Advertisements
Advertisements
प्रश्न
Solve the following L.P.P. by graphical method :
Minimize : Z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0.
Advertisements
उत्तर
To draw the feasible region, construct table as follows:
| Inequality | 5x + y ≥ 5 | x + y ≥ 3 |
| Corresponding equation (of line) | 5x + y = 5 | x + y = 3 |
| Intersection of line with X-axis | (1, 0) | (3, 0) |
| Intersection of line with Y-axis | (0, 5) | (0, 3) |
| Region | Non-origin side | Non-origin side |
Shaded portion XABCY is the feasible region,
whose vertices are A(3, 0), B and C (0, 5).
B is the point of intersection of the lines x + y = 3 and 5x + y = 5
Solving the above equations, we get
x = `(1)/(2), y = (5)/(2)`
∴ B ≡ `(1/2, 5/2)`
Here, the objective function is Z = 7x + y
Z at A(3, 0) = 7(3) + 0 = 21
Z at B`(1/2, 5/2) = 7(1/2) + (5)/(2)`
= `(7)/(2) + (5)/(2)` = 6
Z at C (0, 5) = 7(0) + 5 = 5
∴ Z has minimum value 5 at C(0, 5).
∴ Z is minimum, when x = 0 and y = 5.
APPEARS IN
संबंधित प्रश्न
A company produces two types of goods A and B, that require gold and silver. Each unit of type A requires 3 g of silver and 1 g of golds while that of type B requires 1 g of silver and 2 g of gold. The company can procure a maximum of 9 g of silver and 8 g of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, formulate LPP to maximize profit.
A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry, then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for the production of each unit of A and B, and the number of man-hours the firm has available per week are as follows:
| Gadget | Foundry | Machine-shop |
| A | 10 | 5 |
| B | 6 | 4 |
| Firm's capacity per week | 1000 | 600 |
The profit on the sale of A is Rs 30 per unit as compared with Rs 20 per unit of B. The problem is to determine the weekly production of gadgets A and B, so that the total profit is maximized. Formulate this problem as a LPP.
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.
Amit's mathematics teacher has given him three very long lists of problems with the instruction to submit not more than 100 of them (correctly solved) for credit. The problem in the first set are worth 5 points each, those in the second set are worth 4 points each, and those in the third set are worth 6 points each. Amit knows from experience that he requires on the average 3 minutes to solve a 5 point problem, 2 minutes to solve a 4 point problem, and 4 minutes to solve a 6 point problem. Because he has other subjects to worry about, he can not afford to devote more than
The corner points of the feasible region determined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is
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 = 7x + 11y subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Fill in the blank :
The region represented by the in equations x ≤ 0, y ≤ 0 lines in _______ quadrants.
The region represented by the inequalities x ≥ 0, y ≥ 0 lies in first quadrant.
Solve the following problem :
Minimize Z = 4x + 2y Subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0
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:
If LPP has optimal solution at two point, then
Choose the correct alternative:
The maximum value of Z = 3x + 5y subjected to the constraints x + y ≤ 2, 4x + 3y ≤ 12, x ≥ 0, y ≥ 0 is
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 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:
If the corner points of the feasible region are `(0, 7/3)`, (2, 1), (3, 0) and (0, 0), then the maximum value of Z = 4x + 5y is 12
State whether the following statement is True or False:
If the corner points of the feasible region are (0, 10), (2, 2) and (4, 0), then the minimum value of Z = 3x + 2y is at (4, 0)
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
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
Maximize Z = 2x + 3y subject to constraints
x + 4y ≤ 8, 3x + 2y ≤ 14, x ≥ 0, y ≥ 0.
Maximize Z = 5x + 10y subject to constraints
x + 2y ≤ 10, 3x + y ≤ 12, x ≥ 0, y ≥ 0
Maximize Z = 400x + 500y subject to constraints
x + 2y ≤ 80, 2x + y ≤ 90, x ≥ 0, y ≥ 0
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
______
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 ___.
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 ___
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.
