Advertisements
Advertisements
प्रश्न
In order to supplement daily diet, a person wishes to take some X and some wishes Y tablets. The contents of iron, calcium and vitamins in X and Y (in milligrams per tablet) are given as below:
| Tablets | Iron | Calcium | Vitamin |
| X | 6 | 3 | 2 |
| Y | 2 | 3 | 4 |
The person needs atleast 18 milligrams of iron, 21 milligrams of calcium and 16 milligrams of vitamin. The price of each tablet of X and Y is Rs 2 and Rs 1 respectively. How many tablets of each should the person take in order to satisfy the above requirement at the minimum cost?
Advertisements
उत्तर
Let there be x units of tablet X and y units of tablet Y
So, according to the given information, we have
6x + 2y ≥ 18
⇒ 3x + y ≥ 9 ......(i)
| x | 0 | 3 |
| y | 9 | 0 |
3x + 3y ≥ 21
⇒ x + y ≥ 7 ......(ii)
| x | 0 | 3 |
| y | 9 | 0 |
2x + 4y ≥ 16
⇒ x + 2y ≥ 8 ......(iii)
| x | 0 | 8 |
| y | 4 | 0 |
x ≥ 0, y ≥ 0 ......(iv)
The price of each table of X type is ` 2 and that of y is ` 1.
So, the required LPP is
Minimise Z = 2x + y subject to the constraints
3x + y ≥ 9, x + y ≥ 7, x + 2y ≥ 8, x ≥ 0, y ≥ 0
On solving (ii) and (iii) we get
x = 6 and y = 1
On solving (i) and (ii) we get
x = 1 and y = 6
From the graph, we see that the feasible region ABCD is unbounded whose corner points are A(8, 0), B(6, 1), C(1, 6) and D(0, 9).
Let us evaluate the value of Z
| Corner points | Value of Z = 2x + y | |
| A(8, 0) | Z = 2(8) + 0 = 16 | |
| B(6, 1) | Z = 2(6) + 1 = 13 | |
| C(1, 6) | Z = 2(1) + 6 = 8 | ← Minimum |
| D(0, 9) | Z = 2(0) + 9 = 9 |
Here, we see that 8 is the minimum value of Z at (1, 6) but the feasible region is unbounded.
So, 8 may or may not be the minimum value of Z.
To confirm it, we will draw a graph of inequality 2x + y < 8 and check if it has a common point.
We see from the graph that there is no common point on the line.
Hence, the minimum value of Z is 8 at (1, 6).
Tablet X = 1
Tablet Y = 6.
APPEARS IN
संबंधित प्रश्न
Two tailors, A and B, earn Rs 300 and Rs 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP
Solve the following Linear Programming Problems graphically:
Minimise Z = – 3x + 4 y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
Solve the following Linear Programming Problems graphically:
Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Solve the following Linear Programming Problems graphically:
Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Show that the minimum of Z occurs at more than two points.
Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
Show that the minimum of Z occurs at more than two points.
Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.
A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional elements A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin content of one kg food is given below:
| Food | Vitamin A | Vitamin B | Vitamin C |
| X | 1 | 2 | 3 |
| Y | 2 | 2 | 1 |
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
| Type of toy | Machines | ||
| I | II | III | |
| A | 12 | 18 | 6 |
| B | 6 | 0 | 9 |
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 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 maximize the profit for the airline. What is the maximum profit?
To maintain his health a person must fulfil certain minimum daily requirements for several kinds of nutrients. Assuming that there are only three kinds of nutrients-calcium, protein and calories and the person's diet consists of only two food items, I and II, whose price and nutrient contents are shown in the table below:
| Food I (per lb) |
Food II (per lb) |
Minimum daily requirement for the nutrient |
||||
| Calcium | 10 | 5 | 20 | |||
| Protein | 5 | 4 | 20 | |||
| Calories | 2 | 6 | 13 | |||
| Price (Rs) | 60 | 100 |
What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this as a LPP.
If the feasible region for a linear programming problem is bounded, then the objective function Z = ax + by has both a maximum and a minimum value on R.
Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.
Minimise Z = 13x – 15y subject to the constraints: x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0
In figure, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of Z = x + 2y.
The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y ______.
Compare the quantity in Column A and Column B
| Column A | Column B |
| Maximum of Z | 325 |
The feasible solution for a LPP is shown in Figure. Let Z = 3x – 4y be the objective function. Minimum of Z occurs at ______.
The feasible region for an LPP is shown in the figure. Let F = 3x – 4y be the objective function. Maximum value of F is ______.
Refer to Question 32, Maximum of F – Minimum of F = ______.
A corner point of a feasible region is a point in the region which is the ______ of two boundary lines.
If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = ax + by may or may not exist.
Maximum value of the objective function Z = ax + by in a LPP always occurs at only one corner point of the feasible region.
In a LPP, the maximum value of the objective function Z = ax + by is always finite.
A linear programming problem is as follows:
Minimize Z = 30x + 50y
Subject to the constraints: 3x + 5y ≥ 15, 2x + 3y ≤ 18, x ≥ 0, y ≥ 0
In the feasible region, the minimum value of Z occurs at:
Objective function of a linear programming problem is ____________.
The maximum value of the object function Z = 5x + 10 y subject to the constraints x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0 is ____________.
Z = 7x + y, subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at ____________.
A linear programming problem is one that is concerned with ____________.
In linear programming infeasible solutions
A maximum or a minimum may not exist for a linear programming problem if ____________.
In Corner point method for solving a linear programming problem, one finds the feasible region of the linear programming problem, determines its corner points, and evaluates the objective function Z = ax + by at each corner point. If M and m respectively be the largest and smallest values at corner points then ____________.
If two corner points of the feasible region are both optimal solutions of the same type, i.e., both produce the same maximum or minimum.
Maximize Z = 3x + 5y, subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Maximize Z = 6x + 4y, subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, x ≥ 0, y ≥ 0.
