Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Let x and y be the number of desktop model and portable model respectively.
Number of desktop model and portable model cannot be negative.
Therefore, x ≥ 0 , y ≥ 0
It is given that the monthly demand will not exist 250 units.
∴ \[x + y \leq 250\]
Cost of desktop and portable model is Rs 25,000 and Rs 40,000 respectively. Therefore, cost of x desktop model and y portable model is Rs 25,000 and Rs 40,000 respectively and he does not want to invest more than Rs 70 lakhs.
\[25000x + 40000y \leq 7000000\]
Profit on the desktop model is Rs 4500 and on the portable model is Rs 5000. Therefore, profit made by x desktop model and y portable model is Rs 4500x and Rs 5000y respectively.
Total profit = Z = 4500x + 5000y
The mathematical form of the given LPP is:
Maximize Z = 4500x + 5000y
Subject to constraints: \[x + y \leq 250\]
\[25000x + 40000y \leq 7000000\]
x ≥ 0 , y ≥ 0
First we will convert inequations into equations as follows:
x + y = 250, 25000x + 40000y = 7000000, x = 0 and y = 0
Region represented by x + y ≤ 250:
The line x + y = 250 meets the coordinate axes at A(250, 0) and \[B\left( 0, 250 \right)\] respectively. By joining these points we obtain the line x + y = 250. Clearly (0,0) satisfies the x + y = 250. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 250.
Region represented by 25000x + 40000y ≤ 7000000:
The line 25000x + 40000y = 7000000 meets the coordinate axes at C(280, 0) and \[D\left( 0, 175 \right)\] respectively. By joining these points we obtain the line 25000x + 40000y = 7000000. Clearly (0,0) satisfies the inequation 25000x + 40000y ≤ 7000000. So,the region which contains the origin represents the solution set of the inequation 25000x + 40000y ≤ 7000000.
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 ≤ 250, 25000x + 40000y ≤ 7000000, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), D(0, 175), E(200, 50) and A(250, 0).
The values of the objective function Z at corner points of the feasible region are given in the following table:
|
Corner Points |
Z = 4500x + 5000y |
|
| O(0, 0) | 0 | |
|
D(0, 175) |
875000 |
|
|
E(200, 50) |
1150000 |
← Maximum |
|
A(250, 0) |
1125000 |
Clearly, Z is maximum at x = 200 and y = 50 and the maximum value of Z at this point is 1150000.
Thus, 200 desktop models and 50 portable units should be sold to maximize the profit.
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.
Solve the following L.P.P graphically:
Maximize: Z = 10x + 25y
Subject to: x ≤ 3, y ≤ 3, x + y ≤ 5, x ≥ 0, y ≥ 0
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 L.P.P. graphically Maximise Z = 4x + y
Subject to following constraints x + y ≤ 50
3x + y ≤ 90,
x ≥ 10
x, y ≥ 0
Solve the following LPP graphically :
Maximise Z = 105x + 90y
subject to the constraints
x + y ≤ 50
2x + y ≤ 80
x ≥ 0, y ≥ 0.
Minimize Z = 18x + 10y
Subject to
\[4x + y \geq 20\]
\[2x + 3y \geq 30\]
\[ x, y \geq 0\]
Maximize Z = 50x + 30y
Subject to
\[2x + y \leq 18\]
\[3x + 2y \leq 34\]
\[ x, y \geq 0\]
Maximize Z = 4x + 3y
Subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Minimize Z = x − 5y + 20
Subject to
\[x - y \geq 0\]
\[ - x + 2y \geq 2\]
\[ x \geq 3\]
\[ y \leq 4\]
\[ x, y \geq 0\]
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 the minimum value of 3x + 5y subject to the constraints
− 2x + y ≤ 4, x + y ≥ 3, x − 2y ≤ 2, x, y ≥ 0.
Solved the following linear programming problem graphically:
Maximize Z = 60x + 15y
Subject to constraints
\[x + y \leq 50\]
\[3x + y \leq 90\]
\[ x, y \geq 0\]
A dietician mixes together two kinds of food in such a way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1 kg of food X and 1 kg of food Y are given below:
| Vitamin A |
Vitamin B |
Vitamin |
Vitamin D |
|
| Food X Food Y |
1 2 |
1 1 |
1 3 |
2 1 |
One kg food X costs Rs 5, whereas one kg of food Y costs Rs 8. Find the least cost of the mixture which will produce the desired diet.
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 dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum of vitamin A.
A manufacturer has three machines installed in his factory. machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day. He produces only two items, each requiring the use of three machines. The number of hours required for producing one unit each of the items on the three machines is given in the following table:
| Item | Number of hours required by the machine | ||
A B |
I | II | III |
| 1 2 |
2 1 |
1 5/4 |
|
He makes a profit of Rs 6.00 on item A and Rs 4.00 on item B. Assuming that he can sell all that he produces, how many of each item should he produces so as to maximize his profit? Determine his maximum profit. Formulate this LPP mathematically and then solve it.
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 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 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 products A and B. Each unit of A requires 3 grams of nickel and 1 gram of chromium, while each unit of B requires 1 gram of nickel and 2 grams of chromium. The firm can produce 9 grams of nickel and 8 grams of chromium. The profit is ₹ 40 on each unit of the product of type A and ₹ 50 on each unit of type B. How many units of each type should the company manufacture so as to earn a maximum profit? Use linear programming to find the solution.
A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 hours of work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturer's profit on an item of model A is ₹ 15 and on an item of model B is ₹ 10. How many items of each model should be made per day in order to maximize daily profit? Formulate the above LPP and 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
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 ______.
The maximum value of z = 3x + 10y subjected to the conditions 5x + 2y ≤ 10, 3x + 5y ≤ 15, x, y ≥ 0 is ______.
The maximum of z = 5x + 2y, subject to the constraints x + y ≤ 7, x + 2y ≤ 10, x, y ≥ 0 is ______.
Of all the points of the feasible region for maximum or minimum of objective function the points.
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 ____________.
Minimise z = – 3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 What will be the minimum value of z ?
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?
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 ______.
Minimize z = x + 2y,
Subject to x + 2y ≥ 50, 2x – y ≤ 0, 2x + y ≤ 100, x ≥ 0, y ≥ 0.
Aman has ₹ 1500 to purchase rice and wheat for his grocery shop. Each sack of rice and wheat costs ₹ 180 and Rupee ₹ 120 respectively. He can store a maximum number of 10 bags in his shop. He will earn a profit of ₹ 11 per bag of rice and ₹ 9 per bag of wheat.
- Formulate a Linear Programming Problem to maximise Aman’s profit.
- Calculate the maximum profit.
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.
Which point is an infeasible solution?
The Corner Point Method is based on which theorem?
