Advertisements
Advertisements
प्रश्न
Minimize Z = 5x + 3y
Subject to
\[2x + y \geq 10\]
\[x + 3y \geq 15\]
\[ x \leq 10\]
\[ y \leq 8\]
\[ x, y \geq 0\]
Advertisements
उत्तर
First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 10, x + 3y = 15, x = 10, y = 8
Region represented by 2x + y ≥ 10:
The line 2x + y = 10 meets the coordinate axes at A(5, 0) and B(0, 10) respectively. By joining these points we obtain the line 2x + y = 10.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y≥ 10.
Region represented by x + 3y ≥ 15:
The line x + 3y = 15 meets the coordinate axes at C(15, 0) and D(0, 5) respectively. By joining these points we obtain the line x + 3y = 15.
Clearly (0,0) satisfies the inequation x + 3y ≥ 15. o,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 3y ≥ 15.
The line x = 10 is the line that passes through the point (10, 0) and is parallel to Yaxis.x ≤ 10 is the region to the left of the line x = 10.
The line y = 8 is the line that passes through the point (0, 8) and is parallel to X axis.y ≤ 8 is the region below the line y = 8.
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, 2x + y ≥ 10, x + 3y ≥ 15, x ≤ 10, y ≤ 8, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are E(3, 4), \[H\left( 10, \frac{5}{3} \right)\] F(10, 8) and G(1, 8).
| Corner point | Z = 5x + 3y |
| E(3, 4) | 5 × 3 + 3 × 4 = 27 |
|
\[H\left( 10, \frac{5}{3} \right)\]
|
5 × 10 + 3× \[\frac{5}{3}\] = 55
|
| F(10, 8) | 5 × 10 + 3 × 8 = 74 |
| G(1, 8) | 5 × 1 + 3 × 8 = 29 |
Thus, the optimal value of Z is 27.
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.
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
A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30, respectively. The company makes a profit of Rs 80 on each piece of type A and Rs 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?
Maximize Z = 15x + 10y
Subject to
\[3x + 2y \leq 80\]
\[2x + 3y \leq 70\]
\[ x, y \geq 0\]
Minimize Z = 2x + 4y
Subject to
\[x + y \geq 8\]
\[x + 4y \geq 12\]
\[x \geq 3, y \geq 2\]
Minimize Z = 30x + 20y
Subject to
\[x + y \leq 8\]
\[ x + 4y \geq 12\]
\[5x + 8y = 20\]
\[ x, y \geq 0\]
Maximize Z = x + y
Subject to
\[- 2x + y \leq 1\]
\[ x \leq 2\]
\[ x + y \leq 3\]
\[ x, y \geq 0\]
A diet of two foods F1 and F2 contains nutrients thiamine, phosphorous and iron. The amount of each nutrient in each of the food (in milligrams per 25 gms) is given in the following table:
Nutrients |
Food |
F1 | F2 |
| Thiamine | 0.25 | 0.10 |
|
| Phosphorous | 0.75 | 1.50 | |
| Iron | 1.60 | 0.80 | |
The minimum requirement of the nutrients in the diet are 1.00 mg of thiamine, 7.50 mg of phosphorous and 10.00 mg of iron. The cost of F1 is 20 paise per 25 gms while the cost of F2 is 15 paise per 25 gms. Find the minimum cost of diet.
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1and F2 are available. Food F1 costs Rs 4 per unit and F2 costs Rs 6 per unit one unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements
One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no storage of the other ingredients used in making the cakes.
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.
A manufacturer makes two types A and B of tea-cups. Three machines are needed for the manufacture and the time in minutes required for each cup on the machines is given below:
| Machines | |||
| I | II | III | |
| A B |
12 6 |
18 0 |
6 9 |
Each machine is available for a maximum of 6 hours per day. If the profit on each cup A is 75 paise and that on each cup B is 50 paise, show that 15 tea-cups of type A and 30 of type B should be manufactured in a day to get the maximum profit.
A factory uses three different resources for the manufacture of two different products, 20 units of the resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.
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?
An aeroplane can carry a maximum of 200 passengers. A profit of Rs 400 is made on each first class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats of first class. However, at least 4 times as many passengers prefer to travel by economy class to the first class. Determine how many each type of tickets must be sold in order to maximize the profit for the airline. What is the maximum profit.
A man owns a field of area 1000 sq.m. He wants to plant fruit trees in it. He has a sum of Rs 1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 sq.m of ground per tree and costs Rs 20 per tree and type B requires 20 sq.m of ground per tree and costs Rs 25 per tree. When fully grown, type A produces an average of 20 kg of fruit which can be sold at a profit of Rs 2.00 per kg and type B produces an average of 40 kg of fruit which can be sold at a profit of Rs. 1.50 per kg. How many of each type should be planted to achieve maximum profit when the trees are fully grown? What is the maximum profit?
A producer has 30 and 17 units of labour and capital respectively which he can use to produce two type of goods x and y. To produce one unit of x, 2 units of labour and 3 units of capital are required. Similarly, 3 units of labour and 1 unit of capital is required to produce one unit of y. If x and y are priced at Rs 100 and Rs 120 per unit respectively, how should be producer use his resources to maximize the total revenue? Solve the problem 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 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 and 20 minutes available for cutting and 4 hours available for assembling. The profit is Rs. 50 each for type A and Rs. 60 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize profit? Formulate the above LPP and solve it graphically and also find the maximum profit.
A company manufactures two types of cardigans: type A and type B. It costs ₹ 360 to make a type A cardigan and ₹ 120 to make a type B cardigan. The company can make at most 300 cardigans and spend at most ₹ 72000 a day. The number of cardigans of type B cannot exceed the number of cardigans of type A by more than 200. The company makes a profit of ₹ 100 for each cardigan of type A and ₹ 50 for every cardigan of type B.
Formulate this problem as a linear programming problem to maximize the profit to the company. Solve it graphically and find the maximum profit.
The graph of the inequality 3X − 4Y ≤ 12, X ≤ 1, X ≥ 0, Y ≥ 0 lies in fully in
For L.P.P. maximize z = 4x1 + 2x2 subject to 3x1 + 2x2 ≥ 9, x1 - x2 ≤ 3, x1 ≥ 0, x2 ≥ 0 has ______.
Area of the region bounded by y = cos x, x = 0, x = π and X-axis is ______ sq.units.
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 ______.
The feasible region of an LPP is shown in the figure. If z = 3x + 9y, then the minimum value of z occurs at ______.
The maximum value of z = 3x + 10y subjected to the conditions 5x + 2y ≤ 10, 3x + 5y ≤ 15, x, y ≥ 0 is ______.
The minimum value of z = 7x + 9y subject to 3x + y ≤ 6, 5x + 8y ≤ 40, x ≥ 0, y ≥ 2 is ______.
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 ____________.
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 solution set of the inequality 3x + 5y < 4 is ______.
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:
Maximize: P = 70x + 40y
Subject to: 3x + 2y ≤ 9,
3x + y ≤ 9,
x ≥ 0,y ≥ 0.
Solve the following Linear Programming Problem graphically:
Minimize: Z = 60x + 80y
Subject to constraints:
3x + 4y ≥ 8
5x + 2y ≥ 11
x, y ≥ 0
Draw the rough graph and shade the feasible region for the inequalities x + y ≥ 2, 2x + y ≤ 8, x ≥ 0, y ≥ 0.
If x – y ≥ 8, x ≥ 3, y ≥ 3, x ≥ 0, y ≥ 0 then find the coordinates of the corner points of the feasible region.
Which statement defines a feasible solution?
What is an optimal solution?
What is the result if no feasible region exists?
