Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Let x kg of kind A and y kg of kind B were used.
Quantity cannot be negative.
Therefore,
The given information can be tabulated as follows:
| Nut | Almonds(grams) | Cashewnuts(grams) | Hazel nuts(grams) |
| A(x) | 60 | 30 | 30 |
| B(y) | 30 | 60 | 180 |
| Availability | 240 | 300 | 540 |
Therefore, the constraints are
\[60x + 30y \geq 240\]
\[30x + 60y \geq 300\]
\[30x + 180y \geq 540\]
Mixture A costs Rs 8 per kg. and mixture B costs Rs 12 per kg.
Total cost = Z = \[8x + 12y\]
which is to be minimised.
Thus, the mathematical formulation of the given linear programmimg problem is
Min Z = \[8x + 12y\] subject to
\[2x + y \geq 8\]
\[ x + 2y \geq 10\]
\[ x + 6y \geq 18\]
First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 8, x +2y = 10, x +6y = 18, x = 0 and y = 0
Region represented by 2x + y ≥ 8:
The line 2x + y = 8 meets the coordinate axes at A1(4, 0) and B1(0, 8) respectively. By joining these points we obtain the line 2x + y = 8.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 8. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 8.
Region represented by x +2y ≥ 10:
The line x +2y = 10 meets the coordinate axes at C1(10,0) and D1(0, 5) respectively. By joining these points we obtain the line
x +2y = 10. Clearly (0,0) does not satisfies the inequation x +2y ≥ 10. So,the region which does not contain the origin represents the solution set of the inequation x +2y ≥ 10.
Region represented by x +6y ≥ 18:
The line x +6y = 18 meets the coordinate axes at E1(18,0) and F1(0, 3) respectively. By joining these points we obtain the line x + 6y = 18.Clearly (0,0) does not satisfies the inequation x + 6y ≥ 18. So,the region which does not contain the origin represents the solution set of the inequation x +6y ≥ 18.
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 2x + y ≥ 8, x + 2y ≥ 10,x + 6y ≥ 18, x ≥ 0, and y ≥ 0, are as follows.
The corner points are B1(0, 8), G1(2, 4), H1(6, 2) and E1(18, 0).
The values of Z at these corner points are as follows
| Corner point | Z= 8x + 12y |
| B1 | 96 |
| G1 | 64 |
| H1 | 72 |
| E1 | 144 |
The minimum value of Z is 64 which is attained at G1
Thus, the minimum cost is Rs 64 obtained when 2 units of kind A and 4 units of kind B nuts were used.
APPEARS IN
संबंधित प्रश्न
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.
Maximise Z = x + 2y subject to the constraints
`x + 2y >= 100`
`2x - y <= 0`
`2x + y <= 200`
Solve the above LPP graphically
Solve the following L.P.P. graphically:
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0 and x, y ≥ 0
Maximize Z = 5x + 3y
Subject to
\[3x + 5y \leq 15\]
\[5x + 2y \leq 10\]
\[ x, y \geq 0\]
Minimize Z = 18x + 10y
Subject to
\[4x + y \geq 20\]
\[2x + 3y \geq 30\]
\[ x, y \geq 0\]
Maximize Z = 4x + 3y
subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Maximize Z = 15x + 10y
Subject to
\[3x + 2y \leq 80\]
\[2x + 3y \leq 70\]
\[ x, y \geq 0\]
Maximize Z = 3x + 4y
Subject to
\[2x + 2y \leq 80\]
\[2x + 4y \leq 120\]
Minimize Z = 2x + 4y
Subject to
\[x + y \geq 8\]
\[x + 4y \geq 12\]
\[x \geq 3, y \geq 2\]
Maximize Z = −x1 + 2x2
Subject to
\[- x_1 + 3 x_2 \leq 10\]
\[ x_1 + x_2 \leq 6\]
\[ x_1 - x_2 \leq 2\]
\[ x_1 , x_2 \geq 0\]
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 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 factory uses three different resources for the manufacture of two different products, 20 units of the resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.
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 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 company manufactures two articles A and B. There are two departments through which these articles are processed: (i) assembly and (ii) finishing departments. The maximum capacity of the first department is 60 hours a week and that of other department is 48 hours per week. The product of each unit of article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs 6 for each unit of A and Rs 8 for each unit of B, find the number of units of A and B to be produced per week in order to have maximum 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.
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.
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?
By graphical method, the solution of linear programming problem
\[\text{ Subject } to \text{ 3 } x_1 + 2 x_2 \leq 18\]
\[ x_1 \leq 4\]
\[ x_2 \leq 6\]
\[ x_1 \geq 0, x_2 \geq 0, \text{ is } \]
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 and 20 minutes available for cutting and 4 hours available for assembling. The profit is Rs. 50 each for type A and Rs. 60 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize profit? Formulate the above LPP and solve it graphically and also find the maximum profit.
Find the graphical solution for the system of linear inequation 2x + y ≤ 2, x − y ≤ 1
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 region XOY - plane which is represented by the inequalities -5 ≤ x ≤ 5, -5 ≤ y ≤ 5 is ______
The constraints of an LPP are 7 ≤ x ≤ 12, 8 ≤ y ≤ 13. Determine the vertices of the feasible region formed by them.
The feasible region (shaded) for a L.P.P is shown in the figure. The maximum Z = 5x + 7y is ____________.
Any point in the feasible region that gives the optional value (maximum or minimum) of the objective function is called:-
Solve the following Linear Programming Problem graphically:
Maximize Z = 400x + 300y subject to x + y ≤ 200, x ≤ 40, x ≥ 20, y ≥ 0
The constraints –x1 + x2 ≤ 1, –x1 + 3x2 ≤ 9, x1x2 ≥ 0 define on ______.
The maximum value of z = 5x + 2y, subject to the constraints x + y ≤ 7, x + 2y ≤ 10, x, y ≥ 0 is ______.
The maximum value of 2x + y subject to 3x + 5y ≤ 26 and 5x + 3y ≤ 30, x ≥ 0, y ≥ 0 is ______.
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:
Minimize: Z = 60x + 80y
Subject to constraints:
3x + 4y ≥ 8
5x + 2y ≥ 11
x, 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
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.
