RIGHTARROW y=\FRAC{{{r}^{2}}}{x}\] LET \[S=px+qy=px+\frac{q{{r}^{2}}}{x}\] \[\Rightarrow \frac{dS}{dx}=p-\frac{q{{r}^{2}}}{{{x}^{2}}}\] \[\frac{dS}{dx}=0\] for maximum or minimum. So, \[0=p-\frac{q{{r}^{2}}}{{{x}^{2}}}\] \[\Rightarrow {{x}^{2}}=\frac{q{{r}^{2}}}{p}\Rightarrow x=\pm \sqrt{\frac{q}{p}}.r\] Now, \[\frac{{{d}^{2}}S}{d{{x}^{2}}}=\frac{2q{{r}^{2}}}{{{x}^{3}}}\] At \[x=+\sqrt{\frac{q}{p}}.r\frac{{{d}^{2}}S}{d{{x}^{2}}}>0\] Hence, S is minimum at \[x=\sqrt{\frac{q}{p}}.r\] \[\Rightarrow y=\frac{{{r}^{2}}}{\sqrt{\frac{q}{p}}.r}=\sqrt{\frac{p}{q}}.r\] Minimum value of \[px+qy=p.\sqrt{\frac{q}{p}}.r+q\sqrt{\frac{p}{q}}.r\] \[=\sqrt{PQ}r+\sqrt{pq}\,r=2r\sqrt{pq}\]