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

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 = 1﷮3﷯ 𝜋 𝑟﷮2﷯ℎ V = 1﷮3﷯𝜋 𝑙﷮2﷯sin2𝜃 l cos 𝜃 V= 1﷮3﷯𝜋 𝑙﷮3﷯sin2𝜃 cos 𝜃 Differentiating 𝑑𝑣﷮𝑑𝜃﷯ = 1﷮3﷯𝜋 𝑙﷮3﷯ (2 sin 𝜃 cos 𝜃 . cos 𝜃 + sin2 𝜃 (sin 𝜃)) 𝑑𝑣﷮𝑑𝜃﷯ = 1﷮3﷯𝜋 𝑙﷮3﷯ 2 sin﷮𝜃﷯ cos﷮2﷯﷮𝜃﷯− sin﷮3﷯﷮𝜃﷯﷯ 𝑑𝑣﷮𝑑𝜃﷯ = 1﷮3﷯𝜋 𝑙﷮3﷯ (2 sin 𝜃 cos2 𝜃 sin3 𝜃) 𝑑𝑣﷮𝑑𝜃﷯ = 1﷮3﷯ 𝜋 𝑙﷮3﷯ sin 𝜃 (2 cos2 𝜃 − sin2 𝜃) 𝑑𝑣﷮𝑑𝜃﷯ = 1﷮3﷯ 𝜋 𝑙﷮3﷯ sin 𝜃 ( ﷮2﷯ cos﷮𝜃+ sin﷮𝜃)(﷯﷯ ﷮2﷯ cos 𝜃 − sin 𝜃) 𝑑𝑣﷮𝑑𝜃﷯ = 1﷮3﷯ 𝜋 𝑙﷮3﷯ sin 𝜃 cos 𝜃 ( ﷮2﷯ cos﷮𝜃+ sin﷮𝜃﷯﷯)﷮ cos﷮𝜃﷯﷯ × cos 𝜃 ( ﷮2﷯ cos﷮𝜃− sin﷮𝜃﷯﷯)﷮ cos﷮𝜃﷯﷯ 𝑑𝑣﷮𝑑𝜃﷯ = 1﷮3﷯ 𝜋 𝑙﷮3﷯ sin 𝜃 cos2 𝜃 ( ﷮2﷯﷮+ tan 𝜃﷮ )(﷯﷯ ﷮2﷯ − tan 𝜃) Putting 𝑑𝑣﷮𝑑𝜃﷯ = 0 1﷮3﷯ 𝜋 𝑙﷮3﷯ 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 𝑑𝑣﷮𝑑𝜃﷯ = 1﷮3﷯𝜋 𝑙﷮3﷯ sin 𝜃 cos2 𝜃 ( ﷮2﷯ + tan𝜃) ( ﷮2﷯ − tan 𝜃) 𝑑𝑣﷮𝑑𝜃﷯ = 1﷮3﷯𝜋 𝑙﷮3﷯ sin 𝜃 cos2 𝜃 ( ﷮2﷯ + tan𝜃) ( ﷮2﷯ − tan 𝜃) Since 𝑑𝑣﷮𝑑θ﷯ = changes sign from (+) ve to (−) ve 𝜽 = tan− 1 ﷮𝟐﷯ is the maxima. Hence proved