Question

The corner points of the feasible region are A (50, 50), B (10, 50), C (60, 0) and D (60, 40). The objective function is P = 5/2 x + 3/2 y + 410. The minimum value of P is

a) 510
b) 620
c) 610
d)560
​

Answer 1

$\huge\boxed{Answer}$

Let z0 be the maximum value of z in the feasible region.

Since maximum occurs at both (15,15) and (0,20)$, the value z0 is attained at both (15,15) and (0,20).

⟹z0=p(15)+q(15) and z0

=p(0)+q(20)

⟹p(15)+q(15)=p(0)+q(20)

⟹15p=5q

⟹3p=q

Answer 2

this is right answer 510