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Question
A cone is inscribed in a sphere of radius 12 cm. If the volume of the cone is maximum, find its height
Solution
Let VAB be a cone of greatest volume inscribed in a sphere of radius 12. It is obvious that for
maximum volume the axis of the cone must be along a diameter of the sphere. Let VC be the axis of the cone and O be the centre of the sphere such that OC = x.
Then,
VC = VO + OC = R + x = (12 + x) = height of cone
Applying Pythagoras theorem,
`OA^2 = AC^2 + OC^2`
`AC^2= 12^2 - x^2`
`AC^2 = 144 - x^2`
Let V be the volume of the cone, then
`V = 1/3 pi (AC)^2 (VC)`
`= 1/3pi (144 - x^2) (12 + x)`
`= 1/3 pi [1728 + 144x - 12x^2 - x^3]` .....(i)
`(dV)/dx = 1/3pi [144 - 24x - 3x^2]`
`(d^2V)/(dx^2) = 1/3 pi[-24-6x] = 1/3pi (-6)^2 [4+x]= -2pi(4+x)`
Now, `(dV)/(dx) = 0` given `1/3pi [144 - 24x - 3x^2] = 0`
`i.e. 144 - 24 x - 3x^2 = 0`
i.e `x^2 + 8x - 48 = 0`
i.e (x+12)(x-4) = 0
i.e x = -12 or x = 4
`[(d^2V)/(dx^2)]_(x= 4) = -2pi(4+4) = -16pi <0`
Thus V is maximum when x = 4
Putting x = 4 in (1) we obtain
∴ Height of cone = x + R = 4 + 12 = 16 cm
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