A) \[2r\sqrt{pq}\]
B) \[2pq\sqrt{r}\]
C) \[-2r\sqrt{pq}\]
D) None of these
Correct Answer: A
Solution :
Let\[z=px+qy\]and given\[xy={{r}^{2}}\] \[\Rightarrow \] \[y=\frac{{{r}^{2}}}{x}\] \[\therefore \] \[z=px+q\frac{{{r}^{2}}}{x}\] \[\frac{dz}{dx}=p-\frac{q{{r}^{2}}}{{{x}^{2}}}\] For maxima and minima, \[\frac{dz}{dx}=0\] \[\Rightarrow \] \[p-\frac{q{{r}^{2}}}{{{x}^{2}}}=0\] \[\Rightarrow \] \[p{{x}^{2}}=q{{r}^{2}}\] \[\Rightarrow \] \[x=\sqrt{\frac{q{{r}^{2}}}{p}}\] \[\Rightarrow \] \[x=\sqrt{\frac{q}{p}}.r\] \[\therefore \]At \[x=\sqrt{\frac{p}{q}}.r,\frac{{{d}^{2}}z}{d{{x}^{2}}}>0\,z\]is minimum \[\therefore \] \[{{z}_{\min }}=p\sqrt{\frac{p}{q}}.r+\frac{q{{r}^{2}}}{r}.\sqrt{\frac{p}{q}}\] \[=\frac{pqr+qrp}{\sqrt{pq}}=2r\sqrt{pq}\]
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