Question 51 of 212MediumMCQ
If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is :
Solutions · 1
Official solution
Sphere of radius r = 3 cm
Let b, h be base radius and height of cone respectively.
So, volume of cone = \[{1 \over 2}\] \[\pi \]b2h
In right \[\Delta \]ABC by Pythagoras theorem
(h - r)2 + b2 = r2
\[ \Rightarrow \] b2 = r2 - (h - r)2 = r2 - (h2 - 2hr + r2) = 2hr - h2
\[\therefore\,\,\,\] Volume (v) = \[{1 \over 3}\] \[\pi \]h[2hr - h2] = \[{1 \over 3}\] [ 2h2r - h3]
\[{{dv} \over {dh}}\] = \[{1 \over 3}\] [4hr - 3h2] = 0
\[ \Rightarrow \] h (4r - 3h) = 0
\[{{{d^2}v} \over {d{h^2}}}\] = \[{1 \over 3}\] [4r - 6h]
At h = \[{{4r} \over 3}\], \[{{{d^2}y} \over {d{h^2}}}\] = \[{1 \over 3}\] \[\left[ {4r - {{4r} \over 3} \times 6} \right] = {1 \over 3}\left[ {4r - 8r} \right] < 0\]
\[ \Rightarrow \] maximum volume cours at h = \[{{4r} \over 3}\] = \[{4 \over 3}\] \[ \times \] 3 = 4 cm
As from (1),
(h - r)2 + b2 = r2
\[ \Rightarrow \] b2 = 2hr - h2 = 2.\[{{4r} \over 3}\] r - \[{{16{r^2}} \over 9}\] = \[{{8{r^2}} \over 3}\] - \[{{16{r^2}} \over 9}\]
= \[{{\left( {24 - 16} \right){r^2}} \over 9}\] = \[{{8{r^2}} \over 9}\]
\[ \Rightarrow \] b = \[{{2\sqrt 2 } \over 3}\] r = 2 \[\sqrt 2 \,\,m\]
Therefore curved surface area = \[\pi bl\]
= \[\pi \]b\[\sqrt {{h^2} + {r^2}} \] = \[\pi \]2\[\sqrt 2 \] \[\sqrt {{4^2} + 8} \] = 8\[\sqrt 3 \] π Answer the question to unlock