The corner points of the feasible region determined by the following system of inequalities:
`2 x+y le 10, x+3 y le 15, x, y ge 0` are `(0,0),(5,0),(3,4)`, and `(0,5)`. Let `Z=p x+q y`, where `p, qgt0`.
Condition on `p` and `q` so that the maximum of `Z` occurs at both `(3,4)` and `(0,5)` is a)`p=q` b)`p=2q` c)`p=3q` d)`q=3p`
Topper's Solved these Questions
LINEAR PROGRAMMING
BODY BOOKS PUBLICATION|Exercise EXERCISE|2 Videos
INVERSE TRIGONOMETRIC FUNCTIONS
BODY BOOKS PUBLICATION|Exercise EXERCISE|97 Videos
MATRICES
BODY BOOKS PUBLICATION|Exercise EXERCISE|107 Videos
Similar Questions
Explore conceptually related problems
Solve the following system of linear inequalities 3x + 2y ge 20 , 3x + y le 15 , x ge0 , y ge 0
Solve the following system of linear inequalities graphically. 3x + 4y le 60 , x + 3y le 30 , x ge 0 , y ge 0
Solve the following system of linear inequalities graphically. x + 2y le 10, x+y ge 1, x-y le 0 , x ge 0 , y ge 0
Solve the following system of linear inequalities graphically 3x + 2y le 150 , x + 4y le 80 , x le 15 ,x, y ge 0
Solve the following graphically x + 4y le 8 , 3x - 4y le 12 , x ge 0 , y ge 0
Minimise Z=3 x+2 y ,subject to the constraints: x+y ge 8 , 3 x+5 y le 15 , x ge 0, y ge 0
Maximise Z=5 x+3 y Subject to 3 x+5 y le 15,5 x+2 y le 10, x ge 0, y ge 0
For what values of p and q, the system of equations 2x + py + 6z = 8, x + 2y + qz = 5, x + y + 3z = 4 has no solution
For what values of p and q, the system of equations 2x + py + 6z= 8, x + 2y + qz = 5, x + y + 3z = 4 has a unique solution
BODY BOOKS PUBLICATION-LINEAR PROGRAMMING-EXERCISE