The corner points 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 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

Linear Programming — Class 12 Maths Solution

ncert exercise SA NCERT,Ex.12.2,Q.11,page.520

Question

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) $p = 2q$

(D) $q = 3p$

Step-by-step Solution

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, $3p + 4q = 5q \Rightarrow q = 3p$ .

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Linear Programming. Curated by Sachin Sharma. Free for all students.