Advertisements
Advertisements
Question
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.
Advertisements
Solution
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
RELATED QUESTIONS
Minimize: Z = 6x + 4y
Subject to the conditions:
3x + 2y ≥ 12,
x + y ≥ 5,
0 ≤ x ≤ 4,
0 ≤ y ≤ 4
A dealer in rural area wishes to purchase a number of sewing machines. He has only Rs 5,760 to invest and has space for at most 20 items for storage. An electronic sewing machine cost him Rs 360 and a manually operated sewing machine Rs 240. He can sell an electronic sewing machine at a profit of Rs 22 and a manually operated sewing machine at a profit of Rs 18. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize his profit? Make it as a LPP and solve it graphically.
A cooperative society of farmers has 50 hectares of land to grow two crops A and B. The profits from crops A and B per hectare are estimated as Rs 10,500 and Rs 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops A and B at the rate of 20 litres and 10 litres per hectare, respectively. Further not more than 800 litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. Keeping in mind that the protection of fish and other wildlife is more important than earning profit, how much land should be allocated to each crop so as to maximize the total profit? Form an LPP from the above and solve it graphically. Do you agree with the message that the protection of wildlife is utmost necessary to preserve the balance in environment?
A retired person wants to invest an amount of Rs. 50, 000. His broker recommends investing in two type of bonds ‘A’ and ‘B’ yielding 10% and 9% return respectively on the invested amount. He decides to invest at least Rs. 20,000 in bond ‘A’ and at least Rs. 10,000 in bond ‘B’. He also wants to invest at least as much in bond ‘A’ as in bond ‘B’. Solve this linear programming problem graphically to maximise his returns.
Minimum and maximum z = 5x + 2y subject to the following constraints:
x-2y ≤ 2
3x+2y ≤ 12
-3x+2y ≤ 3
x ≥ 0,y ≥ 0
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?
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
Solve the following L.P.P graphically: Maximise Z = 20x + 10y
Subject to the following constraints x + 2y ≤ 28,
3x + y ≤ 24,
x ≥ 2,
x, y ≥ 0
Maximize Z = 15x + 10y
Subject to
\[3x + 2y \leq 80\]
\[2x + 3y \leq 70\]
\[ x, y \geq 0\]
Minimize Z = 5x + 3y
Subject to
\[2x + y \geq 10\]
\[x + 3y \geq 15\]
\[ x \leq 10\]
\[ y \leq 8\]
\[ x, y \geq 0\]
Minimize Z = x − 5y + 20
Subject to
\[x - y \geq 0\]
\[ - x + 2y \geq 2\]
\[ x \geq 3\]
\[ y \leq 4\]
\[ x, y \geq 0\]
Minimize Z = 3x1 + 5x2
Subject to
\[x_1 + 3 x_2 \geq 3\]
\[ x_1 + x_2 \geq 2\]
\[ x_1 , x_2 \geq 0\]
Show the solution zone of the following inequalities on a graph paper:
\[5x + y \geq 10\]
\[ x + y \geq 6\]
\[x + 4y \geq 12\]
\[x \geq 0, y \geq 0\]
Find x and y for which 3x + 2y is minimum subject to these inequalities. Use a graphical method.
Two tailors, A and B earn Rs 15 and Rs 20 per day respectively. A can stitch 6 shirts and 4 pants while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to produce (at least) 60 shirts and 32 pants at a minimum labour cost?
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 chemical company produces two compounds, A and B. The following table gives the units of ingredients, C and D per kg of compounds A and B as well as minimum requirements of C and D and costs per kg of A and B. Find the quantities of A and B which would give a supply of C and D at a minimum cost.
| Compound | Minimum requirement | ||
| A | B | ||
| Ingredient C Ingredient D |
1 3 |
2 1 |
80 75 |
| Cost (in Rs) per kg | 4 | 6 | - |
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.
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.
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 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. If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the number of tennis rackets and cricket bats that the factory must manufacture to earn the maximum profit. Make it as an LPP and solve it graphically.
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at ₹7 profit and that of B at a profit of ₹4. Find the production level per day for maximum profit graphically.
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 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 ______.
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.
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.
For L.P.P. maximize z = 4x1 + 2x2 subject to 3x1 + 2x2 ≥ 9, x1 - x2 ≤ 3, x1 ≥ 0, x2 ≥ 0 has ______.
The maximum value of z = 3x + 10y subjected to the conditions 5x + 2y ≤ 10, 3x + 5y ≤ 15, x, y ≥ 0 is ______.
Maximise and Minimise Z = 3x – 4y subject to x – 2y ≤ 0, – 3x + y ≤ 4, x – y ≤ 6, x, y ≥ 0
Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities,
Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and ____________.
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 ____________.
Solve the following Linear Programming Problem graphically:
Minimize: Z = 60x + 80y
Subject to constraints:
3x + 4y ≥ 8
5x + 2y ≥ 11
x, 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.
