**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.

#### Solution

The draw the feasible region, construct table as follows:

Inequality | 3x + 5y ≤ 26 | 5x + 3y ≤ 26 |

Corresponding equation (of line) | 3x + 5y 26 | 5x + 3y = 30 |

Intersection of line with X-axis | `(26/3, 0)` | (6, 0) |

Intersection of line withY-axis | `(0, 26/5)` | (0, 10) |

Region | Origin side | Origin side |

Shaded portion OABC is the feasible region,

whose vertices are O(0, 0), A(6, 0), B and C `(0, 26/5)`.

B is the point of intersection of the lines 5x + 3y = 30 and 3x + 5y = 26.

Solving the above equations, we get

x = `(9)/(2), y = (5)/(2)`

∴ B = `(9/2 , 5/2)` ≡ (4.5, 2.5)

Here, the objective function is Z = 7x + 11y

∴ Z at O(0, 0) = 7(0) + 11(0) = 0

Z at A(6, 0) = 7(6) + 11(0) = 42

Z at B `(9/2, 5/2) = 7(9/2) + 11(5/2) = (63 + 55)/(2)` = 59

Z at C`(0, 26/5) = 7(0) + 11(26/5) (286)/(5)` = 57.2

∴ Z has maximum value 59 at B`(9/2, 5/2)`.

i.e. at B(4.5, 2.5)

∴ Z is maximum, when `x = (9)/(2) "and" y = (5)/(2)`

i.e. when x = 4.5 and y = 2.5