Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let z = px = qy, where p, q 0. Condition on p and q so that the minimum of z occurs at (3, 0) and (1, 1) is

Question

Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let z = px = qy, where p, q > 0. Condition on p and q so that the minimum of z occurs at (3, 0) and (1, 1) is (A) p = 2q (B) p = (q/2) (C) p = 3q (D) p = q

Tardigrade Admin · Accepted Answer

Given corner points are (0,3),(1,1),(3,0) z=p x+q y At (3,0), z=3 p At (1,1), z=p+q It is given that the minimum of z occurs at (3,0) and (1,1) ⇒ 3 p=p+q ⇒ 2 p = q ⇒ p = q / 2

Q. Corner points of the feasible region determined by the system of linear constraints are $(0, 3), (1, 1)$ and $(3, 0)$ . Let $z = p x = q y$ , where $p, q > 0$ . Condition on $p$ and $q$ so that the minimum of $z$ occurs at $(3, 0)$ and $(1, 1)$ is

$p = 2 q$

$p = \frac{q}{2}$

$p = 3 q$

$p = q$

Given corner points are $(0, 3), (1, 1), (3, 0)$
$z = p x + q y$
At $(3, 0), z = 3 p$
At $(1, 1), z = p + q$
It is given that the minimum of $z$ occurs at $(3, 0)$ and $(1, 1)$
$\Rightarrow 3 p = p + q$
$\Rightarrow 2 p = q$
$\Rightarrow p = q /2$

Q. Corner points of the feasible region determined by the system of linear constraints are (0,3),(1,1) and (3,0). Let z=px=qy, where p,q>0. Condition on p and q so that the minimum of z occurs at (3,0) and (1,1) is

p=2q

p=2q​

p=3q

p=q