Advertisements
Advertisements
प्रश्न
A company produces two types of leather belts, say type A and B. Belt A is a superior quality and belt B is of a lower quality. Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Each belt of type A requires twice as much time as required by a belt of type B. If all belts were of type B, the company could produce 1000 belts per day. But the supply of leather is sufficient only for 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. For belt of type B, only 700 buckles are available per day.
How should the company manufacture the two types of belts in order to have a maximum overall profit?
Advertisements
उत्तर
Let the company produces x belts of type A and y belts of type B.
Number of belts cannot be negative.
Therefore,
x + y ≤ 800
It is given that the rate of production of belts of type B is 1000 per day.Hence, the time taken to produce y belts of type B is \[\frac{y}{1000}\] And, since each belt of type A requires twice as much time as a belt of type B, the rate of production of belts of type A is 500 per day and therefore, total time taken to produce x belts of type A is \[\frac{x}{500}\] Thus, we have \[\frac{x}{500} + \frac{y}{1000} \leq 1\]
\[ \Rightarrow 2x + y \leq 1000\]
Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. x ≤ 400
For belt of type B, only 700 buckles are available per day.
y ≤ 700
Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Therefore, profit gained on x belts of type A and y belts of type B is Rs 2x and
Rs 1.50y respectively.Hence, the total profit would be Rs (2x + 1.50y).
Let Z denote the total profit.
Thus, the mathematical formulation of the given linear programming problem is Max Z = \[2x + 1 . 5y\] subject to \[x + y \leq 800\]
\[2x + y \leq 1000\]
\[x \leq 400\]
\[y \leq 700\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows :
x + y = 800, 2x + y = 1000, x = 400, y = 700, x = 0 and y = 0
Region represented by x + y ≤ 800:
The line x + y = 800 meets the coordinate axes at A1(800, 0) and B1(0, 800) respectively. By joining these points we obtain the line x + y = 800. Clearly (0,0) satisfies the x + y = 800. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 800.
Region represented by 2x + y ≤ 1000:
The line 2x + y = 1000 meets the coordinate axes at C1(500, 0) and D1(0, 1000) respectively. By joining these points we obtain the line 2x + y = 1000. Clearly (0,0) satisfies the inequation 2x + y ≤ 1000. So,the region which contains the origin represents the solution set of the inequation 2x + y ≤ 1000.
Region represented by x ≤ 400:
The line x = 400 will pass through E1(400, 0). The region to the left of the line x = 400 will satisfy the inequation x ≤ 400.
Region represented by y ≤ 700:
The line y = 700 will pass through F1(0, 700). The region below the line y = 700 will satisfy the inequation y ≤ 700.
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 x + y ≤ 800, 2x + y ≤ 1000, x≤ 400, y ≤ 700, x ≥ 0, and y ≥ 0 are as follows.
The feasible region determined by the system of constraints is
The corner points are F1(0, 700), G1(200, 600), H1(400, 200) and E1(400, 0). The values of Z at these corner points are as follows
| Corner point | Z= 2x +1.5y |
| F1(0, 700) | 1050 |
| G1(200, 600) | 1300 |
| H1(400, 200) | 1100 |
| E1(400, 0) | 800 |
The maximum value of Z is 1300 which is attained at G1(200, 600).
Thus, the maximum profit is Rs 1300 obtained when 200 belts of type A and 600 belts of type B were produced.
APPEARS IN
संबंधित प्रश्न
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 Rs 7 profit and B at a profit of Rs 4. Find the production level per day for maximum profit graphically.
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
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 LPP by graphical method:
Maximize: z = 3x + 5y
Subject to: x + 4y ≤ 24
3x + y ≤ 21
x + y ≤ 9
x ≥ 0, y ≥ 0
Also find the maximum value of z.
Solve the following LPP graphically :
Maximise Z = 105x + 90y
subject to the constraints
x + y ≤ 50
2x + y ≤ 80
x ≥ 0, y ≥ 0.
