Please select a subject first
Advertisements
Advertisements
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
Concept: undefined >> undefined
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
Concept: undefined >> undefined
Advertisements
A company manufactures two types of ladies dresses C and D. The raw material and labour available per day is given in the table.
| Resources | Dress C(x) | Dress D(y) | Max. availability |
| Raw material | 5 | 4 | 60 |
| Labour | 5 | 3 | 50 |
P is the profit, if P = 50x + 100y, solve this LPP to find x and y to get the maximum profit
Concept: undefined >> undefined
Smita is a diet conscious house wife, wishes to ensure certain minimum intake of vitamins A, B and C for the family. The minimum daily needs of vitamins A, B, and C for the family are 30, 20, and 16 units respectively. For the supply of the minimum vitamin requirements Smita relies on 2 types of foods F1 and F2. F1 provides 7, 5 and 2 units of A, B, C vitamins per 10 grams and F2 provides 2, 4 and 8 units of A, B and C vitamins per 10 grams. F1 costs ₹ 3 and F2 costs ₹ 2 per 10 grams. How many grams of each F1 and F2 should buy every day to keep her food bill minimum
Concept: undefined >> undefined
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
Concept: undefined >> undefined
A wholesale dealer deals in two kinds of mixtures A and B of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew and 30 grams of hazel nuts. Each kg of mixture B contains 30 grams of almonds, 60 grams of cashew and 180 grams of hazel nuts. A 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 and 540 grams of hazel nuts. Mixture A costs ₹ 8 and B costs ₹ 12 per kg. How many kgs of each mixture should he use to minimize the cost of the kgs
Concept: undefined >> undefined
Maximize Z = 2x + 3y subject to constraints
x + 4y ≤ 8, 3x + 2y ≤ 14, x ≥ 0, y ≥ 0.
Concept: undefined >> undefined
Maximize Z = 5x + 10y subject to constraints
x + 2y ≤ 10, 3x + y ≤ 12, x ≥ 0, y ≥ 0
Concept: undefined >> undefined
Maximize Z = 400x + 500y subject to constraints
x + 2y ≤ 80, 2x + y ≤ 90, x ≥ 0, y ≥ 0
Concept: undefined >> undefined
Minimize Z = 24x + 40y subject to constraints
6x + 8y ≥ 96, 7x + 12y ≥ 168, x ≥ 0, y ≥ 0
Concept: undefined >> undefined
Minimize Z = x + 4y subject to constraints
x + 3y ≥ 3, 2x + y ≥ 2, x ≥ 0, y ≥ 0
Concept: undefined >> undefined
Minimize Z = 2x + 3y subject to constraints
x + y ≥ 6, 2x + y ≥ 7, x + 4y ≥ 8, x ≥ 0, y ≥ 0
Concept: undefined >> undefined
Amartya wants to invest ₹ 45,000 in Indira Vikas Patra (IVP) and in Public Provident fund (PPF). He wants to invest at least ₹ 10,000 in PPF and at least ₹ 5000 in IVP. If the rate of interest on PPF is 8% per annum and that on IVP is 7% per annum. Formulate the above problem as LPP to determine maximum yearly income.
Solution: Let x be the amount (in ₹) invested in IVP and y be the amount (in ₹) invested in PPF.
x ≥ 0, y ≥ 0
As per the given condition, x + y ______ 45000
He wants to invest at least ₹ 10,000 in PPF.
∴ y ______ 10000
Amartya wants to invest at least ₹ 5000 in IVP.
∴ x ______ 5000
Total interest (Z) = ______
The formulated LPP is
Maximize Z = ______ subject to
______
Concept: undefined >> undefined
Solve the following LPP graphically:
Maximize Z = 9x + 13y subject to constraints
2x + 3y ≤ 18, 2x + y ≤ 10, x ≥ 0, y ≥ 0
Solution: Convert the constraints into equations and find the intercept made by each one of it.
| Inequation | Equation | X intercept | Y intercept | Region |
| 2x + 3y ≤ 18 | 2x + 3y = 18 | (9, 0) | (0, ___) | Towards origin |
| 2x + y ≤ 10 | 2x + y = 10 | ( ___, 0) | (0, 10) | Towards origin |
| x ≥ 0, y ≥ 0 | x = 0, y = 0 | X axis | Y axis | ______ |
The feasible region is OAPC, where O(0, 0), A(0, 6),
P( ___, ___ ), C(5, 0)
The optimal solution is in the following table:
| Point | Coordinates | Z = 9x + 13y | Values | Remark |
| O | (0, 0) | 9(0) + 13(0) | 0 | |
| A | (0, 6) | 9(0) + 13(6) | ______ | |
| P | ( ___,___ ) | 9( ___ ) + 13( ___ ) | ______ | ______ |
| C | (5, 0) | 9(5) + 13(0) | ______ |
∴ Z is maximum at __( ___, ___ ) with the value ___.
Concept: undefined >> undefined
Solve the LPP graphically:
Minimize Z = 4x + 5y
Subject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0
Solution: Convert the constraints into equations and find the intercept made by each one of it.
| Inequations | Equations | X intercept | Y intercept | Region |
| 5x + y ≥ 10 | 5x + y = 10 | ( ___, 0) | (0, 10) | Away from origin |
| x + y ≥ 6 | x + y = 6 | (6, 0) | (0, ___ ) | Away from origin |
| x + 4y ≥ 12 | x + 4y = 12 | (12, 0) | (0, 3) | Away from origin |
| x, y ≥ 0 | x = 0, y = 0 | x = 0 | y = 0 | 1st quadrant |
∵ Origin has not satisfied the inequations.
∴ Solution of the inequations is away from origin.
The feasible region is unbounded area which is satisfied by all constraints.
In the figure, ABCD represents
The set of the feasible solution where
A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).
The coordinates of B are obtained by solving equations
x + 4y = 12 and x + y = 6
The coordinates of C are obtained by solving equations
5x + y = 10 and x + y = 6
Hence the optimum solution lies at the extreme points.
The optimal solution is in the following table:
| Point | Coordinates | Z = 4x + 5y | Values | Remark |
| A | (12, 0) | 4(12) + 5(0) | 48 | |
| B | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | ______ |
| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
∴ Z is minimum at ___ ( ___, ___ ) with the value ___
Concept: undefined >> undefined
Choose the correct alternative:
The cost matrix of an unbalanced assignment problem is not a ______
Concept: undefined >> undefined
An unbalanced assignment problems can be balanced by adding dummy rows or columns with ______ cost
Concept: undefined >> undefined
A ______ assignment problem does not allow some worker(s) to be assign to some job(s)
Concept: undefined >> undefined
State whether the following statement is True or False:
To convert the assignment problem into maximization problem, the smallest element in the matrix is to deducted from all other elements
Concept: undefined >> undefined
Find the assignments of salesman to various district which will yield maximum profit
| Salesman | District | |||
| 1 | 2 | 3 | 4 | |
| A | 16 | 10 | 12 | 11 |
| B | 12 | 13 | 15 | 15 |
| C | 15 | 15 | 11 | 14 |
| D | 13 | 14 | 14 | 15 |
Concept: undefined >> undefined
