Home
Class 12
MATHS
Minimise and Maximise Z=x+2y Subject t...

Minimise and Maximise `Z=x+2y`
Subject to `x+2yge100,2x-yle0,2x+yle200,x,yge0`.

Text Solution

Verified by Experts

The correct Answer is:
`Z_(max) = 400 at (0, 200)` and `Z_(min) = 100` at all points on the line segment joining the points B (0, 50) and E(20,40).
Promotional Banner

Topper's Solved these Questions

  • LINEAR PROGRAMMING

    MODERN PUBLICATION|Exercise Exercise 12.2|11 Videos

  • LINEAR PROGRAMMING

    MODERN PUBLICATION|Exercise Miscellaneous Exercise|10 Videos

  • LINEAR PROGRAMMING

    MODERN PUBLICATION|Exercise EXERCISE 12 (C ) (Very Short Answer Type Questions)|15 Videos

  • INVERSE - TRIGONOMETRIC FUNCTIONS

    MODERN PUBLICATION|Exercise CHAPTER TEST (2)|11 Videos

  • MATRICES

    MODERN PUBLICATION|Exercise CHAPTER TEST (3)|12 Videos

Similar Questions

Explore conceptually related problems

Maximise Z=3x+2y Subject to x+2yle10,3x+yle15, x,yge0

Minimise and Maximise Z=5x+10y Subject to x+2yle120,x+yge60,x-2yge0,yge0 .

Maximise Z=x+y Subject to x-yle -1, -x+yle0,x,yge0 .

The minimum value of z=x+2y subject to x+2yge50,2x-yle0,2x+yle100,xge0,yge0 is

Maximise Z=5x+3y Subject to 3x+5yle15, 5x+2yle10,xge0,yge0 .

Minimise Z=-3x+4y Subject to x+2yle8, 3x+2yle12,xge0,yge0 .

The objective function z=x+2y subject to x+2yge50,2x-yle0,2x+yle100,x,yge0 can be minimized

Minimise z=3x+y , subject to 2x+3yle6,x+yge1, xge0,yge0

Minimize z = 30x+20y subject to x+yle8,x+2yge4,6x+4yge12,xge,yge0