Question

The minimum value of $(px+qy)$ when $xy=n^2$is equal to

• A. $2n\sqrt{pq}$
• B. $2pq\sqrt{n}$
• C. $2\sqrt{npq}$
• D. $2pqn$

Answer 1

The correct option is A $2n\sqrt{pq}$

$f\left(x\right)=px+\frac{q{n}^2}{x}$
$f^{\prime }\left(x\right)=p-\frac{q{n}^2}{x^2}$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow x=\pm \sqrt{\frac{q}{p}}n$
$f^{\prime \prime }\left(x\right)=\frac{2q{n}^2}{x^3}$
$f^{\prime \prime }\left(x\right)>0$ at $x=\sqrt{\frac{q}{p}}n$
Hence f(x) has a minimum at $x=\sqrt{\frac{q}{p}}n$
$f\left(\sqrt{\frac{q}{p}}n\right)=2n\sqrt{pq}$