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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
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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
For the following assignment problem minimize total man hours:
| Subordinates | Required hours for task | |||
| I | II | III | IV | |
| A | 7 | 25 | 26 | 10 |
| B | 12 | 27 | 3 | 25 |
| C | 37 | 18 | 17 | 14 |
| D | 18 | 25 | 23 | 9 |
Subtract the `square` element of each `square` from every element of that `square`
| Subordinates | Required hours for task | |||
| I | II | III | IV | |
| A | 0 | 18 | 19 | 3 |
| B | 9 | 24 | 0 | 22 |
| C | 23 | 4 | 3 | 0 |
| D | 9 | 16 | 14 | 0 |
Subtract the smallest element in each column from `square` of that column.
| Subordinates | Required hours for task | |||
| I | II | III | IV | |
| A | `square` | `square` | 19 | `square` |
| B | `square` | `square` | 0 | `square` |
| C | `square` | `square` | 3 | `square` |
| D | `square` | `square` | 14 | `square` |
The lines covering all zeros is `square` to the order of matrix `square`
The assignment is made as follows:
| Subordinates | Required hours for task | |||
| I | II | III | IV | |
| A | 0 | 14 | 19 | 3 |
| B | 9 | 20 | 0 | 22 |
| C | 23 | 0 | 3 | 0 |
| D | 9 | 12 | 14 | 0 |
Optimum solution is shown as follows:
A → `square, square` → III, C → `square, square` → IV
Minimum hours required is `square` hours
Concept: undefined >> undefined
Use quantifiers to convert the following open sentence defined on N, into a true statement.
3x - 4 < 9
Concept: undefined >> undefined
`int 1/sqrt(x^2 - 9) dx` = ______.
Concept: undefined >> undefined
The slope of a tangent to the curve y = 3x2 – x + 1 at (1, 3) is ______.
Concept: undefined >> undefined
The area of the region bounded by the curve y = x2, x = 0, x = 3, and the X-axis is ______.
Concept: undefined >> undefined
State whether the following statement is true or false.
If `int (4e^x - 25)/(2e^x - 5)` dx = Ax – 3 log |2ex – 5| + c, where c is the constant of integration, then A = 5.
Concept: undefined >> undefined
`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`
Concept: undefined >> undefined
Find the area between the two curves (parabolas)
y2 = 7x and x2 = 7y.
Concept: undefined >> undefined
Divide 20 into two ports, so that their product is maximum.
Concept: undefined >> undefined
State whether the following statement is true or false:
To convert a maximization-type assignment problem into a minimization problem, the smallest element in the matrix is deducted from all elements of the matrix.
Concept: undefined >> undefined
Calculate the cost of living index number for the following data by aggregative expenditure method:
| Group | Base year | Current year | |
| Price | Quantity | Price | |
| Food | 120 | 15 | 170 |
| Clothing | 150 | 20 | 190 |
| Fuel and lighting | 130 | 30 | 220 |
| House rent | 160 | 10 | 180 |
| Miscellaneous | 200 | 11 | 220 |
Concept: undefined >> undefined
