The volume of the largest right circular cylinder is [tex]=\frac{4\pi r^3}{3\sqrt{3} }[/tex] cu. unit
Now, According to the question:
The given sphere is of radius R.
Let h be the height and r be the radius of the cylinder inscribed in the sphere.
We know that:
Volume of cylinder
V = [tex]\pi R^2h[/tex] .....(1)
In right Triangle OBA
[tex]AB^2 + OB^2 = OA^2[/tex]
[tex]R^2 + \frac{h^2}{4} = r^2[/tex]
So, [tex]R^2 = r^2 - \frac{h^2}{4}[/tex]
Putting the value of [tex]R^2[/tex] in equation (1), We get
V = [tex]\pi (r^2 - \frac{h^2}{4} )h[/tex]
V = [tex]\pi (r^2h - \frac{h^3}{4} )[/tex] ....(2)
dV/dh = [tex]\pi (r^2 - \frac{3h^2}{4} )[/tex] .....(3)
For, Stationary point, dV/dh = 0
[tex]\pi (r^2 - \frac{3h^2}{4} )[/tex] = 0
[tex](r^2 - \frac{3h}{4} )[/tex] => [tex]h^2 - \frac{4r^2}{3}[/tex] => [tex]h - \frac{2r}{\sqrt{3} }[/tex]
Now, [tex]\frac{d^2V}{dh^2} = \pi (-\frac{6}{4}h )[/tex]
[tex][\frac{d^2V}{dh^2}]_a_t_h_=_\frac{2r}{\sqrt{3} }[/tex] = x[-3/2 , [tex]2r/\sqrt{3}[/tex]]< 0
Volume is maximum at h = 2r/[tex]\sqrt{3}[/tex]
Maximum volume is :
[tex]= \pi (r^2.\frac{2r}{\sqrt{3} }- \frac{1}{4}.\frac{8r^3}{3\sqrt{3} } )[/tex]
[tex]=\pi (\frac{2r^3}{\sqrt{3} }-\frac{2r^3}{3\sqrt{3} } )[/tex]
[tex]=\pi (\frac{6r^3-2r^3}{3\sqrt{3} } )[/tex]
[tex]=\frac{4\pi r^3}{3\sqrt{3} }[/tex] cu. unit
Learn more about Volume of Cylinder at:
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