The maximum value of the function z = 3x + 3y, subject to the constraints x + 2y ≤ 30, 2x + y ≤ 50, x ≥ 0, y ≥ 0 is :

80

x + 2y = 30

2x + y = 50

So points of feasible region are 

O(0, 0), A(25, 0), B(0, 15) and $C(\frac{70}{3},\frac{10}{3})$

Value of Z = 3x + 3y

O = 3 × 0 + 3 × 0 = 0       [0, 0]

A = 3 × 25 + 3 × 0 = 75   [25, 0]

B = 3 × 0 + 3 × 15 = 45   [0, 15]

C = $3 ×\frac{70}{3} + 3 ×\frac{10}{3} = 80$  $(\frac{70}{3},\frac{10}{3})$

So, the maximum value of Z = 80