Advertisements
Advertisements
प्रश्न
A firm manufacturing two types of electric items, A and B, can make a profit of Rs 20 per unit of A and Rs 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers. Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month. Formulate the linear programing problem for maximum profit and solve it graphically.
Advertisements
उत्तर
Let x units of item A and y units of item B were manufactured.
Number of items cannot be negative.
Therefore, \[x, y \geq 0\]
The given information can be tabulated as follows:
| Product | Motors | Transformers |
| A(x) | 3 | 4 |
| B(y) | 2 | 4 |
| Availability | 210 | 300 |
Further, it is given that type B is an export model, whose supply is restricted to 65 units per month.
Therefore, the constraints are
\[3x + 2y \leq 210\]
\[4x + 4y \leq 300\]
\[y \leq 65\]
A and B can make a profit of Rs 20 per unit of A and Rs 30 per unit of B.Therefore, profit gained from x units of item A and y units of item B is Rs 20x and Rs 30y respectively.
Total profit = Z = \[20x + 30y\] which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = \[20x + 30y\]
subject to
\[3x + 2y \leq 210\]
\[4x + 4y \leq 300\]
\[y \leq 65\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows:
3x + 2y = 210, 4x + 4y = 300, y = 65, x = 0 and y = 0
Region represented by 3x + 2y ≤ 210:
The line 3x + 2y = 210 meets the coordinate axes at A1(70, 0) and B1(0, 105) respectively. By joining these points we obtain the line 3x + 2y = 210. Clearly (0,0) satisfies the 3x + 2y = 210. So,the region which contains the origin represents the solution set of the inequation 3x + 2y ≤ 210.
Region represented by 4x + 4y ≤ 300:
The line 4x + 4y = 300 meets the coordinate axes at C1(75, 0) and D1(0, 75) respectively. By joining these points we obtain the line
4x + 4y = 300.Clearly (0,0) satisfies the inequation 4x + 4y ≤ 300. So,the region which contains the origin represents the solution set of the inequation
4x + 4y ≤ 300.
y = 65 is the line passing through the point E1(0, 65) and is parallel to X axis.
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 3x + 2y ≤ 210, 4x + 4y ≤ 300, y ≤ 65, x ≥ 0, and y ≥ 0 are as follows
The corner points are O(0, 0), E1(0, 65), G1
The values of Z at these corner points are as follows
| Corner point | Z= 20x + 30y |
| O | 0 |
| E1 | 1950 |
| G1 | 2150 |
| F1 | 1650 |
| A1 | 1400 |
The maximum value of Z is 2150 which is attained at G1
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.
Find graphically, the maximum value of z = 2x + 5y, subject to constraints given below :
2x + 4y ≤ 83
x + y ≤ 6
x + y ≤ 4
x ≥ 0, y≥ 0
Maximize Z = 50x + 30y
Subject to
\[2x + y \leq 18\]
\[3x + 2y \leq 34\]
\[ x, y \geq 0\]
Maximize Z = 10x + 6y
Subject to
\[3x + y \leq 12\]
\[2x + 5y \leq 34\]
\[ x, y \geq 0\]
Minimize Z = x − 5y + 20
Subject to
\[x - y \geq 0\]
\[ - x + 2y \geq 2\]
\[ x \geq 3\]
\[ y \leq 4\]
\[ x, y \geq 0\]
Solved the following linear programming problem graphically:
Maximize Z = 60x + 15y
Subject to constraints
\[x + y \leq 50\]
\[3x + y \leq 90\]
\[ x, y \geq 0\]
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 linear programming problem graphically:
Minimize z = 6 x + 3 y
Subject to the constraints:
4 x + \[y \geq\] 80
x + 5 \[y \geq\] 115
3 x + 2 \[y \leq\] 150
\[x \geq\] 0 , \[y \geq\] 0
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum of vitamin A.
If a young man drives his vehicle at 25 km/hr, he has to spend Rs 2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to Rs 5/per km. He has Rs 100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
A manufacturer produces two types of steel trunks. He has two machines A and B. For completing, the first types of the trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of the trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit?
A company manufactures two articles A and B. There are two departments through which these articles are processed: (i) assembly and (ii) finishing departments. The maximum capacity of the first department is 60 hours a week and that of other department is 48 hours per week. The product of each unit of article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs 6 for each unit of A and Rs 8 for each unit of B, find the number of units of A and B to be produced per week in order to have maximum profit.
A firm makes items A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and half an hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs 300 and on one item of B is Rs 160. How many items of each type should be produced to maximize the profit? Solve the problem graphically.
If a young man drives his vehicle at 25 km/hr, he has to spend ₹2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to ₹5 per km. He has ₹100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
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.
A manufacturer has three machine I, II, III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines.
The number of hours required for producing 1 unit each of M and N on the three machines are given in the following table:
| Items | Number of hours required on machines | ||
| I | II | III | |
| M | 1 | 2 | 1 |
| N | 2 | 1 | 1.25 |
She makes a profit of ₹600 and ₹400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?
A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs 100 and that on a bracelet is Rs 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit?
It is being given that at least one of each must be produced.
Tow godowns, A and B, have grain storage capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F, whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:
| Transportation cost per quintal(in Rs.) | ||
| From-> | A | B |
| To | ||
| D | 6.00 | 4.00 |
| E | 3.00 | 2.00 |
| F | 2.50 | 3.00 |
How should the supplies be transported in order that the transportation cost is minimum?
By graphical method, the solution of linear programming problem
\[\text{ Subject } to \text{ 3 } x_1 + 2 x_2 \leq 18\]
\[ x_1 \leq 4\]
\[ x_2 \leq 6\]
\[ x_1 \geq 0, x_2 \geq 0, \text{ is } \]
The value of objective function is maximum under linear constraints ______.
A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 6% phosphoric acid and of type B which contains 5% nitrogen and 10% phosphoric acid. After the soil test, it is found that at least 7 kg of nitrogen and the same quantity of phosphoric acid is required for a good crop. The fertilizer of type A costs ₹ 5.00 per kg and the type B costs ₹ 8.00 per kg. Using Linear programming, find how many kilograms of each type of fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph.
Maximum value of 4x + 13y subject to constraints x ≥ 0, y ≥ 0, x + y ≤ 5 and 3x + y ≤ 9 is ______.
Area of the region bounded by y = cos x, x = 0, x = π and X-axis is ______ sq.units.
The constraints of an LPP are 7 ≤ x ≤ 12, 8 ≤ y ≤ 13. Determine the vertices of the feasible region formed by them.
If 4x + 5y ≤ 20, x + y ≥ 3, x ≥ 0, y ≥ 0, maximum 2x + 3y is ______.
The minimum value of z = 7x + 9y subject to 3x + y ≤ 6, 5x + 8y ≤ 40, x ≥ 0, y ≥ 2 is ______.
A feasible region in the set of points which satisfy ____________.
Of all the points of the feasible region for maximum or minimum of objective function the points.
A feasible solution to a linear programming problem
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 ______.
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.
The solution set of constraints x + 2y ≥ 11, 3x + 4y ≤ 30, 2x + 5y ≤ 30 and x ≥ 0, y ≥ 0, includes the point ______.
For a bounded feasible region, where do the maximum and minimum values occur?
For \[x+y\leq50\], \[3x+y\leq90\], \[x\geq0\], and \[y\geq0\], which is a corner point of the feasible region?
