The maximum value of the function z = 3x + 3y, subject to the constraints x + 2y ≤ 30, 2x + y ≤ 50, x ≥ 0, y ≥ 0 is :
Answer & explanation
Correct answer: option 3
x + 2y = 30
|
x |
0 |
30 |
|
y |
15 |
0 |
2x + y = 50
|
x |
0 |
25 |
|
y |
50 |
0 |
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
