Answer :
The volume of the largest cylinder inscribed in a sphere of radius r cm
= [tex]\frac{4\pi r^{3} }{3\sqrt{3} }[/tex]
Radius of sphere.
- The radius of a sphere is the shortest distance from its center to any point on its surface.
- It is half the length of the diameter of the sphere.
- The radius, being a measure of length or distance is expressed in linear units such as mm, cm, m, in, or ft.
Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.
From the drawn diagram , we know that,
[tex]R^{2} + (\frac{h}{2} )^{2} = r^{2}[/tex]
∴ [tex]R^2 = r^2 - \frac{h^2}{4}[/tex] ⇒ (1)
Let V be the volume of the cylinder
Then V = π[tex]R^2[/tex] h
= [tex]\pi (r^2 - \frac{h^2}{4} ) h[/tex] [substitute [tex]R^2[/tex] value ]
= [tex]\pi (r^2 h - \frac{h^3}{4} )[/tex]
[tex]\frac{dV}{dh} = \pi \frac{d}{dh} (r^2h - \frac{h^3}{4} )[/tex]
= [tex]\pi (r^2 * 1 - \frac{1}{4} * 3h^2 )[/tex]
= [tex]\pi (r^2 - \frac{3}{4} h^2 )[/tex]
[tex]\frac{d^2V}{dh^2}[/tex] = [tex]\pi \frac{d}{dh}(r^2 - \frac{3}{4}h^2 )[/tex]
= -[tex]\frac{3}{2} \pi h[/tex]
Now , [tex]\frac{dV}{dh}[/tex] = 0 , [tex]\pi (r^2 - \frac{3}{4} h^2 )[/tex] = 0
By solving we get
[tex]h^2 = \frac{4r^2}{3}[/tex]
h = [tex]\frac{2r}{\sqrt{3} }[/tex] [ h > 0]
[tex]\frac{d^2V}{dh^2}[/tex] at h= [tex]\frac{2r}{\sqrt{3} }[/tex] = [tex]-\frac{3}{2} \pi * \frac{2r}{\sqrt{3} } < 0[/tex]
∴ V is maximum at h = [tex]\frac{2r}{\sqrt{3} }[/tex]
If h = [tex]\frac{2r}{\sqrt{3} }[/tex] , from (1) we get
[tex]R^2 = r^2 - \frac{1}{4} *\frac{4r^2}{3}[/tex]
= [tex]\frac{2r^2}{3}[/tex]
Therefore, volume of the largest cylinder
= [tex]\frac{4\pi r^3}{3\sqrt{3} }[/tex] cu cm.
Hence, the volume of the largest cylinder inscribed in a sphere of radius ‘r’ cm = [tex]\frac{4\pi r^3}{3\sqrt{3} }[/tex] cu cm.
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