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Application Of Derivatives

Question
CBSEENMA12035570

Find the altitude of a right circular cone of maximum curved surface which can be inscribed in a sphere of radius r.

Solution

Let R be radius and l be the slanted height of cone.
Let OD = x, OA = OB = OC = r, where r is radius of sphere.
In space rt. space angle straight d space increment ODC comma space space space space space space straight R space equals space square root of straight r squared minus straight x squared end root
In space rt. space angle straight d space increment ADC comma
space space space space space space space space space space straight l space equals space square root of straight R squared plus left parenthesis straight x plus straight r right parenthesis squared end root space equals space square root of straight r squared minus straight x squared plus straight x squared plus straight r squared plus 2 space straight r space straight x end root
space space space space space space space space space space space equals space square root of 2 straight r squared plus 2 xr end root space equals space square root of 2 straight r end root space square root of straight x plus straight r end root

Let S be curved surface area of cone. 
therefore space space space space straight S space equals space πRl space equals space straight pi square root of straight r squared minus straight x squared end root space square root of 2 straight r end root space square root of straight x plus straight r end root space equals space straight pi square root of 2 straight r end root left parenthesis straight x plus straight r right parenthesis space square root of straight r minus straight x end root
space space space space space space dS over dx space equals space straight pi square root of 2 straight r end root open square brackets left parenthesis straight x plus straight r right parenthesis space fraction numerator negative 1 over denominator 2 square root of straight r minus straight x end root end fraction plus square root of straight r minus straight x end root. space 1 close square brackets space equals straight pi space square root of 2 straight r end root space open square brackets fraction numerator negative straight x minus straight r plus 2 straight r minus 2 straight x over denominator 2 square root of straight r minus straight x end root end fraction close square brackets
space space space space space space space space space space space space space space space space space equals space straight pi square root of 2 straight r end root open square brackets fraction numerator straight r minus 3 straight x over denominator 2 square root of straight r minus straight x end root end fraction close square brackets
For S to be maximum or minimum,  dS over dx space equals space 0
therefore space space space space space straight pi square root of 2 straight r end root space open square brackets fraction numerator straight r minus 3 straight x over denominator 2 square root of straight r minus straight x end root end fraction close square brackets space equals space 0 space space space space rightwards double arrow space space space space straight x space equals space straight r over 3
When  straight x less than straight r over 3 (slightly), dS over dx space equals space plus ve
and space space When space straight x greater than straight r over 3 space left parenthesis slightly right parenthesis comma space ds over dx space equals space minus ve
therefore space space space at space straight x space equals space straight r over 3 comma space dS over dx space changes space from space plus ve space to space minus ve
therefore space space space space straight S space is space maximum space at space straight x space equals space straight r over 3
therefore space space space altitude space AD space equals space straight r over 3 plus straight r space equals space fraction numerator 4 straight r over denominator 3 end fraction

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