KCET · Maths · Probability
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
- A \(p=2 q\)
- B \(p=\frac{q}{2}\)
- C \(p=3 q\)
- D \(p=q\)
Answer & Solution
Correct Answer
(B) \(p=\frac{q}{2}\)
Step-by-step Solution
Detailed explanation
The minimum value of \(z\) is unique.
It is given that the minimum value of \(z\) occurs at two points \((3,0)\) and \((1,1)\)
\(\begin{array}{lrl}
\Rightarrow & p(3)+q(0) & =p(1)+q(\mathbf{l}) \\
\Rightarrow & & 3 p=p+q \\
\Rightarrow & & 2 p=q \\
\Rightarrow & & p=q / 2
\end{array}\)
It is given that the minimum value of \(z\) occurs at two points \((3,0)\) and \((1,1)\)
\(\begin{array}{lrl}
\Rightarrow & p(3)+q(0) & =p(1)+q(\mathbf{l}) \\
\Rightarrow & & 3 p=p+q \\
\Rightarrow & & 2 p=q \\
\Rightarrow & & p=q / 2
\end{array}\)
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