The corner points of the bounded feasible region for a linear programming problem (LPP) are (0, 3/2), (1, 2) and (4, 0). If the objective function is $Z = ax + by$, where 'a' and 'b' are positive, then the condition on 'a' and 'b' so that the maximum of Z occurs at (1, 2) and (4, 0) is:

$3a=2b$

The correct answer is Option (3) → $3a=2b$

Corner points: $(0,\frac{3}{2}),\ (1,2),\ (4,0)$

Objective function: $Z=ax+by$

At $(1,2):\ Z=a+2b$

At $(4,0):\ Z=4a$

At $(0,\frac{3}{2}):\ Z=\frac{3}{2}b$

For maximum to occur at both $(1,2)$ and $(4,0)$, values of $Z$ at these two points must be equal and greater than or equal to value at $(0,\frac{3}{2})$.

Condition of equality:

$a+2b=4a \ \Rightarrow\ 3a=2b \ \Rightarrow\ \frac{a}{b}=\frac{2}{3}$