Solve the following LPP by using graphical method.

Maximize : Z = 6x + 4y

Subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, x ≥ 0, y ≥ 0.

Also find maximum value of Z.

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#### Solution

Inequalities | x ≤ 2 | x+y≤3 | -2x+y≤1 |

Equalities | x=2 | x+y=3 | -2x+y=1 |

Intercept form |
`x/2=1` |
`x/3+y/3=1` |
`x/(-1/2)+y/1=1` |

Origin Test | 0≤2 | 0+0≤3 | -2(0)+0≤1 |

True Origin Side | True Origin Side | True Origin Side |

Shaded portion OABC is the feasible region, Where O(0,0) A(2, 0) D(0, 1), B(2, 1)

For C :

x + y = 3

– 2x + y = 1

– – –

----------------------

3x = 2

∴ x = 2/3

2/3+y=3 i.e y=7/3

`∴ c(2/3,7/3)`

Z = 6x + 4y

Z at O(0, 0) = 6(0) + 4(0) = 0

Z at A(2, 0) = 6(2) + 4(0) = 12

Z at B(2, 1) = 6(2) + 4(1) = 16

Z at `c(2/3,7/3)=6(2/3)+(7/3)4=40/3`

Z at D(0,1) = 6(0) + 4(1) = 4 Thus, Z is maximized at B(2, 1) and its maximum value is 16.

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