Advertisements
Advertisements
प्रश्न
A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3 gm of silver and 1 gm of gold while that of type B requires 1 gm of silver and 2 gm of gold. The company can produce 9 gm of silver and 8 gm of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, find the number of units of each type that the company should produce to maximize the profit. What is the maximum profit?
Advertisements
उत्तर
Let x goods of type A and y goods of type B were produced.
Number of goods cannot be negative.
Therefore, \[x, y \geq 0\]
The given information can be tabulated as follows:
| Silver(gm) | Gold white(gm) | |
| Type A | 3 | 1 |
| Type B | 1 | 2 |
| Availability | 9 | 8 |
Therefore, the constraints are
\[3x + y \leq 9\]
\[x + 2y \leq 8\]
If each unit of type A brings a profit of Rs 40 and that of type B Rs 50.Then, x goods of type A and y goods of type B brings a profit of Rs 40x and Rs 50y.
Total profit = Z = \[40x + 50y\] which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = \[40x + 50y\]
subject to
\[3x + y \leq 9\]
\[x + 2y \leq 8\]
First we will convert inequations into equations as follows :
3x + y = 9, x + 2y = 8, x = 0 and y = 0
Region represented by 3x + y ≤ 9:
The line 3x + y = 9 meets the coordinate axes at A1(3, 0) and B1(0, 9) respectively. By joining these points we obtain the line
3x + y = 9. Clearly (0,0) satisfies the 3x + y = 9. So,the region which contains the origin represents the solution set of the inequation 3x + y ≤ 9.
Region represented by x + 2y ≤ 8:
The line x + 2y = 8 meets the coordinate axes at C1(8, 0) and D1(0, 4) respectively. By joining these points we obtain the line x + 2y = 8. Clearly (0,0) satisfies the inequation x + 2y ≤ 8. So,the region which contains the origin represents the solution set of the inequation x + 2y ≤ 8.
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 3x + y ≤ 9, x + 2y ≤ 8, x ≥ 0, and y ≥ 0 are as follows.
The corner points are O(0, 0), D1(0, 4), E1(2, 3), A1(3, 0)
The values of Z at these corner points are as follows
| Corner point | Z = 40x + 50y |
| O | 0 |
| D1 | 200 |
| E1 | 230 |
| A1 | 120 |
The maximum value of Z is 230 which is attained at E1(2, 3).
Thus, the maximum profit is of Rs 230 obtained when 2 units of type A and 3 units of type B produced.
APPEARS IN
संबंधित प्रश्न
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?
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.
Minimum and maximum z = 5x + 2y subject to the following constraints:
x-2y ≤ 2
3x+2y ≤ 12
-3x+2y ≤ 3
x ≥ 0,y ≥ 0
Solve the following L. P. P. graphically:Linear Programming
Minimize Z = 6x + 2y
Subject to
5x + 9y ≤ 90
x + y ≥ 4
y ≤ 8
x ≥ 0, 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
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 = 3x + 4y
Subject to
\[2x + 2y \leq 80\]
\[2x + 4y \leq 120\]
Minimize Z = 3x1 + 5x2
Subject to
\[x_1 + 3 x_2 \geq 3\]
\[ x_1 + x_2 \geq 2\]
\[ x_1 , x_2 \geq 0\]
Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs Rs 5 per kg and rice costs Rs 4 per kg
A manufacturer of Furniture makes two products : chairs and tables. processing of these products is done on two machines A and B. A chair requires 2 hrs on machine A and 6 hrs on machine B. A table requires 4 hrs on machine A and 2 hrs on machine B. There are 16 hrs of time per day available on machine A and 30 hrs on machine B. Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively. Find with the help of graph what should be the daily production of each of the two products so as to maximize his 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 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.
By graphical method, the solution of linear programming problem
\[\text{ Subject } to \text{ 3 } x_1 + 2 x_2 \leq 18\]
\[ x_1 \leq 4\]
\[ x_2 \leq 6\]
\[ x_1 \geq 0, x_2 \geq 0, \text{ is } \]
From the details given below, calculate the five-year moving averages of the number of students who have studied in a school. Also, plot these and original data on the same graph paper.
| Year | 1993 | 1994 | 1995 | 1996 | 1997 | 1998 | 1999 | 2000 | 2001 | 2002 |
| Number of Students | 332 | 317 | 357 | 392 | 402 | 405 | 410 | 427 | 405 | 438 |
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.
For L.P.P. maximize z = 4x1 + 2x2 subject to 3x1 + 2x2 ≥ 9, x1 - x2 ≤ 3, x1 ≥ 0, x2 ≥ 0 has ______.
The maximum value of z = 3x + 10y subjected to the conditions 5x + 2y ≤ 10, 3x + 5y ≤ 15, x, y ≥ 0 is ______.
The point which provides the solution to the linear programming problem: Max P = 2x + 3y subject to constraints: x ≥ 0, y ≥ 0, 2x + 2y ≤ 9, 2x + y ≤ 7, x + 2y ≤ 8, is ______
Maximise and Minimise Z = 3x – 4y subject to x – 2y ≤ 0, – 3x + y ≤ 4, x – y ≤ 6, x, y ≥ 0
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 objective function Z has both a maximum and a minimum value on R and ____________.
In Corner point method for solving a linear programming problem the first step is to ____________.
A feasible solution to a linear programming problem
The feasible region (shaded) for a L.P.P is shown in the figure. The maximum Z = 5x + 7y is ____________.
Which of the statements describe the solution set for `-2(x + 8) = - 2x + 20`?
Solve the following Linear Programming Problem graphically:
Maximize Z = 400x + 300y subject to x + y ≤ 200, x ≤ 40, x ≥ 20, y ≥ 0
The maximum value of z = 5x + 2y, subject to the constraints x + y ≤ 7, x + 2y ≤ 10, x, 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.
The corner points of the feasible region of a linear programming problem are (0, 4), (8, 0) and `(20/3, 4/3)`. If Z = 30x + 24y is the objective function, then (maximum value of Z – minimum value of Z) is equal to ______.
Solve the following Linear Programming Problem graphically:
Minimize: Z = 60x + 80y
Subject to constraints:
3x + 4y ≥ 8
5x + 2y ≥ 11
x, y ≥ 0
Minimize z = x + 2y,
Subject to x + 2y ≥ 50, 2x – y ≤ 0, 2x + y ≤ 100, 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
If x – y ≥ 8, x ≥ 3, y ≥ 3, x ≥ 0, y ≥ 0 then find the coordinates of the corner points of the feasible region.
If two corner points give the same optimum value, which points are also optimal?
