Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
Let x pieces of type A and y pieces of type B be manufactured per week.
Let Z denote the total profit. Then,
Z = 80x + 120y
Since each piece of A requires 9 labour hours of fabricating and each piece of B requires 12 labour hours of fabricating, the total labour hours required for fabrication are 9x + 12y. This must be less than or equal to the total hours available for fabrication. Hence,
9x + 12y ≤180
i.e. 3x + 4y ≤60 ... (i)
Also, each piece of A requires 1 labour hour for finishing and each piece of B requires 3 labour hours for finishing. Hence, the total labour hours required for finishing are 1x + 3y. But the maximum number of hours available for finishing is 30.
∴ 1x + 3y ≤30 ... (ii)
Since, the number of pieces cannot be negative, therefore,
x≥0, y≥0
Hence, the linear programming problem for the given problem is as follows:
Maximise Z = 80x + 120y, subject to the constraints
3x + 4y ≤60
x + 3y ≤30
and x≥0, y≥0
To solve this LPP graphically, we have to first convert the inequalities into equations and draw the corresponding lines.
The feasible region of the LPP is shaded in the figure. The coordinates of the corner points of the feasible region are A (0, 10), B (12, 6), C (20, 0).
The values of the objective function at these points are given in the following table:
| Point (x, y) | Value of objective function Z = 80x + 120y |
| A (0, 10) | Z = 80 × 0 + 120 × 10 = 1200 |
| B (12, 6) | Z = 80 × 12 + 120 × 6 = 1680 |
| C (20, 0) | Z = 80 × 20 + 120 × 0 = 1600 |
Clearly, Z is maximum at (12, 6). The maximum value of Z is 1680.
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.
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 linear programming problem graphically :
Maximise Z = 7x + 10y subject to the constraints
4x + 6y ≤ 240
6x + 3y ≤ 240
x ≥ 10
x ≥ 0, y ≥ 0
Maximize Z = 9x + 3y
Subject to
2x + 3y ≤ 13
3x + y ≤ 5
x, y ≥ 0
Minimize Z = 2x + 4y
Subject to
\[x + y \geq 8\]
\[x + 4y \geq 12\]
\[x \geq 3, y \geq 2\]
Minimize Z = 5x + 3y
Subject to
\[2x + y \geq 10\]
\[x + 3y \geq 15\]
\[ x \leq 10\]
\[ y \leq 8\]
\[ x, y \geq 0\]
Minimize Z = 30x + 20y
Subject to
\[x + y \leq 8\]
\[ x + 4y \geq 12\]
\[5x + 8y = 20\]
\[ x, y \geq 0\]
Maximize Z = 4x + 3y
Subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 of calories. Two foods A and B, are available at a cost of Rs 4 and Rs 3 per unit respectively. If one unit of A contains 200 units of vitamin, 1 unit of mineral and 40 calories and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost?
A factory manufactures two types of screws, A and B, each type requiring the use of two machines - an automatic and a hand-operated. It takes 4 minute on the automatic and 6 minutes on the hand-operated machines to manufacture a package of screws 'A', while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws 'A' at a profit of 70 P and screws 'B' at a profit of Rs 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum 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 manufacturer produces two types of steel trunks. He has two machines A and B. For completing, the first types of the trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of the trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit?
A manufacturer of patent medicines is preparing a production plan on medicines, A and B. There are sufficient raw materials available to make 20000 bottles of A and 40000 bottles of B, but there are only 45000 bottles into which either of the medicines can be put. Further, it takes 3 hours to prepare enough material to fill 1000 bottles of A, it takes 1 hour to prepare enough material to fill 1000 bottles of B and there are 66 hours available for this operation. The profit is Rs 8 per bottle for A and Rs 7 per bottle for B. How should the manufacturer schedule his production in order to maximize his profit?
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp while it takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at most 20 hours and the grinding/cutting machine for at most 12 hours. The profit from the sale of a lamp is ₹5.00 and a shade is ₹3.00. Assuming that the manufacturer sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his 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.
(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.
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 |
The point at which the maximum value of x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is obtained, is ______.
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
Sketch the graph of inequation x ≥ 5y in xoy co-ordinate system
The maximum value of z = 3x + 10y subjected to the conditions 5x + 2y ≤ 10, 3x + 5y ≤ 15, x, y ≥ 0 is ______.
A feasible region in the set of points which satisfy ____________.
Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities,
In the Corner point method for solving a linear programming problem the second step after finding the feasible region of the linear programming problem and determining its corner points is ____________.
A feasible solution to a linear programming problem
Any point in the feasible region that gives the optional value (maximum or minimum) of the objective function is called:-
The shaded part of given figure indicates in feasible region, then the constraints are:
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?
Solve the following Linear Programming Problem graphically:
Maximize: P = 70x + 40y
Subject to: 3x + 2y ≤ 9,
3x + y ≤ 9,
x ≥ 0,y ≥ 0.
