A) \[2r\sqrt{pq}\]
B) \[2pq\sqrt{r}\]
C) \[-2r\sqrt{pq}\]
D) \[\sqrt{pq}\]
Correct Answer: A
Solution :
| \[f(x)=px+qy\Rightarrow f(x)=px+\frac{q{{r}^{2}}}{x}\] |
| \[f'(x)=P-\frac{q{{r}^{2}}}{x}=0\Rightarrow x=\pm r\sqrt{\frac{q}{p}}\] |
| \[f''(x)>0\operatorname{for}\,x=r\sqrt{\frac{q}{p}}\] |
| \[\therefore \]\[\,\,f{{(x)}_{\min }}=pr\sqrt{\frac{q}{p}}+\frac{q{{r}^{2}}}{r\frac{q}{p}}\] |
| \[=\sqrt{pq}.r+\sqrt{pq}.r=2r\sqrt{pq}\] |
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