Advertisements
Advertisements
प्रश्न
A medical company has factories at two places, A and B. From these places, supply is made to each of its three agencies situated at P, Q and R. The monthly requirements of the agencies are respectively 40, 40 and 50 packets of the medicines, while the production capacity of the factories, A and B, are 60 and 70 packets respectively. The transportation cost per packet from the factories to the agencies are given below:
| Transportation Cost per packet(in Rs.) | ||
| From-> | A | B |
| To | ||
| P | 5 | 4 |
| Q | 4 | 2 |
| R | 3 | 5 |
Advertisements
उत्तर
Let x and y packets be transported from factory A to the agencies P and Q respectively. Then, [60 − (x + y)] packets be transported to the agency R.
The requirement at agency P is 40 packets. Since, x packets are transported from factory A,
Therefore, the remaining (40 − x) packets are transported from factory B.
Similarly, (40 − y) packets are transported by B to Q and 50− [60 − (x + y)] i.e. (x + y − 10) packets will be transported from factory B to agency R respectively.
Number of packets cannot be negative.Therefore,
\[x \geq 0, y \geq 0 \text{ and } 60 - x - y \geq 0\]
\[ \Rightarrow x \geq 0, y \geq 0 \text{ and } x + y \leq 60\]
\[40 - x \geq 0, 40 - y \geq 0 \text{ and } x + y - 10 \geq 0\]
\[ \Rightarrow x \leq 40, y \leq 40 \text{ and } x + y \geq 10\]
Total transportation cost Z is given by,
\[Z = 5x + 4y + 3\left[ 60 - \left( x + y \right) \right] + 4\left( 40 - x \right) + 2\left( 40 - y \right) + 5\left( x + y - 10 \right)\]
\[ = 3x + 4y + 10\]
\[x + y \leq 60\]
\[x \leq 40\]
\[y \leq 40\]
\[x + y \geq 10\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows:x + y = 60, x = 40, y = 40, x + y = 10, x = 0 and y = 0
Region represented by x + y ≤ 60:
The line x + y = 60 meets the coordinate axes at A1(60, 0) and B1(0, 60) respectively. By joining these points we obtain the line x + y = 60. Clearly (0,0) satisfies the x + y = 60. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 60.
Region represented by x ≤ 40:
x = 40 is the line that passes C1(40, 0) and is parallel to the Y axis.The region to the left of the line x = 40 will satisfy the inequation x ≤ 40.
Region represented by y ≤ 40:
y = 40 is the line that passes D1(0, 40) and is parallel to the X axis . The region below the line y = 40 will satisfy the inequation y ≤ 40.
Region represented by x + y ≥ 10:
The line x + y = 10 meets the coordinate axes at E1(10, 0) and \[F_1 \left( 0, 10 \right)\] respectively. By joining these points we obtain the line x + y = 10. Clearly (0,0) does not satisfies the inequation x + y ≥ 10. So,the region which does not contain the origin represents the solution set of the inequation x + y ≥ 10.
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 ≤ 60, x ≤ 40, y ≤ 40, x + y ≥ 10, x ≥ 0 and y ≥ 0 are as follows.
The values of Z at these corner points are as follows
| Corner point | Z= 3x + 4y + 370 |
| D1 | 530 |
| H1 | 590 |
| G1 | 570 |
| C1 | 490 |
| E1 | 400 |
| F1 | 410 |
The minimum value of Z is 400 which is at E1(10, 0).
Thus, the minimum cost is Rs 400.
Hence,
From A: 10 packets, 0 packets and 50 packets to P, Q and R respectively
From B: 30 packets, 40 packets and 0 packets to P, Q and R respectively
APPEARS IN
संबंधित प्रश्न
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?
Minimize :Z=6x+4y
Subject to : 3x+2y ≥12
x+y ≥5
0 ≤x ≤4
0 ≤ y ≤ 4
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 Rs 10 per kg and 'B' cost Rs 8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost
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.
