Advertisements
Advertisements
प्रश्न
A chemist has a compound to be made using 3 basic elements X, Y, Z so that it has at least 10 litres of X, 12 litres of Y and 20 litres of Z. He makes this compound by mixing two compounds (I) and (II). Each unit compound (I) had 4 litres of X, 3 litres of Y. Each unit compound (II) had 1 litre of X, 2 litres of Y and 4 litres of Z. The unit costs of compounds (I) and (II) are ₹ 400 and ₹ 600 respectively. Find the number of units of each compound to be produced so as to minimize the cost
Advertisements
उत्तर
Let the chemist produce x units of compound I and y units of compound II.
Since x and y cannot be negative, x ≥ 0, y ≥ 0
The unit costs of compounds I and II are ₹ 400 and ₹ 600 respectively.
Total Cost = Z = 400x + 600y
We construct a table with constraints of X, Y and Z as follows:
| Compound I | Compound II | Least value | |
| X | 4 | 1 | 10 |
| Y | 3 | 2 | 12 |
| Z | – | 4 | 20 |
From the table, the constraints are
4x + y ≥ 10
3x + 2y ≥ 12
4y ≥ 20
∴ Given problem can be formulated as follows:
Minimize Z = 400x + 600y
Subject to 4x + y ≥ 10
3x + 2y ≥ 12
4y ≥ 20, x ≥ 0, y ≥ 0
To draw the feasible region, construct table as follows:
| Inequality | 4x + y ≥ 10 | 3x + 2y ≥ 12 | 4y ≥ 20 |
| Corresponding equation (of line) | 4x + y = 10 | 3x + 2y = 12 | 4y = 20 |
| Intersection of line with X-axis | `(5/2, 0)` | (4, 0) | – |
| Intersection of line with Y-axis | (0, 10) | (0, 6) | (0, 5) |
| Region | Non-Origin side | Non-Origin side | Non-Origin side |
Shaded portion EABY is the feasible region, whose vertices are A and B(0, 10).
A is the point of intersection of the lines 4y = 20 and 4x + y = 10,
Solving the above equations, we get
x = `5/4`, y = 5
∴ A ≡ `(5/4, 5)`
Here, the objective function is
Z = 400x + 600y
∴ Z at A`(5/4, 5) = 400(5/4) + 600(5)`
= 500 + 3000
= 3500
Z at B (0, 10) = 400(0) + 600(10)
= 6000
∴ Z has minimum value 3500 at x = `5/4` and y = 5.
∴ The chemist should produce `5/4` units of compound I and 5 units of compound II to minimize the cost.
APPEARS IN
संबंधित प्रश्न
The postmaster of a local post office wishes to hire extra helpers during the Deepawali season, because of a large increase in the volume of mail handling and delivery. Because of the limited office space and the budgetary conditions, the number of temporary helpers must not exceed 10. According to past experience, a man can handle 300 letters and 80 packages per day, on the average, and a woman can handle 400 letters and 50 packets per day. The postmaster believes that the daily volume of extra mail and packages will be no less than 3400 and 680 respectively. A man receives Rs 225 a day and a woman receives Rs 200 a day. How many men and women helpers should be hired to keep the pay-roll at a minimum ? Formulate an LPP and solve it graphically.
A firm manufactures 3 products A, B and C. The profits are Rs 3, Rs 2 and Rs 4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product :
| Machine | Products | ||
| A | B | C | |
| M1 M2 |
4 | 3 | 5 |
| 2 | 2 | 4 | |
Machines M1 and M2 have 2000 and 2500 machine minutes respectively. The firm must manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's. Set up a LPP to maximize the profit.
A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:
| Product A | Product B | Weekly capacity | |
| Department 1 | 3 | 2 | 130 |
| Department 2 | 4 | 6 | 260 |
| Selling price per unit | ₹ 25 | ₹ 30 | |
| Labour cost per unit | ₹ 16 | ₹ 20 | |
| Raw material cost per unit | ₹ 4 | ₹ 4 |
The problem is to determine the number of units to produce each product so as to maximize total contribution to profit. Formulate this as a LPP.
