The corner points of the feasible region determined by the system of linear inequalities are (0, 0), (4, 0), (2, 4) and (0, 5). If the maximum value of z=ax+by, where a, b>0 occurs at both (2, 4) and (4, 0), then

Given: Corner points, (0, 0), (4, 0), (2, 4), (0, 5), z=ax+by, a, b>0
Solve for z where (x, y)=(0, 0)
∴z=ax+by=a(0)+b(0)=0
Similarly, for (4, 0), (2, 4) and (0, 5) respectively z=4a, z=2a+4b and z=5b
The maximum occurs at both point (2, 4) and (4, 0)
Hence the value of z at (2, 4) and (4, 0) should be same,
∴ 2a+ 4b = 4a
∴ 4b = 2a
∴ a = 2b
Therefore option (A) is correct.