, 75), Z = 0 + 30 (75) = 2250
The maximum value of Z is 2400 and it occurs at R (30, 60)
Thus 30 units of A and 60 units of B must be manufactured to get maximum profit of ₹ 2400.
APPEARS IN
संबंधित प्रश्न
The postmaster of a local post office wishes to hire extra helpers during the Deepawali season, because of a large increase in the volume of mail handling and delivery. Because of the limited office space and the budgetary conditions, the number of temporary helpers must not exceed 10. According to past experience, a man can handle 300 letters and 80 packages per day, on the average, and a woman can handle 400 letters and 50 packets per day. The postmaster believes that the daily volume of extra mail and packages will be no less than 3400 and 680 respectively. A man receives Rs 225 a day and a woman receives Rs 200 a day. How many men and women helpers should be hired to keep the pay-roll at a minimum ? Formulate an LPP and solve it graphically.
A company produces two types of goods A and B, that require gold and silver. Each unit of type A requires 3 g of silver and 1 g of golds while that of type B requires 1 g of silver and 2 g of gold. The company can procure a maximum of 9 g of silver and 8 g of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, formulate LPP to maximize profit.
A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry, then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for the production of each unit of A and B, and the number of man-hours the firm has available per week are as follows:
| Gadget | Foundry | Machine-shop |
| A | 10 | 5 |
| B | 6 | 4 |
| Firm's capacity per week | 1000 | 600 |
The profit on the sale of A is Rs 30 per unit as compared with Rs 20 per unit of B. The problem is to determine the weekly production of gadgets A and B, so that the total profit is maximized. Formulate this problem as a LPP.
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 L.P.P. by graphical method:
Minimize: Z = 6x + 2y subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
Choose the correct alternative :
The maximum value of z = 5x + 3y. subject to the constraints
Choose the correct alternative :
The maximum value of z = 10x + 6y, subjected to the constraints 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, y ≥ 0 is.
Fill in the blank :
The region represented by the in equations x ≤ 0, y ≤ 0 lines in _______ quadrants.
State whether the following is True or False :
The region represented by the inqualities x ≤ 0, y ≤ 0 lies in first quadrant.
Graphical solution set of x ≤ 0, y ≥ 0 in xy system lies in second quadrant.
Solve the following problem :
Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, x2 ≥ 0
Solve the following problem :
Minimize Z = 4x + 2y Subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0
Solve the following problem :
Minimize Z = 2x + 3y Subject to x – y ≤ 1, x + y ≥ 3, x ≥ 0, y ≥ 0
Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
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 the availability of each machine is given by the following table.
| Product/Machines | Chair (x) |
Table (y) |
Available time (hours) |
| Assembling | 3 | 3 | 36 |
| Finishing | 5 | 2 | 50 |
| Polishing | 2 | 6 | 60 |
Formulate and solve the following Linear programming problems using graphical method.
Solve the following problem :
A company manufactures bicyles and tricycles, each of which must be processed through two machines A and B Maximum availability of machine A and B is respectively 120 and 180 hours. Manufacturing a bicycle requires 6 hours on machine A and 3 hours on machine B. Manufacturing a tricycle requires 4 hours on machine A and 10 hours on machine B. If profits are ₹ 180 for a bicycle and ₹ 220 on a tricycle, determine the number of bicycles and tricycles that should be manufacturing in order to maximize the profit.
Solve the following problem :
A factory produced two types of chemicals A and B The following table gives the units of ingredients P & Q (per kg) of Chemicals A and B as well as minimum requirements of P and Q and also cost per kg. of chemicals A and B.
| Ingredients per kg. /Chemical Units | A (x) |
B (y) |
Minimum requirements in |
| P | 1 | 2 | 80 |
| Q | 3 | 1 | 75 |
| Cost (in ₹) | 4 | 6 |
Find the number of units of chemicals A and B should be produced so as to minimize the cost.
Choose the correct alternative:
If LPP has optimal solution at two point, then
Choose the correct alternative:
The minimum value of Z = 4x + 5y subjected to the constraints x + y ≥ 6, 5x + y ≥ 10, x, y ≥ 0 is
Choose the correct alternative:
The corner points of the feasible region are (0, 3), (3, 0), (8, 0), `(12/5, 38/5)` and (0, 10), then the point of maximum Z = 6x + 4y = 48 is at
Choose the correct alternative:
The corner points of the feasible region are (4, 2), (5, 0), (4, 1) and (6, 0), then the point of minimum Z = 3.5x + 2y = 16 is at
State whether the following statement is True or False:
If the corner points of the feasible region are `(0, 7/3)`, (2, 1), (3, 0) and (0, 0), then the maximum value of Z = 4x + 5y is 12
State whether the following statement is True or False:
If the corner points of the feasible region are (0, 10), (2, 2) and (4, 0), then the minimum value of Z = 3x + 2y is at (4, 0)
State whether the following statement is True or False:
The point (6, 4) does not belong to the feasible region bounded by 8x + 5y ≤ 60, 4x + 5y ≤ 40, 0 ≤ x, y
The feasible region represented by the inequations x ≥ 0, y ≤ 0 lies in ______ quadrant.
A company manufactures two types of ladies dresses C and D. The raw material and labour available per day is given in the table.
| Resources | Dress C(x) | Dress D(y) | Max. availability |
| Raw material | 5 | 4 | 60 |
| Labour | 5 | 3 | 50 |
P is the profit, if P = 50x + 100y, solve this LPP to find x and y to get the maximum profit
A chemist has a compound to be made using 3 basic elements X, Y, Z so that it has at least 10 litres of X, 12 litres of Y and 20 litres of Z. He makes this compound by mixing two compounds (I) and (II). Each unit compound (I) had 4 litres of X, 3 litres of Y. Each unit compound (II) had 1 litre of X, 2 litres of Y and 4 litres of Z. The unit costs of compounds (I) and (II) are ₹ 400 and ₹ 600 respectively. Find the number of units of each compound to be produced so as to minimize the cost
Amartya wants to invest ₹ 45,000 in Indira Vikas Patra (IVP) and in Public Provident fund (PPF). He wants to invest at least ₹ 10,000 in PPF and at least ₹ 5000 in IVP. If the rate of interest on PPF is 8% per annum and that on IVP is 7% per annum. Formulate the above problem as LPP to determine maximum yearly income.
Solution: Let x be the amount (in ₹) invested in IVP and y be the amount (in ₹) invested in PPF.
x ≥ 0, y ≥ 0
As per the given condition, x + y ______ 45000
He wants to invest at least ₹ 10,000 in PPF.
∴ y ______ 10000
Amartya wants to invest at least ₹ 5000 in IVP.
∴ x ______ 5000
Total interest (Z) = ______
The formulated LPP is
Maximize Z = ______ subject to
______
