Advertisements
Advertisements
प्रश्न
To maintain one's health, a person must fulfil certain minimum daily requirements for the following three nutrients: calcium, protein and calories. The diet consists of only items I and II whose prices and nutrient contents are shown below:
| Food I | Food II | Minimum daily requirement | |
| Calcium Protein Calories |
10 5 2 |
4 6 6 |
20 20 12 |
| Price | Rs 0.60 per unit | Rs 1.00 per unit |
Find the combination of food items so that the cost may be minimum.
उत्तर
Let the person takes x units and y units of food I and II respectively that were taken in the diet.
Since, per unit of food I costs Rs 0.60 and that of food II costs Rs 1.00.
Therefore, x lbs of food I costs Rs 0.60x and y lbs of food II costs Rs 1.00y.
Total cost per day = Rs (0.60x +1.00y)
Let Z denote the total cost per day
Then, Z = 0.60x +1.00y
Since, each unit of food I contains 10 units of calcium.Therefore, x units of food I contains 10x units of calcium.
Each unit of food II contains 4 units of calcium.So, y units of food II contains 4y units of calcium.
Thus, x units of food I and y units of food II contains (10x + 4y) units of calcium.
But, the minimum requirement is 20 units of calcium.
∴ \[10x + 4y \geq 20\]
Since, each unit of food I contains 5 units of protein.Therefore, x units of food I contains 5x units of protein.
Each unit of food II contains 6 units of protein.So,y units of food II contains 6y units of protein.
Thus, x units of food I and y units of food II contains (5x + 6y) units of protein.
But, the minimum requirement is 20 lbs of protein.
∴ \[5x + 6y \geq 20\]
Since, each unit of food I contains 2 units of calories.Therefore, x units of food I contains 2x units of calories.
Each unit of food II contains 6 units of calories.So,y units of food II contains 6y units of calories.
Thus, x units of food I and y units of food II contains (2x + 6y) units of calories.
But, the minimum requirement is 12 lbs of calories.\[\therefore 2x + 6y \geq 12\]
Finally, the quantities of food I and food II are non negative values.
So,
\[x, y \geq 0\]
Hence, the required LPP is as follows:
Min Z = 0.60x + 1.00y
subject to
\[10x + 4y \geq 20\]
\[5x + 6y \geq 20\]
\[2x + 6y \geq 12\]
\[x, y \geq 0\]
First, we will convert the given inequations into equations, we obtain the following equations:
10x + 4y = 20, 5x +6y = 20, 2x + 6y =12, x = 0 and y = 0
Region represented by 10x + 4y ≥ 20:
The line 10x + 4y = 20 meets the coordinate axes at A(2, 0) and B(0, 5) respectively. By joining these points we obtain the line
10x + 4y = 20.Clearly (0,0) does not satisfies the inequation 10x + 4y ≥ 20. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 10x + 4y ≥ 20.
Region represented by 5x +6y ≥ 20:
The line 5x +6y = 20 meets the coordinate axes at
\[C\left( 4, 0 \right)\] and \[D\left( 0, \frac{10}{3} \right)\] respectively. By joining these points we obtain the line
5x +6y = 20.Clearly (0,0) does not satisfies the 5x +6y ≥ 20. So,the region which does not contains the origin represents the solution set of the inequation 5x +6y ≥ 20.
Region represented by 2x + 6y ≥ 12:
The line 2x + 6y =12 meets the coordinate axes at \[E\left( 6, 0 \right)\] and \[E\left( 6, 0 \right)\] respectively. By joining these points we obtain the line
2x + 6y =12.Clearly (0,0) does not satisfies the inequation 2x + 6y ≥ 12. So,the region which does not contains the origin represents the solution set of the inequation 2x + 6y≥ 12.
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 10x + 4y ≥ 20, 5x +6y ≥ 20, 2x + 6y ≥ 12, x ≥ 0, and y ≥ 0 are as follows.
The set of all feasible solutions of the above LPP is represented by the feasible region shaded in the graph.
The corner points of the feasible region are B(0, 5), G \[\left( 1, \frac{5}{2} \right)\],H \[\left( \frac{8}{3}, \frac{10}{9} \right)\] and E \[\left( 6, 0 \right)\]
The value of the objective function at these points are given by the following table
| Points | Value of Z |
| B |
\[0 . 6\left( 0 \right) + 5 = 5\]
|
| G |
\[0 . 6\left( 1 \right) + \frac{5}{2} = 3 . 1\]
|
| H |
\[0 . 6\left( \frac{8}{3} \right) + \left( \frac{10}{9} \right) = 1 . 6 + 1 . 1 = 2 . 7\]
|
| E |
\[0 . 6\left( 6 \right) + \left( 0 \right) = 3 . 6\]
|
We see that the minimum cost is 2.7 which is at \[\left( \frac{8}{3}, \frac{10}{9} \right)\].Thus, at minimum cost, \[\frac{8}{3}\] units of food I and \[\frac{10}{9}\] units of food II should be included in the diet.
