# Ex 6.5,25 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 6.5,25 Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan –1 2 Let 𝑙 be the slant height & θ be the semi vertical angle of the cone. Now, Height of cone = h = 𝑙 cos θ Radius of cone = r = 𝑙 sin θ We need to maximize volume of cone V = 13 𝜋 𝑟2ℎ V = 13𝜋 𝑙2sin2𝜃 l cos 𝜃 V= 13𝜋 𝑙3sin2𝜃 cos 𝜃 Differentiating 𝑑𝑣𝑑𝜃 = 13𝜋 𝑙3 (2 sin 𝜃 cos 𝜃 . cos 𝜃 + sin2 𝜃 (sin 𝜃)) 𝑑𝑣𝑑𝜃 = 13𝜋 𝑙3 2 sin𝜃 cos2𝜃− sin3𝜃 𝑑𝑣𝑑𝜃 = 13𝜋 𝑙3 (2 sin 𝜃 cos2 𝜃 sin3 𝜃) 𝑑𝑣𝑑𝜃 = 13 𝜋 𝑙3 sin 𝜃 (2 cos2 𝜃 − sin2 𝜃) 𝑑𝑣𝑑𝜃 = 13 𝜋 𝑙3 sin 𝜃 ( 2 cos𝜃+ sin𝜃)( 2 cos 𝜃 − sin 𝜃) 𝑑𝑣𝑑𝜃 = 13 𝜋 𝑙3 sin 𝜃 cos 𝜃 ( 2 cos𝜃+ sin𝜃) cos𝜃 × cos 𝜃 ( 2 cos𝜃− sin𝜃) cos𝜃 𝑑𝑣𝑑𝜃 = 13 𝜋 𝑙3 sin 𝜃 cos2 𝜃 ( 2+ tan 𝜃 )( 2 − tan 𝜃) Putting 𝑑𝑣𝑑𝜃 = 0 13 𝜋 𝑙3 sin 𝜃 cos2 𝜃 ( 2+ tan 𝜃 )( 2 − tan 𝜃) = 0 sin 𝜃 cos2 𝜃 ( 2+ tan 𝜃 )( 2 − tan 𝜃) = 0 The value is either maxima or minima. Finding maxima or minima by first derivative test 𝑑𝑣𝑑𝜃 = 13𝜋 𝑙3 sin 𝜃 cos2 𝜃 ( 2 + tan𝜃) ( 2 − tan 𝜃) 𝑑𝑣𝑑𝜃 = 13𝜋 𝑙3 sin 𝜃 cos2 𝜃 ( 2 + tan𝜃) ( 2 − tan 𝜃) Since 𝑑𝑣𝑑θ = changes sign from (+) ve to (−) ve 𝜽 = tan− 1 𝟐 is the maxima. Hence proved

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Ex 6.5,25 You are here

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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.