Points |
Value of z |
(0, 0) |
0 |
(0, 4, 0) |
120 |
(20, 40) |
200 |
(60, 20) |
300 |
(60, 0) |
240 |
Given, z = 4x + 3y
Given feasible points are;
(0, 0), (0, 40), (20, 40), (60, 20), (60, 0)
z(0, 0) = 4(0) + 3(0) = 0
z(0, 40) = 4(0) + 3(40) = 120
z(20, 40) = 4(20) + 3(40) = 80 + 120 = 200
z(60, 20) = 4(60) + 3(20)
= 240 + 60
= 300
z(60, 0) = 4(60) + 3(0)
z(60, 0) = 240
∴ max z = 300
i.e. column A is 300 and column B is 325
∴ The quantity of B is greater than A.
Hence, the correct answer is option B.