A) \[\frac{3a}{2}\]
B) \[\frac{\sqrt{2}a}{3}\]
C) \[\frac{a}{\sqrt{3}}\]
D) \[\frac{2}{\sqrt{3a}}\]
Correct Answer: D
Solution :
Let a be the raidius and \[h\]the height from figure \[{{r}^{2}}+\frac{{{h}^{2}}}{4}={{a}^{2}}\] \[\therefore \] \[{{h}^{2}}=4({{a}^{2}}-{{r}^{2}})\] Now \[\upsilon =\pi {{r}^{2}}h=\pi \left( {{a}^{2}}-\frac{{{h}^{2}}}{4} \right)h\] \[=\pi \left( {{a}^{2}}h-\frac{{{h}^{2}}}{4} \right)\] \[\therefore \] \[\frac{d\upsilon }{dh}=\pi \left( {{a}^{2}}-\frac{3{{h}^{2}}}{4} \right)=0\] for maximum or minimum \[\Rightarrow \]\[h=2/\sqrt{3}a\Rightarrow \frac{{{d}^{2}}\upsilon }{d{{h}^{2}}}=-\frac{6h}{4}<0\] \[\therefore \]\[\upsilon \]is maximum when \[h=\frac{2}{\sqrt{3}a}\]You need to login to perform this action.
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