**Choose the correct alternative :**

The maximum value of z = 10x + 6y, subjected to the constraints 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, y ≥ 0 is.

#### Options

56

`65

55

66

#### Solution

Z = 10x + 6y

The given inequalities are 3x + y ≤ 12 and 2x + 5y ≤ 34.

Consider line L_{1} and L_{2} where

L_{1} : 3x + y = 12, L_{2} : 2x + 5y = 34

For line L_{1}, plot A (0, 12) and B (4, 0)

For line L_{2}, plot P (0, 6.8) and Q (17, 0)

Solving both lines, we get x = 2, y = 6.

The coordinates of origin O (0, 0) satisfies both the inequalities.

∴ The required region is on the origin side of both the lines L_{1} and L_{2}.

As x ≥ 0, y ≥ 0, the feasible region is in the 1^{st} quadrant.

OBRPO is the required feasible region.

At O (0, 0), Z = 0

At B (4, 0), Z = 10 (4) + 0 = 40

At R (2, 6), Z = 10 (2) + 6 (6) = 56

At P (0, 6.8), Z = 0 + 6 (6.8) = 40.8

The maximum value of Z is **56** and it occurs at R (2, 6).