Advertisements
Advertisements
प्रश्न
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
______
Advertisements
उत्तर
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) = 0.07x + 0.08y
The formulated LPP is
Maximize Z = 0.07x + 0.08y subject to
x + y ≤ 45000
y ≥ 10000
x ≥ 5000
x ≥ 0, y ≥ 0
APPEARS IN
संबंधित प्रश्न
Choose the correct alternative:
The value of objective function is maximize under linear constraints.
Choose the correct alternative :
The maximum value of z = 10x + 6y, subjected to the constraints 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, y ≥ 0 is.
Solve the following problem :
Maximize Z = 5x1 + 6x2 Subject to 2x1 + 3x2 ≤ 18, 2x1 + x2 ≤ 12, x ≥ 0, x2 ≥ 0
Solve the following problem :
Minimize Z = 4x + 2y Subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0
Solve the following problem:
Maximize Z = 4x1 + 3x2 Subject to 3x1 + x2 ≤ 15, 3x1 + 4x2 ≤ 24, x1 ≥ 0, x2 ≥ 0
A carpenter makes chairs and tables, profits are ₹ 140 per chair and ₹ 210 per table. Both products are processed on three machines, Assembling, Finishing and Polishing. The time required for each product in hours and the availability of each machine is given by the following table.
| Product/Machines | Chair (x) |
Table (y) |
Available time (hours) |
| Assembling | 3 | 3 | 36 |
| Finishing | 5 | 2 | 50 |
| Polishing | 2 | 6 | 60 |
Formulate and solve the following Linear programming problems using graphical method.
Solve the following problem :
A factory produced two types of chemicals A and B The following table gives the units of ingredients P & Q (per kg) of Chemicals A and B as well as minimum requirements of P and Q and also cost per kg. of chemicals A and B.
| Ingredients per kg. /Chemical Units | A (x) |
B (y) |
Minimum requirements in |
| P | 1 | 2 | 80 |
| Q | 3 | 1 | 75 |
| Cost (in ₹) | 4 | 6 |
Find the number of units of chemicals A and B should be produced so as to minimize the cost.
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:
The maximum value of Z = 5x + 3y subjected to constraints 3x + y ≤ 12, 2x + 3y ≤ 18, 0 ≤ x, y is 20
State whether the following statement is True or False:
If LPP has two optimal solutions, then the LPP has infinitely many solutions
State whether the following statement is True or False:
Corner point method is most suitable method for solving the LPP graphically
State whether the following statement is True or False:
Of all the points of feasible region, the optimal value is obtained at the boundary of the feasible region
State whether the following statement is True or False:
The point (6, 4) does not belong to the feasible region bounded by 8x + 5y ≤ 60, 4x + 5y ≤ 40, 0 ≤ x, y
State whether the following statement is True or False:
The graphical solution set of the inequations 0 ≤ y, x ≥ 0 lies in second quadrant
A set of values of variables satisfying all the constraints of LPP is known as ______
If the feasible region is bounded by the inequations 2x + 3y ≤ 12, 2x + y ≤ 8, 0 ≤ x, 0 ≤ y, then point (5, 4) is a ______ of the feasible region
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
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
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
Minimize Z = 24x + 40y subject to constraints
6x + 8y ≥ 96, 7x + 12y ≥ 168, x ≥ 0, y ≥ 0
Minimize Z = x + 4y subject to constraints
x + 3y ≥ 3, 2x + y ≥ 2, x ≥ 0, y ≥ 0
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 ___
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.
Shraddho wants to invest at most ₹ 25,000/- in saving certificates and fixed deposits. She wants to invest at least ₹ 10,000/- in saving certificate and at least ₹ 15,000/- in fixed deposits. The rate of interest on saving certificate is 5% and that on fixed deposits is 7% per annum. Formulate the above problem as LPP to determine maximum income yearly.