APPEARS IN
संबंधित प्रश्न
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
Solve the following L.P.P. graphically:
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0 and x, y ≥ 0
Solve the following L.P.P. graphically Maximise Z = 4x + y
Subject to following constraints x + y ≤ 50
3x + y ≤ 90,
x ≥ 10
x, y ≥ 0
Maximize Z = 9x + 3y
Subject to
\[2x + 3y \leq 13\]
\[ 3x + y \leq 5\]
\[ x, y \geq 0\]
Minimize Z = 18x + 10y
Subject to
\[4x + y \geq 20\]
\[2x + 3y \geq 30\]
\[ x, y \geq 0\]
Maximize Z = 4x + 3y
subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Maximize Z = 10x + 6y
Subject to
\[3x + y \leq 12\]
\[2x + 5y \leq 34\]
\[ x, y \geq 0\]
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
\[x + y \leq 8\]
\[ x + 4y \geq 12\]
\[5x + 8y = 20\]
\[ x, y \geq 0\]
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
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 for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 of calories. Two foods A and B, are available at a cost of Rs 4 and Rs 3 per unit respectively. If one unit of A contains 200 units of vitamin, 1 unit of mineral and 40 calories and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost?
A fruit grower can use two types of fertilizer in his garden, brand P and Q. The amounts (in kg) of nirogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicates that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
| kg per bag | ||
| Brand P | Brand P | |
| Nitrogen | 3 | 3.5 |
| Phosphoric acid | 1 | 2 |
| Potash | 3 | 1.5 |
| Chlorine | 1.5 | 2 |
If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
A manufacturer of Furniture makes two products : chairs and tables. processing of these products is done on two machines A and B. A chair requires 2 hrs on machine A and 6 hrs on machine B. A table requires 4 hrs on machine A and 2 hrs on machine B. There are 16 hrs of time per day available on machine A and 30 hrs on machine B. Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively. Find with the help of graph what should be the daily production of each of the two products so as to maximize his 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 firm manufactures headache pills in two sizes A and B. Size A contains 2 grains of aspirin, 5 grains of bicarbonate and 1 grain of codeine; size B contains 1 grain of aspirin, 8 grains of bicarbonate and 66 grains of codeine. It has been found by users that it requires at least 12 grains of aspirin, 7.4 grains of bicarbonate and 24 grains of codeine for providing immediate effects. Determine graphically the least number of pills a patient should have to get immediate relief. Determine also the quantity of codeine consumed by patient.
A small firm manufacturers items A and B. The total number of items A and B that it can manufacture in a day is at the most 24. Item A takes one hour to make while item B takes only half an hour. The maximum time available per day is 16 hours. If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160, how many of each type of item be produced to maximize the profit? Solve the problem graphically.
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 manufacturing company makes two models A and B of a product. Each piece of model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of ₹8000 on each piece of model A and ₹12000 on each piece of model B. How many pieces of model A and model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week?
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
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?
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?
An aeroplane can carry a maximum of 200 passengers. A profit of ₹1000 is made on each executive class ticket and a profit of ₹600 is made on each economy class ticket. The airline reserves atleast 20 seats for executive class. However, atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit of the airline. What is the maximum profit?
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 ₹10 per kg and 'B' cost ₹8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requiremnets are met at a minimum cost
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?
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 manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 hours of work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturer's profit on an item of model A is ₹ 15 and on an item of model B is ₹ 10. How many items of each model should be made per day in order to maximize daily profit? Formulate the above LPP and solve it graphically and find the maximum profit.
The maximum value of z = 6x + 8y subject to x - y ≥ 0, x + 3y ≤ 12, x ≥ 0, y ≥ 0 is ______.
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.
A set of values of decision variables which satisfies the linear constraints and nn-negativity conditions of an L.P.P. is called its ____________.
Z = 20x1 + 20x2, subject to x1 ≥ 0, x2 ≥ 0, x1 + 2x2 ≥ 8, 3x1 + 2x2 ≥ 15, 5x1 + 2x2 ≥ 20. The minimum value of Z occurs at ____________.
In linear programming feasible region (or solution region) for the problem is ____________.
In the Corner point method for solving a linear programming problem the second step after finding the feasible region of the linear programming problem and determining its corner points is ____________.
The solution set of the inequality 3x + 5y < 4 is ______.
Solve the following Linear Programming Problem graphically:
Maximize: P = 70x + 40y
Subject to: 3x + 2y ≤ 9,
3x + y ≤ 9,
x ≥ 0,y ≥ 0.
Draw the rough graph and shade the feasible region for the inequalities x + y ≥ 2, 2x + y ≤ 8, x ≥ 0, y ≥ 0.
