A) \[2r\sqrt{pq}\]
B) \[2pq\sqrt{r}\]
C) \[-2r\sqrt{pq}\]
D) \[2rpq\]
Correct Answer: A
Solution :
| [a] Given that \[xy={{r}^{2}}\] |
| \[\Rightarrow y=\frac{{{r}^{2}}}{x}\] |
| 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 \[px+qy=p.\sqrt{\frac{q}{p}}.r+q\sqrt{\frac{p}{q}}.r\] |
| \[=\sqrt{pq}r+\sqrt{pq}\,r=2r\sqrt{pq}\] |
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