Solve the following LPP by graphical method:
Maximize z = 11x + 8y, subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0
Solve the following L.P.P. by graphical method:
Maximize: Z = 10x + 25y
subject to 0 ≤ x ≤ 3,
0 ≤ y ≤ 3,
x + y ≤ 5.
Also find the maximum value of z.
Solve the following L.P.P. by graphical method:
Minimize: Z = 6x + 2y subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
The constraint that a factory has to employ more women (y) than men (x) is given by _______
Solve the following problem :
Minimize Z = 2x + 3y Subject to x – y ≤ 1, x + y ≥ 3, x ≥ 0, y ≥ 0
Solve the following problem:
Maximize Z = 4x1 + 3x2 Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0
Maximize Z = 60x + 50y Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
Solve the following problem :
A company manufactures bicyles and tricycles, each of which must be processed through two machines A and B Maximum availability of machine A and B is respectively 120 and 180 hours. 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 ₹ 180 for a bicycle and ₹ 220 on a tricycle, determine the number of bicycles and tricycles that should be manufacturing in order to maximize the profit.
Solve the following problem :
A Company produces mixers and processors Profit on selling one mixer and one food processor is ₹ 2000 and ₹ 3000 respectively. Both the products are processed through three machines A, B, C The time required in hours by each product and total time available in hours per week on each machine are as follows:
| Machine/Product | Mixer per unit | Food processor per unit | Available time |
| A | 3 | 3 | 36 |
| B | 5 | 2 | 50 |
| C | 2 | 6 | 60 |
How many mixers and food processors should be produced to maximize the profit?
Choose the correct alternative:
The point at which the minimum value of Z = 8x + 12y subject to the constraints 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0 is obtained at the point
Choose the correct alternative:
The point at which the maximum value of Z = 4x + 6y subject to the constraints 3x + 2y ≤ 12, x + y ≥ 4, x ≥ 0, y ≥ 0 is obtained at the point
Choose the correct alternative:
The corner points of the feasible region are (0, 3), (3, 0), (8, 0), `(12/5, 38/5)` and (0, 10), then the point of maximum Z = 6x + 4y = 48 is at
State whether the following statement is True or False:
A convex set includes the points but not the segment joining the points
State whether the following statement is True or False:
Corner point method is most suitable method for solving the LPP graphically
A company manufactures 2 types of goods P and Q that requires copper and brass. Each unit of type P requires 2 grams of brass and 1 gram of copper while one unit of type Q requires 1 gram of brass and 2 grams of copper. The company has only 90 grams of brass and 80 grams of copper. Each unit of types P and Q brings profit of ₹ 400 and ₹ 500 respectively. Find the number of units of each type the company should produce to maximize its profit
A dealer deals in two products X and Y. He has ₹ 1,00,000/- to invest and space to store 80 pieces. Product X costs ₹ 2500/- and product Y costs ₹ 1000/- per unit. He can sell the items X and Y at respective profits of ₹ 300 and ₹ 90. Construct the LPP and find the number of units of each product to be purchased to maximize its profit
Maximize Z = 400x + 500y subject to constraints
x + 2y ≤ 80, 2x + y ≤ 90, x ≥ 0, y ≥ 0
Minimize Z = x + 4y subject to constraints
x + 3y ≥ 3, 2x + y ≥ 2, x ≥ 0, y ≥ 0
Minimize Z = 2x + 3y subject to constraints
x + y ≥ 6, 2x + y ≥ 7, x + 4y ≥ 8, x ≥ 0, y ≥ 0
A linear function z = ax + by, where a and b are constants, which has to be maximised or minimised according to a set of given condition is called a:-
Maximised value of z in z = 3x + 4y, subject to constraints : x + y ≤ 4, x ≥ 0. y ≥ 0
If z = 200x + 500y .....(i)
Subject to the constraints:
x + 2y ≥ 10 .......(ii)
3x + 4y ≤ 24 ......(iii)
x, 0, y ≥ 0 ......(iv)
At which point minimum value of Z is attained.
Graphical solution set of the inequations x ≥ 0 and y ≤ 0 lies in ______ quadrant.
