Advertisements
Advertisements
प्रश्न
A factory manufactures two types of screws, A and B, each type requiring the use of two machines - an automatic and a hand-operated. It takes 4 minute on the automatic and 6 minutes on the hand-operated machines to manufacture a package of screws 'A', while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws 'A' at a profit of 70 P and screws 'B' at a profit of Rs 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
Advertisements
उत्तर
Let the factory manufacture x screws of type A and y screws of type B on each day. Therefore,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.
| Screw A | Screw B | Availability | |
| Automatic Machine (min) | 4 | 6 | 4 × 60 = 240 |
| Hand Operated Machine (min) | 6 | 3 | 4 × 60 = 240 |
\[4x + 6y \geq 240\]
The manufacturer can sell a package of screws 'A' at a profit of Rs 0.7 and screws 'B' at a profit of Re 1.
Total profit, Z = 0.7x + 1y
The mathematical formulation of the given problem is
Maximize Z = 0.7x + 1y
subject to the constraints,
x, y ≥ 0
First we will convert inequations into equations as follows:
4x + 6y = 240, 6x + 3y = 240, x = 0 and y = 0
Region represented by 4x + 6y ≤ 240:
The line 4x + 6y = 240 meets the coordinate axes at A1(60, 0) and B1(0, 40) respectively. By joining these points we obtain the line 4x + 6y = 240. Clearly (0,0) satisfies the 4x + 6y = 240. So,the region which contains the origin represents the solution set of the inequation 4x + 6y ≤ 240.
Region represented by 6x + 3y ≤ 240:
The line 6x + 3y = 240 meets the coordinate axes at C1(40, 0) and D1(0, 80) respectively. By joining these points we obtain the line 6x + 3y = 240. Clearly (0,0) satisfies the inequation 6x + 3y ≤ 240. So,the region which contains the origin represents the solution set of the inequation 6x + 3y ≤ 240.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 4x + 6y ≤ 240, 6x + 3y ≤ 240, x ≥ 0, and y ≥ 0 are as follows.
The corner points are C1(40, 0), E1(30, 20) and B1(0, 40).
The values of Z at these corner points are as follows.
| Corner point | Z = 7x + 10y |
| C1(40, 0) | 280 |
| E1(30, 20) | 410 |
| B1(0, 40) | 400 |
The maximum value of Z is 410 at (30, 20).
Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.
APPEARS IN
संबंधित प्रश्न
Minimize `z=4x+5y ` subject to `2x+y>=7, 2x+3y<=15, x<=3,x>=0, y>=0` solve using graphical method.
Minimize: Z = 6x + 4y
Subject to the conditions:
3x + 2y ≥ 12,
x + y ≥ 5,
0 ≤ x ≤ 4,
0 ≤ y ≤ 4
Minimize :Z=6x+4y
Subject to : 3x+2y ≥12
x+y ≥5
0 ≤x ≤4
0 ≤ y ≤ 4
There are two types of fertilisers 'A' and 'B'. 'A' consists of 12% nitrogen and 5% phosphoric acid whereas 'B' consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops. If 'A' costs Rs 10 per kg and 'B' cost Rs 8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost
Solve the following LPP by graphical method:
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.
Maximise Z = x + 2y subject to the constraints
`x + 2y >= 100`
`2x - y <= 0`
`2x + y <= 200`
Solve the above LPP graphically
Solve the following LPP by graphical method:
Minimize Z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
Maximize Z = 4x + 3y
subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Minimize Z = 2x + 4y
Subject to
\[x + y \geq 8\]
\[x + 4y \geq 12\]
\[x \geq 3, y \geq 2\]
Maximize Z = 2x + 3y
Subject to
\[x + y \geq 1\]
\[10x + y \geq 5\]
\[x + 10y \geq 1\]
\[ x, y \geq 0\]
Find the maximum and minimum value of 2x + y subject to the constraints:
x + 3y ≥ 6, x − 3y ≤ 3, 3x + 4y ≤ 24, − 3x + 2y ≤ 6, 5x + y ≥ 5, x, y ≥ 0.
Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below:
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, y ≥ 0
Solve the following LPP graphically:
Maximize Z = 20 x + 10 y
Subject to the following constraints
\[x +\]2\[y \leq\]28
3x+ \[y \leq\]24
\[x \geq\] 2x.
\[y \geq\] 0
A company manufactures two types of novelty Souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours available for assembling. The profit is 50 paise each for type A and 60 paise each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?
A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300 and that on a chain is Rs 190, find the number of rings and chains that should be manufactured per day, so as to earn the maximum profit. Make it as an LPP and solve it graphically.
A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and his profit on the desktop model is Rs 4500 and on the portable model is Rs 5000. Make an LPP and solve it graphically.
There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:
| From \ To | Cost (in ₹) | ||
| A | B | C | |
| P | 160 | 100 | 150 |
| Q | 100 | 120 | 100 |
How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost?
The point at which the maximum value of x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is obtained, is ______.
A company manufactures two types of products A and B. Each unit of A requires 3 grams of nickel and 1 gram of chromium, while each unit of B requires 1 gram of nickel and 2 grams of chromium. The firm can produce 9 grams of nickel and 8 grams of chromium. The profit is ₹ 40 on each unit of the product of type A and ₹ 50 on each unit of type B. How many units of each type should the company manufacture so as to earn a maximum profit? Use linear programming to find the solution.
Find the graphical solution for the system of linear inequation 2x + y ≤ 2, x − y ≤ 1
Maximum value of 4x + 13y subject to constraints x ≥ 0, y ≥ 0, x + y ≤ 5 and 3x + y ≤ 9 is ______.
The minimum value of z = 2x + 9y subject to constraints x + y ≥ 1, 2x + 3y ≤ 6, x ≥ 0, y ≥ 0 is ______.
For the function z = 19x + 9y to be maximum under the constraints 2x + 3y ≤ 134, x + 5y ≤ 200, x ≥ 0, y ≥ 0; the values of x and y are ______.
For the LPP, maximize z = x + 4y subject to the constraints x + 2y ≤ 2, x + 2y ≥ 8, x, y ≥ 0 ______.
A feasible solution to a linear programming problem
The corner points of the bounded feasible region of a LPP are A(0,50), B(20, 40), C(50, 100) and D(0, 200) and the objective function is Z = x + 2y. Then the maximum value is ____________.
The maximum value of Z = 3x + 4y subjected to contraints x + y ≤ 40, x + 2y ≤ 60, x ≥ 0 and y ≥ 0 is ____________.
Which of the statements describe the solution set for `-2(x + 8) = - 2x + 20`?
Minimise z = – 3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 What will be the minimum value of z ?
The maximum value of z = 5x + 2y, subject to the constraints x + y ≤ 7, x + 2y ≤ 10, x, y ≥ 0 is ______.
The objective function Z = x1 + x2, subject to the constraints are x1 + x2 ≤ 10, – 2x1 + 3x2 ≤ 15, x1 ≤ 6, x1, x2 ≥ 0, has maximum value ______ of the feasible region.
The objective function Z = ax + by of an LPP has maximum vaiue 42 at (4, 6) and minimum value 19 at (3, 2). Which of the following is true?
The corner points of the feasible region of a linear programming problem are (0, 4), (8, 0) and `(20/3, 4/3)`. If Z = 30x + 24y is the objective function, then (maximum value of Z – minimum value of Z) is equal to ______.
Minimize z = x + 2y,
Subject to x + 2y ≥ 50, 2x – y ≤ 0, 2x + y ≤ 100, x ≥ 0, y ≥ 0.
Solve the following Linear Programming Problem graphically.
Maximise Z = 5x + 2y subject to:
x – 2y ≤ 2,
3x + 2y ≤ 12,
– 3x + 2y ≤ 3,
x ≥ 0, y ≥ 0
Aman has ₹ 1500 to purchase rice and wheat for his grocery shop. Each sack of rice and wheat costs ₹ 180 and Rupee ₹ 120 respectively. He can store a maximum number of 10 bags in his shop. He will earn a profit of ₹ 11 per bag of rice and ₹ 9 per bag of wheat.
- Formulate a Linear Programming Problem to maximise Aman’s profit.
- Calculate the maximum profit.
What is called the feasible region?
What is a corner point of a feasible region?
Which description represents an unbounded region?
The Corner Point Method is based on which theorem?
