Advertisements
Advertisements
Question
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.
Advertisements
Solution
Let young man drives x km at a speed of 25 km/hr and y km at a speed of \[40 km/hr\] Clearly, \[x, y \geq 0\]
It is given that, he spends Rs 2 per km if he drives at a speed of \[25 km/hr\] and Rs 5 per km if he drives at a speed of \[40 km/hr\] . Therefore, money spent by him when he travelled x km and y km is Rs 2x and Rs 5y respectively.
It is given that he has a maximum of Rs 100 to spend.
Thus, \[2x + 5y \leq 100\]
\[\text{ Time spent by him when travelling with a speed of 25 km/hr } = \frac{x}{25}hr\]
\[\text{ Time spent by him when travelling with a speed of 40 km/hr} = \frac{x}{40}hr\]
Also, the available time is of 1 hour.
\[\frac{x}{25} + \frac{y}{40} \leq 1\]
\[ \Rightarrow 40x + 25y \leq 1000\]
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = \[x + y\]
\[2x + 5y \leq 100\]
\[40x + 25y \leq 1000\]
2x + 5y = 100, 40x + 25y = 1000, x = 0 and y = 0
Region represented by 2x + 5y ≤ 100:
The line 2x + 5y = 100 meets the coordinate axes at \[A\left( 50, 0 \right)\] and \[B\left( 0, 20 \right)\] respectively. By joining these points we obtain the line 2x + 5y = 100. Clearly (0,0) satisfies the 2x + 5y = 100. So,the region which contains the origin represents the solution set of the inequation 2x + 5y ≤ 100.
Region represented by 40x + 25y ≤ 1000:
The line 40x + 25y = 1000 meets the coordinate axes at
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 + 5y ≤ 100, 40x + 25y ≤ 1000, x ≥ 0, and y ≥ 0 are as follows
The corner points are O(0, 0), B(0, 20),
The values of Z at these corner points are as follows
| Corner point | Z = x + y |
| O | 0 |
| B | 20 |
| E | 30 |
| C | 25 |
The maximum value of Z is 30 which is attained at E.
Thus, the maximum distance travelled by the young man is 30 kms, if he drives
APPEARS IN
RELATED QUESTIONS
Maximize Z = 10x + 6y
Subject to
\[3x + y \leq 12\]
\[2x + 5y \leq 34\]
\[ x, y \geq 0\]
Minimize Z = 2x + 4y
Subject to
\[x + y \geq 8\]
\[x + 4y \geq 12\]
\[x \geq 3, y \geq 2\]
Maximize Z = 3x + 5y
Subject to
\[x + 2y \leq 20\]
\[x + y \leq 15\]
\[ y \leq 5\]
\[ x, y \geq 0\]
Maximize Z = −x1 + 2x2
Subject to
\[- x_1 + 3 x_2 \leq 10\]
\[ x_1 + x_2 \leq 6\]
\[ x_1 - x_2 \leq 2\]
\[ x_1 , x_2 \geq 0\]
Find the minimum value of 3x + 5y subject to the constraints
− 2x + y ≤ 4, x + y ≥ 3, x − 2y ≤ 2, x, 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
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.
One kind of cake requires 300 gm of flour and 15 gm of fat, another kind of cake requires 150 gm of flour and 30 gm of fat. Find the maximum number of cakes which can be made from 7.5 kg of flour and 600 gm of fat, assuming that there is no shortage of the other ingradients used in making the cake. Make it as an LPP and solve it graphically.
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 firm manufactures two products A and B. Each product is processed on two machines M1 and M2. Product A requires 4 minutes of processing time on M1 and 8 min. on M2 ; product B requires 4 minutes on M1 and 4 min. on M2. The machine M1 is available for not more than 8 hrs 20 min. while machine M2 is available for 10 hrs. during any working day. The products A and B are sold at a profit of Rs 3 and Rs 4 respectively.
