Question

The comer point of the feasible region determined by the following system of linear inequalities:

2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let x = Px + qx where P, q > 0 condition on P and Q so that the maximum of z occurs at both (3, 4) and (0, 5) is

Options

• P = a

• P = 2q

• P = 3a

• Q = 3P

Answer 1

Q = 3P

Explanation:

Maximum value of z = px + qy occurs at (3, 4) and (0, 5)

At (3, 4)	z = Pz + Qy = 3P + 4Q
At (0, 5)	z = 0 + q5 = 5Q

Both are the maximum values

⇒ 3P + 4Q = 5Q or a = 3P.