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
Maximise z = 8x + 9y subject to the constraints given below :
2x + 3y ≤ 6
3x − 2y ≤6
y ≤ 1
x, y ≥ 0
Maximize Z = 9x + 3y
Subject to
2x + 3y ≤ 13
3x + y ≤ 5
x, y ≥ 0
Minimize Z = 3x1 + 5x2
Subject to
\[x_1 + 3 x_2 \geq 3\]
\[ x_1 + x_2 \geq 2\]
\[ x_1 , x_2 \geq 0\]
Maximize Z = −x1 + 2x2
Subject to
\[- x_1 + 3 x_2 \leq 10\]
\[ x_1 + x_2 \leq 6\]
\[ x_1 - x_2 \leq 2\]
\[ x_1 , x_2 \geq 0\]
Maximize Z = 3x1 + 4x2, if possible,
Subject to the constraints
\[x_1 - x_2 \leq - 1\]
\[ - x_1 + x_2 \leq 0\]
\[ x_1 , x_2 \geq 0\]
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 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
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1and F2 are available. Food F1 costs Rs 4 per unit and F2 costs Rs 6 per unit one unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements
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.
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹60/kg and food Q costs ₹80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
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 man owns a field of area 1000 sq.m. He wants to plant fruit trees in it. He has a sum of Rs 1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 sq.m of ground per tree and costs Rs 20 per tree and type B requires 20 sq.m of ground per tree and costs Rs 25 per tree. When fully grown, type A produces an average of 20 kg of fruit which can be sold at a profit of Rs 2.00 per kg and type B produces an average of 40 kg of fruit which can be sold at a profit of Rs. 1.50 per kg. How many of each type should be planted to achieve maximum profit when the trees are fully grown? What is the maximum profit?
An oil company has two depots, A and B, with capacities of 7000 litres and 4000 litres respectively. The company is to supply oil to three petrol pumps, D, E, F whose requirements are 4500, 3000 and 3500 litres respectively. The distance (in km) between the depots and petrol pumps is given in the following table:
Figure
Assuming that the transportation cost per km is Rs 1.00 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?
A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300 and that on a chain is Rs 190, find the number of rings and chains that should be manufactured per day, so as to earn the maximum profit. Make it as an LPP and solve it graphically.
A cooperative society of farmers has 50 hectares of land to grow two crops X and Y. The profits from crops X and Y per hectare are estimated as ₹10,500 and ₹9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y 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. How much land should be allocated to each crop so as to maximise the total profit of the society?
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.
There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds the she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs ₹6/kg and F2 costs ₹5/kg, determine how much of each type of fertilizer should be used so that the nutrient requirements are met at minimum cost. What is the minimum cost?
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?
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.
A company manufactures two types of cardigans: type A and type B. It costs ₹ 360 to make a type A cardigan and ₹ 120 to make a type B cardigan. The company can make at most 300 cardigans and spend at most ₹ 72000 a day. The number of cardigans of type B cannot exceed the number of cardigans of type A by more than 200. The company makes a profit of ₹ 100 for each cardigan of type A and ₹ 50 for every cardigan of type B.
Formulate this problem as a linear programming problem to maximize the profit to the company. Solve it graphically and find the maximum profit.
Draw the graph of inequalities x ≤ 6, y −2 ≤ 0, x ≥ 0, y ≥ 0 and indicate the feasible region
The maximum value of Z = 5x + 4y, Subject to y ≤ 2x, x ≤ 2y, x + y ≤ 3, x ≥ 0, y ≥ 0 is ______.
The region XOY - plane which is represented by the inequalities -5 ≤ x ≤ 5, -5 ≤ y ≤ 5 is ______
Maximise and Minimise Z = 3x – 4y subject to x – 2y ≤ 0, – 3x + y ≤ 4, x – y ≤ 6, x, y ≥ 0
Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) 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 ____________.
A manufacturer wishes to produce two commodities A and B. The number of units of material, labour and equipment needed to produce one unit of each commodity is shown in the table given below. Also shown is the available number of units of each item, material, labour, and equipment.
| Items | Commodity A | Commodity B | Available no. of Units |
| Material | 1 | 2 | 8 |
| Labour | 3 | 2 | 12 |
| Equipment | 1 | 1 | 10 |
Find the maximum profit if each unit of commodity A earns a profit of ₹ 2 and each unit of B earns a profit of ₹ 3.
The maximum value of 2x + y subject to 3x + 5y ≤ 26 and 5x + 3y ≤ 30, x ≥ 0, 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.
Solve the following Linear Programming Problem graphically:
Minimize: Z = 60x + 80y
Subject to constraints:
3x + 4y ≥ 8
5x + 2y ≥ 11
x, y ≥ 0