Maximize Z = 5x + 3y
Subject to
\[3x + 5y \leq 15\]
\[5x + 2y \leq 10\]
\[ x, y \geq 0\]
Maximize Z = 50x + 30y
Subject to
\[2x + y \leq 18\]
\[3x + 2y \leq 34\]
\[ x, y \geq 0\]
Maximize Z = 4x + 3y
subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Minimize Z = x − 5y + 20
Subject to
\[x - y \geq 0\]
\[ - x + 2y \geq 2\]
\[ x \geq 3\]
\[ y \leq 4\]
\[ x, y \geq 0\]
Maximize Z = 2x + 3y
Subject to
\[x + y \geq 1\]
\[10x + y \geq 5\]
\[x + 10y \geq 1\]
\[ x, y \geq 0\]
Solved the following linear programming problem graphically:
Maximize Z = 60x + 15y
Subject to constraints
\[x + y \leq 50\]
\[3x + y \leq 90\]
\[ x, y \geq 0\]
Solve the following LPP graphically:
Maximize Z = 20 x + 10 y
Subject to the following constraints
\[x +\]2\[y \leq\]28
3x+ \[y \leq\]24
\[x \geq\] 2x.
\[y \geq\] 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.
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.
A dietician mixes together two kinds of food in such a way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1 kg of food X and 1 kg of food Y are given below:
| Vitamin A |
Vitamin B |
Vitamin |
Vitamin D |
|
| Food X Food Y |
1 2 |
1 1 |
1 3 |
2 1 |
One kg food X costs Rs 5, whereas one kg of food Y costs Rs 8. Find the least cost of the mixture which will produce the desired diet.
A wholesale dealer deals in two kinds, A and B (say) of mixture of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew nuts and 30 grams of hazel nuts. Each kg of mixture B contains 30 grams of almonds, 60 grams of cashew nuts and 180 grams of hazel nuts. The remainder of both mixtures is per nuts. The dealer is contemplating to use mixtures A and B to make a bag which will contain at least 240 grams of almonds, 300 grams of cashew nuts and 540 grams of hazel nuts. Mixture A costs Rs 8 per kg. and mixture B costs Rs 12 per kg. Assuming that mixtures A and B are uniform, use graphical method to determine the number of kg. of each mixture which he should use to minimise the cost of the bag.
A farmer mixes two brands P and Q of cattle feed. Brand P, costing ₹250 per bag, contains 2 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ₹200 per bag contains 1.5 units of nutritional element 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 firm manufacturing two types of electric items, A and B, can make a profit of Rs 20 per unit of A and Rs 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers. Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month. Formulate the linear programing problem for maximum profit and solve it graphically.
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 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 sells two different products, A and B. The two products are produced in a common production process, which has a total capacity of 500 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of unit of A that can be sold is 70 and that for B is 125. If the profit is Rs 20 per unit for the product A and Rs 15 per unit for the product B, how many units of each product should be sold to maximize profit?
A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 sq. metre of cardboard is in stock, and if the profits on the large and small boxes are Rs 3 and Rs 2 per box, how many of each should be made in order to maximize the total 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.
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?
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 |
For L.P.P. maximize z = 4x1 + 2x2 subject to 3x1 + 2x2 ≥ 9, x1 - x2 ≤ 3, x1 ≥ 0, x2 ≥ 0 has ______.
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 ______.
For the LPP, maximize z = x + 4y subject to the constraints x + 2y ≤ 2, x + 2y ≥ 8, x, y ≥ 0 ______.
The minimum value of z = 7x + 9y subject to 3x + y ≤ 6, 5x + 8y ≤ 40, x ≥ 0, y ≥ 2 is ______.
In linear programming feasible region (or solution region) for the problem is ____________.
Which of the statements describe the solution set for `-2(x + 8) = - 2x + 20`?
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.
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.
If two corner points give the same optimum value, which points are also optimal?
