Advertisements
Advertisements
प्रश्न
Minimize Z = x + 4y subject to constraints
x + 3y ≥ 3, 2x + y ≥ 2, x ≥ 0, y ≥ 0
उत्तर
To draw the feasible region, construct table as follows:
Inequality | x + 3y ≥ 3 | 2x + y ≥ 2 |
Corresponding equation (of line) | x + 3y = 3 | 2x + y = 2 |
Intersection of line with X-axis | (3, 0) | (1, 0) |
Intersection of line with Y-axis | (0, 1) | (0, 2) |
Region | Non-origin side | Non-origin side |
Shaded portion XABCY is the feasible region, whose vertices are A(3, 0), B and C(0, 2).
B is the point of intersection of the lines 2x + y = 2 and x + 3y = 3.
∴ B ≡ `(3/5, 4/5)`
Here, the objective function is
Z = x + 4y
∴ Z at A(3, 0) = 3 + 4(0)
= 3
Z at B`(3/5, 4/5) = 3/5 + 4(4/5)`
= `19/5`
= 3.8
Z at C(0, 2) = 0 + 4(2)
= 8
∴ Z has minimum value 3 at x = 3 and y = 0.
APPEARS IN
संबंधित प्रश्न
Solve the following L.P.P. by graphical method :
Maximize: Z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of Z.
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.
Fill in the blank :
Graphical solution set of the in equations x ≥ 0, y ≥ 0 is in _______ quadrant
Fill in the blank :
The region represented by the in equations x ≤ 0, y ≤ 0 lines in _______ quadrants.
The region represented by the inequality y ≤ 0 lies in _______ quadrants.
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.
Choose the correct alternative:
The minimum value of Z = 4x + 5y subjected to the constraints x + y ≥ 6, 5x + y ≥ 10, x, y ≥ 0 is
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 feasible region for the inequations, x + y ≤ 5, x + 2y ≤ 6, x ≥ 0, y ≥ 0 are
Choose the correct alternative:
The corner points of the feasible region are (4, 2), (5, 0), (4, 1) and (6, 0), then the point of minimum Z = 3.5x + 2y = 16 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
A set of values of variables satisfying all the constraints of LPP is known as ______
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
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 ___
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.