Advertisements
Advertisements
प्रश्न
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.
उत्तर
Let x number of tennis rackets and y number of cricket bats were sold.
Number of tennis rackets and cricket balls cannot be negative.
Therefore , \[x \geq 0, y \geq 0\]
It is given that 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.
Also, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time. Therefore,
\[1 \cdot 5x + 3y \leq 42\]
\[3x + y \leq 24\]
If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively.Therefore, profit made on x tennis rackets and y cricket bats is Rs 20x and Rs 10y respectively.
Total profit = Z = 20x + 10y
The mathematical form of the given LPP is:
Maximize Z = 20x + 10y
Subject to constraints:
\[1 \cdot 5x + 3y \leq 42\]
\[3x + y \leq 24\]
First we will convert inequations into equations as follows:
1.5x + 3y = 42, 3x + y = 24, x = 0 and y = 0
Region represented by 1.5x + 3y ≤ 42:
The line 1.5x + 3y = 42 meets the coordinate axes at A(28, 0) and B(0, 14) respectively. By joining these points we obtain the line 1.5x + 3y = 42. Clearly, (0, 0) satisfies the 1.5x + 3y = 42. So, the region which contains the origin represents the solution set of the inequation 1.5x + 3y ≤ 42.
Region represented by 3x + y ≤ 24:
The line 3x + y = 24 meets the coordinate axes at C(8, 0) and D(0, 24) respectively. By joining these points we obtain the line 3x + y = 24. Clearly, (0, 0) satisfies the inequation 3x + y ≤ 24. So, the region which contains the origin represents the solution set of the inequation 3x + y ≤ 24.
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 1.5x + 3y ≤ 42, 3x + y ≤ 24, x ≥ 0 and y ≥ 0 are as follows.
In the above graph, the shaded region is the feasible region.
The corner points are O(0, 0), B(0, 14), E(4, 12), and C(8, 0).
The values of the objective function Z at corner points of the feasible region are given in the following table:
|
Corner Points |
Z = 20x +10y |
|
| O(0, 0) | 0 | |
|
B(0, 14) |
140 |
|
|
E(4, 12) |
200 |
← Maximum |
|
C(8, 0) |
160 |
Clearly, Z is maximum at x = 4 and y = 12 and the maximum value of Z at this point is 200.
(i) 4 tennis rackets and 12 cricket bats must be made if the factory is to work at full capacity.
(ii) The maximum profit of the factory when it works at full capacity is Rs 200.
APPEARS IN
संबंधित प्रश्न
Solve the following LPP by using graphical method.
Maximize : Z = 6x + 4y
Subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, x ≥ 0, y ≥ 0.
Also find maximum value of Z.
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?
Solve the following L.P.P graphically:
Maximize: Z = 10x + 25y
Subject to: x ≤ 3, y ≤ 3, x + y ≤ 5, x ≥ 0, y ≥ 0
Minimize :Z=6x+4y
Subject to : 3x+2y ≥12
x+y ≥5
0 ≤x ≤4
0 ≤ y ≤ 4
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.
Maximize Z = 4x + 3y
subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Maximize Z = 3x + 4y
Subject to
\[2x + 2y \leq 80\]
\[2x + 4y \leq 120\]
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
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?
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 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?
A furniture manufacturing company plans to make two products : chairs and tables. From its available resources which consists of 400 square feet to teak wood and 450 man hours. It is known that to make a chair requires 5 square feet of wood and 10 man-hours and yields a profit of Rs 45, while each table uses 20 square feet of wood and 25 man-hours and yields a profit of Rs 80. How many items of each product should be produced by the company so that the profit is maximum?
A company manufactures two types of novelty Souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours available for assembling. The profit is 50 paise each for type A and 60 paise each for type B souvenirs. How many souvenirs of each type should the company manufacture in order 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 of product A and 16 of product B. A working week consists of 40 hours of production and weekly turnover must not be less than Rs 10000. If the profit on each of product A is Rs 20 and on product B is Rs 30, then how many of each should be produced so that the profit is maximum. Also, find the maximum profit.
An aeroplane can carry a maximum of 200 passengers. A profit of Rs 400 is made on each first class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats of first class. However, at least 4 times as many passengers prefer to travel by economy class to the first class. Determine how many each type of tickets must be sold in order to maximize the profit for the airline. What is the maximum profit.
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 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 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 firm makes items A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and half an hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs 300 and on one item of B is Rs 160. How many items of each type should be produced to maximize the profit? Solve the problem graphically.
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 manufacturer has three machine I, II, III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines.
The number of hours required for producing 1 unit each of M and N on the three machines are given in the following table:
| Items | Number of hours required on machines | ||
| I | II | III | |
| M | 1 | 2 | 1 |
| N | 2 | 1 | 1.25 |
She makes a profit of ₹600 and ₹400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?
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.
The graph of the inequality 3X − 4Y ≤ 12, X ≤ 1, X ≥ 0, Y ≥ 0 lies in fully in
Find the solution set of inequalities 0 ≤ x ≤ 5, 0 ≤ 2y ≤ 7
Draw the graph of inequalities x ≤ 6, y −2 ≤ 0, x ≥ 0, y ≥ 0 and indicate the feasible region
For L.P.P. maximize z = 4x1 + 2x2 subject to 3x1 + 2x2 ≥ 9, x1 - x2 ≤ 3, x1 ≥ 0, x2 ≥ 0 has ______.
Area of the region bounded by y = cos x, x = 0, x = π and X-axis is ______ sq.units.
The minimum value of z = 2x + 9y subject to constraints x + y ≥ 1, 2x + 3y ≤ 6, x ≥ 0, y ≥ 0 is ______.
The maximum value of z = 3x + 10y subjected to the conditions 5x + 2y ≤ 10, 3x + 5y ≤ 15, x, y ≥ 0 is ______.
In linear programming feasible region (or solution region) for the problem is ____________.
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 maximum value of Z = 3x + 4y subjected to contraints x + y ≤ 40, x + 2y ≤ 60, x ≥ 0 and y ≥ 0 is ____________.
Any point in the feasible region that gives the optional value (maximum or minimum) of the objective function is called:-
The corner points of the shaded unbounded feasible region of an LPP are (0, 4), (0.6, 1.6) and (3, 0) as shown in the figure. The minimum value of the objective function Z = 4x + 6y occurs at ______.
The constraints –x1 + x2 ≤ 1, –x1 + 3x2 ≤ 9, x1x2 ≥ 0 define on ______.
Solve the following Linear Programming Problem graphically:
Maximize: P = 70x + 40y
Subject to: 3x + 2y ≤ 9,
3x + y ≤ 9,
x ≥ 0,y ≥ 0.
Solve the following Linear Programming Problem graphically.
Maximise Z = 5x + 2y subject to:
x – 2y ≤ 2,
3x + 2y ≤ 12,
– 3x + 2y ≤ 3,
x ≥ 0, y ≥ 0