Formulate the problem as a linear programming problem and find how many products of each type should be produced by the firm each day in order to get maximum profit.
A publisher sells a hard cover edition of a text book for Rs 72.00 and paperback edition of the same ext for Rs 40.00. Costs to the publisher are Rs 56.00 and Rs 28.00 per book respectively in addition to weekly costs of Rs 9600.00. Both types require 5 minutes of printing time, although hardcover requires 10 minutes binding time and the paperback requires only 2 minutes. Both the printing and binding operations have 4,800 minutes available each week. How many of each type of book should be produced in order to maximize profit?
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 types of toys A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs 50 each on type A and Rs 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?
A manufacturer makes two products, A and B. Product A sells at Rs 200 each and takes 1/2 hour to make. Product B sells at Rs 300 each and takes 1 hour to make. There is a permanent order for 14 units of product A and 16 units of product B. A working week consists of 40 hours of production and the weekly turn over must not be less than Rs 10000. If the profit on each of product A is Rs 20 and an product B is Rs 30, then how many of each should be produced so that the profit is maximum? Also find the maximum profit.
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:
| Types of Toys | Machines | ||
| I | II | III | |
| A | 12 | 18 | 6 |
| B | 6 | 0 | 9 |
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
By graphical method, the solution of linear programming problem
\[\text{ Subject } to \text{ 3 } x_1 + 2 x_2 \leq 18\]
\[ x_1 \leq 4\]
\[ x_2 \leq 6\]
\[ x_1 \geq 0, x_2 \geq 0, \text{ is } \]
From the details given below, calculate the five-year moving averages of the number of students who have studied in a school. Also, plot these and original data on the same graph paper.
| Year | 1993 | 1994 | 1995 | 1996 | 1997 | 1998 | 1999 | 2000 | 2001 | 2002 |
| Number of Students | 332 | 317 | 357 | 392 | 402 | 405 | 410 | 427 | 405 | 438 |
The minimum value of z = 10x + 25y subject to 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≥ 5 is ______.
The region XOY - plane which is represented by the inequalities -5 ≤ x ≤ 5, -5 ≤ y ≤ 5 is ______
The maximum value of z = 3x + 10y subjected to the conditions 5x + 2y ≤ 10, 3x + 5y ≤ 15, x, y ≥ 0 is ______.
If 4x + 5y ≤ 20, x + y ≥ 3, x ≥ 0, y ≥ 0, maximum 2x + 3y is ______.
A feasible region in the set of points which satisfy ____________.
A set of values of decision variables which satisfies the linear constraints and nn-negativity conditions of an L.P.P. is called its ____________.
In linear programming feasible region (or solution region) for the problem 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 ____________.
The feasible region (shaded) for a L.P.P is shown in the figure. The maximum Z = 5x + 7y is ____________.

The solution set of the inequality 3x + 5y < 4 is ______.
The maximum value of z = 5x + 2y, subject to the constraints x + y ≤ 7, x + 2y ≤ 10, x, y ≥ 0 is ______.
The objective function Z = x1 + x2, subject to the constraints are x1 + x2 ≤ 10, – 2x1 + 3x2 ≤ 15, x1 ≤ 6, x1, x2 ≥ 0, has maximum value ______ of the feasible region.
The objective function Z = ax + by of an LPP has maximum vaiue 42 at (4, 6) and minimum value 19 at (3, 2). Which of the following is true?
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:
Minimize: Z = 5x + 10y
Subject to constraints:
x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, y ≥ 0.
Solve the following linear programming problem graphically:
Maximize: Z = x + 2y
Subject to constraints:
x + 2y ≥ 100,
2x – y ≤ 0
2x + y ≤ 200,
x ≥ 0, y ≥ 0.
Minimize z = x + 2y,
Subject to x + 2y ≥ 50, 2x – y ≤ 0, 2x + y ≤ 100, 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 ______.
