Advertisements
Advertisements
प्रश्न
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.
उत्तर
Let the firm manufacture x items of A and y items of B per day.
Number of items cannot be negative.
Therefore, x ≥0 and y ≥0.
It is given that the total number of items manufactured per day is at most 24.
∴ x + y ≤ 24
Item A takes 1 hour to make and item B takes 0.5 hour to make.
The maximum number of hours available per day is 16 hours.
∴ x + 0.5y ≤ 16
If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160.Therefore, profit gained on x items of A and y items of B is
Rs 300x and Rs 160y respectively.
∴ Total profit, Z = 300x + 160y
Thus, the mathematical formulation of the given problem is:
Maximise Z = 300x + 160y
subject to the constraints
x+ y ≤ 24
x + 0.5y ≤ 16
x, y ≥0
First we will convert inequations into equations as follows:
x + y = 24, x + 0.5y = 16, x = 0 and y = 0
Region represented by x + y ≤ 24:
The line x + y = 24 meets the coordinate axes at A1(24, 0) and B1(0, 24) respectively. By joining these points we obtain the line x + y = 24. Clearly (0,0) satisfies the x + y = 24. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 24.
Region represented by x + 0.5y ≤ 16:
The line x + 0.5y = 16 meets the coordinate axes at C1(16, 0) and D1(0, 32) respectively. By joining these points we obtain the line x + 0.5y = 16. Clearly (0,0) satisfies the inequation x + 0.5y ≤ 16. So,the region which contains the origin represents the solution set of the inequation x + 0.5y ≤ 16.
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 ≤ 24, x + 0.5y ≤ 16, x ≥ 0and y ≥ 0 are as follows.
The corner points are O(0, 0) ,C1(16, 0), E1(8, 16) and B1(0, 24).
The value of Z at these corner points are as follows:
| Corner point | Z = 300x + 160y |
| O(0, 0) | 0 |
| C1(16, 0) | 4800 |
| E1(8, 16) | 4960 |
| B1(0, 24) | 3840 |
Thus, the maximum value of Z is 4960 at E1(8, 16).
Thus, 8 units of item A and 16 units of item B should be manufactured per day to maximise the profits.
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 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.
A company manufactures bicycles and tricycles each of which must be processed through machines A and B. Machine A has maximum of 120 hours available and machine B has maximum of 180 hours available. Manufacturing a bicycle requires 6 hours on machine A and 3 hours on machine B. Manufacturing a tricycle requires 4 hours on machine A and 10 hours on machine B.
If profits are Rs. 180 for a bicycle and Rs. 220 for a tricycle, formulate and solve the L.P.P. to determine the number of bicycles and tricycles that should be manufactured in order to maximize the profit.
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
Solve the following LPP by graphical method:
Minimize Z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
A dietician wishes to mix 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 arid 8 units of vitamin C. The vitamin contents of one kg food is given below:
| Food | Vitamin A | Vitamin.B | Vitamin C |
| X | 1 unit | 2 unit | 3 unit |
| Y | 2 unit | 2 unit | 1 unit |
Orie kg of food X costs Rs 24 and one kg of food Y costs Rs 36. Using Linear Programming, find the least cost of the total mixture. which will contain the required vitamins.
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
\[- x_1 + 3 x_2 \leq 10\]
\[ x_1 + x_2 \leq 6\]
\[ x_1 - x_2 \leq 2\]
\[ x_1 , x_2 \geq 0\]
Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below:
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, 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
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 contents 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 ₹16 and one kg of food Y costs ₹20. Find the least cost of the mixture which will produce the required diet?
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.
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 gardener has supply of fertilizer of type I which consists of 10% nitrogen and 6% phosphoric acid and type II fertilizer which consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type I fertilizer costs 60 paise per kg and type II fertilizer costs 40 paise per kg, determine how many kilograms of each fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
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?
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 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. If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the number of tennis rackets and cricket bats that the factory must manufacture to earn the maximum profit. Make it as an LPP and solve it graphically.
A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and his profit on the desktop model is Rs 4500 and on the portable model is Rs 5000. Make an LPP and solve it graphically.
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 |
Maximize: z = 3x + 5y Subject to
x +4y ≤ 24 3x + y ≤ 21
x + y ≤ 9 x ≥ 0 , y ≥0
A carpenter has 90, 80 and 50 running feet respectively of teak wood, plywood and rosewood which is used to product A and product B. Each unit of product A requires 2, 1 and 1 running feet and each unit of product B requires 1, 2 and 1 running feet of teak wood, plywood and rosewood respectively. If product A is sold for Rs. 48 per unit and product B is sold for Rs. 40 per unit, how many units of product A and product B should be produced and sold by the carpenter, in order to obtain the maximum gross income? Formulate the above as a Linear Programming Problem and solve it, indicating clearly the feasible region in the graph.
A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 6% phosphoric acid and of type B which contains 5% nitrogen and 10% phosphoric acid. After the soil test, it is found that at least 7 kg of nitrogen and the same quantity of phosphoric acid is required for a good crop. The fertilizer of type A costs ₹ 5.00 per kg and the type B costs ₹ 8.00 per kg. Using Linear programming, find how many kilograms of each type of fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph.
Sketch the graph of inequation x ≥ 5y in xoy co-ordinate system
Find the solution set of inequalities 0 ≤ x ≤ 5, 0 ≤ 2y ≤ 7
Maximum value of 4x + 13y subject to constraints x ≥ 0, y ≥ 0, x + y ≤ 5 and 3x + y ≤ 9 is ______.
For L.P.P. maximize z = 4x1 + 2x2 subject to 3x1 + 2x2 ≥ 9, x1 - x2 ≤ 3, x1 ≥ 0, x2 ≥ 0 has ______.
The minimum value of z = 2x + 9y subject to constraints x + y ≥ 1, 2x + 3y ≤ 6, x ≥ 0, y ≥ 0 is ______.
Maximise and Minimise Z = 3x – 4y subject to x – 2y ≤ 0, – 3x + y ≤ 4, x – y ≤ 6, x, y ≥ 0
A feasible region in the set of points which satisfy ____________.
Of all the points of the feasible region for maximum or minimum of objective function the points.
A set of values of decision variables which satisfies the linear constraints and nn-negativity conditions of an L.P.P. is called its ____________.
Z = 20x1 + 20x2, subject to x1 ≥ 0, x2 ≥ 0, x1 + 2x2 ≥ 8, 3x1 + 2x2 ≥ 15, 5x1 + 2x2 ≥ 20. The minimum value of Z occurs at ____________.
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 ____________.
In Corner point method for solving a linear programming problem the first step is to ____________.
The feasible region (shaded) for a L.P.P is shown in the figure. The maximum Z = 5x + 7y is ____________.
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:
Minimize: z = x + 2y,
Subject to the constraints: x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200, x, y ≥ 0.
A linear programming problem is given by Z = px + qy where p, q > 0 subject to the constraints: x + y ≤ 60, 5x + y ≤ 100, x ≥ 0 and y ≥ 0
- Solve graphically to find the corner points of the feasible region.
- If Z = px + qy is maximum at (0, 60) and (10, 50), find the relation of p and q. Also mention the number of optimal solution(s) in this case.
