Question

The minimum value of $px+qy$ when $xy=r^2$ is

• A. $2r\sqrt{pq}$
• B. $r\sqrt{pq}$
• C. $r\sqrt{q/p}$
• D. None of the above

Answer 1

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

Let $A=px+qy$ where $xy=r^2$

$\Rightarrow A=px+q\left[\frac{r^2}{x}\right]=px+q{r}^2.\frac{1}{x}$

Differentiate both side, w.r.t.$x$

$\frac{dA}{dx}=p+q{r}^2\left(\frac{-1}{x^2}\right)$

Now, for maxima or minima,

Put $\frac{dA}{dx}=0$

$\Rightarrow p-\frac{q{r}^2}{x^2}=0$

$\Rightarrow x^2=\frac{q{r}^2}{p}$

$\Rightarrow x=\sqrt{\frac{q}{p}}r$

Now,$\frac{d^{2}A}{d{x}^2}=\frac{2q{r}^2}{x^3}>0$

$\therefore A=px+qy$ has minima and minimum value of

$A=pr\sqrt{\frac{q}{p}}+q{r}^2\frac{\sqrt{p}}{r\sqrt{q}}$

when $x=r\sqrt{\frac{q}{p}}$

minimum value of $A=r\sqrt{pq}+r\sqrt{pq}=2r\sqrt{pq}$