Example 38 - Chapter 6 Class 12 Application of Derivatives
Last updated at March 11, 2017 by Teachoo
Transcript
Example 38 Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. Let OC = r be the radius of cone & OA = h be height of cone & OAQ = be the semi vertical angle of cone Let OE = x be the radius of cylinder Height of cylinder = OO From (1) & (2) = = = = = Now, Curved Surface Area Of cylinder = 2 Radius of cylinder Height of cylinder S = 2 S = 2 S = 2 2 S = 2 We need to minimize S, So, finding S (x) S = 2 S = 2 S = 2 Putting S = 0 0 = 2 2 = 0 = 2 Now, Finding S (x) at x = 2 S = 2 S = 2 S = 0 2 S = 2 So, S = 2 Hence, at = 2 = 2 <0 = is maxima of S. Hence, radius of cylinder with greatest curved surface area which can be inscribed in a given cone is half of that cone.
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Chapter 6 Class 12 Application of Derivatives
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.