The corner points of the feasible region determined by the following system of linear inequalities:
$2x + y \le 10,x + 3y \le 15,x,y \ge 0\;are\left( {0,0} \right),\left( {5,0} \right),\left( {3,4} \right)$ and (0, 5). Let $Z = px + qy,$ where $p,q > 0$. Condition on p and q so that maximum of Z occurs at both (3, 4) and (0, 5) is
(A) $p = q$
(B) $p = 2q$
(C) $P = 3q$
(D) $q = 3p$
The corner points of the feasible region determined by the following system of linear inequalities:
$2x + y \le 10,x + 3y \le 15,x,y \ge 0\;are\left( {0,0} \right),\left( {5,0} \right),\left( {3,4} \right)$ and (0, 5). Let $Z = px + qy,$ where $p,q > 0$. Condition on p and q so that maximum of Z occurs at both (3, 4) and (0, 5) is
(A) $p = q$
(B) $p = 2q$
(C) $P = 3q$
(D) $q = 3p$
Official Solution
VVidaara Team
✓ Verified solution
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Option D is correct
${Z_{\max }}$ occurs at (3, 4) and (0, 5).
At (3, 4), $Z = px + qy = 3p + 4q.$ At
$(0,5),Z = 0 + q(5) = 5q$
Since both are the maximum values, $3p + 4q = 5q \Rightarrow q = 3p$ .
Miscellaneous Exercise
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