Ex 6.5,24 - Chapter 6 Class 12 Application of Derivatives

Transcript

Ex 6.5,24 Show that the right circular cone of least curved surface and given volume has an altitude equal to ﷮2﷯ time the radius of the base Let r & h be the radius & height of a cone respectively And V & S be the volume & curved surface area of cone respectively Given volume of cone is constant Volume of cone = 1﷮3﷯𝜋 𝑟𝑎𝑑𝑖𝑢𝑠﷯﷮2﷯ ℎ𝑒𝑖𝑔ℎ𝑡﷯ V = 1﷮3﷯𝜋 𝑟﷯﷮2﷯ℎ 3𝑉﷮𝜋﷯= 𝑟﷮2﷯ℎ ℎ= 3𝑉﷮𝜋﷯﷯ 1﷮ 𝑟﷮2﷯﷯ ℎ= 𝐾﷮ 𝑟﷮2﷯﷯ We need to show Curved surface Area is least & Altitude = ﷮2﷯ times of radius of base i.e. ℎ= ﷮2﷯ 𝑟 Curved Surface Area of Cone = π 𝑟𝑙 where 𝑙 = ﷮ ℎ﷯﷮2﷯+ 𝑟﷮2﷯﷯ Now, S = π 𝑟 ﷮ ℎ﷮2﷯+ 𝑟﷮2﷯﷯ S = π 𝑟 ﷮ 𝑘﷮ 𝑟﷮2﷯﷯﷯﷮2﷯+ 𝑟﷮2﷯﷯ S = π 𝑟 ﷮ 𝑘﷮2﷯﷮ 𝑟﷮4﷯﷯+ 𝑟﷮2﷯﷯ S = π 𝑟 ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷮ 𝑟﷮4﷯﷯﷯ S = π 𝑟﷮ 𝑟﷮2﷯﷯ ﷮ 𝑘﷮2﷯+ 𝑟﷮6﷯﷯ S = 𝜋﷮𝑟﷯ ﷮ 𝑘﷮2﷯+ 𝑟﷮6﷯﷯ S = π ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯ ﷮𝑟﷯﷯ Diff w.r.t r 𝑑𝑠﷮𝑑𝑟﷯=𝜋 𝑑﷮𝑑𝑟﷯ ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯ ﷮𝑟﷯﷯ =𝜋 ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷯﷮′﷯ 𝑟﷯ − 𝑟﷯﷮′﷯ ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷯﷮ 𝑟﷮2﷯﷯﷯ =𝜋 𝑟﷮2 ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷯ × 𝑑 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷮ 𝑑𝑟﷯ . 𝑟 −1 . ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷮ 𝑟﷮2﷯﷯﷯ =𝜋 𝑟﷮2 ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷯ . 0 + 6 𝑟﷮5﷯﷯ − ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷮ 𝑟﷮2﷯﷯﷯ =𝜋 6 𝑟﷮6﷯﷮2 ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷯ − ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷮ 𝑟﷮2﷯﷯﷯ =𝜋 6 𝑟﷮6﷯ − 2 ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷯﷮2﷯﷮2 ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯ × 𝑟﷮2﷯﷯﷯ =𝜋 6 𝑟﷮6﷯− 2 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷮2 𝑟﷮2﷯ ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷯﷯ =𝜋 6 𝑟﷮6﷯− 2 𝑘﷮2﷯− 2 𝑟﷮6﷯﷮2 𝑟﷮2﷯ ﷮ 𝑘﷮2﷯+ 𝑟﷮6﷯﷯﷯﷯ =𝜋 4 𝑟﷮6﷯− 2 𝑘﷮2﷯﷮2 𝑟﷮2﷯ ﷮ 𝑘﷮2﷯ + 𝑟﷮6﷯﷯﷯﷯ =𝜋 2 2 𝑟﷮6﷯− 𝑘﷮2﷯﷯﷮2 𝑟﷮2﷯ ﷮ 𝑘﷮2﷯+ 𝑟﷮2﷯﷯ ﷯﷯ =𝜋 2 𝑟﷮6﷯− 𝑘﷮2﷯﷯﷮ 𝑟﷮2﷯ ﷮ 𝑘﷮2﷯+ 𝑟﷮2﷯﷯ ﷯﷯ Putting 𝑑𝑠﷮𝑑𝑟﷯=0 𝜋 2 𝑟﷮6﷯− 𝑘﷮2﷯﷮ 𝑟﷮2﷯ ﷮ 𝑘﷮2﷯+ 𝑟﷮2﷯﷯﷯﷯=0 2 𝑟﷮6﷯− 𝑘﷮2﷯=0 ×𝜋 × 𝑟﷮2﷯ ﷮ 𝑘﷮2﷯+ 𝑟﷮2﷯﷯ 2 𝑟﷮6﷯− 𝑘﷮2﷯=0 𝑟﷮6﷯= 𝑘﷮2﷯﷮2﷯ Now, Finding 𝑑﷮2﷯𝑠﷮𝑑 𝑟﷮2﷯﷯ 𝑑𝑠﷮𝑑𝑟﷯=𝜋 2 𝑟﷮6﷯− 𝑘﷮2﷯﷯﷮ 𝑟﷮2﷯ ﷮ 𝑘﷮2﷯+ 𝑟﷮6﷯﷯﷯﷯ 𝑑𝑠﷮𝑑𝑟﷯=𝜋 2 𝑟﷮6﷯− 𝑘﷮2﷯﷮ ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷯﷯ Diff Again w.r.t 𝑟 𝑑﷮2﷯𝑠﷮𝑑 𝑟﷮2﷯﷯=𝜋 𝑑﷮𝑑𝑟﷯ 2 𝑟﷮6﷯− 𝑘﷮2﷯﷯ . ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷯ − 𝑑﷮ 𝑑𝑟﷯ ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯ . 2 𝑟﷮6﷯− 𝑘﷮2﷯﷯﷯ ﷮ ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷯﷮2﷯﷯﷯ =𝜋 12 𝑟﷮5﷯+ 0﷯ ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯ − 1﷮2 ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯+ 𝑟﷮10﷯﷯﷯ 𝑑﷮𝑑𝑥﷯ 𝑘﷮2﷯ 𝑟﷮4﷯+ 𝑟﷮10﷯﷯ 2 𝑟﷮6﷯ − 𝑘﷮2﷯﷯﷮ ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷯﷮2﷯﷯﷯ =𝜋 12 𝑟﷮5﷯ ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯ − 1﷮2 ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷯ 4 𝑘﷮2﷯ 𝑟﷮3﷯ + 10 𝑟﷮9﷯﷯ 2 𝑟﷮6﷯ − 𝑘﷮2﷯﷯﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷯﷯ =𝜋 12 𝑟﷮5﷯ ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯ − 4 𝑘﷮2﷯ 𝑟﷮3﷯ + 10 𝑟﷮9﷯﷯ 2 𝑟﷮6﷯ − 𝑘﷮2﷯﷯﷮2 ﷮ 𝑘﷮2﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷯﷯ Putting 𝑘﷮2﷯=2 𝑟﷮6﷯ =𝜋 12 𝑟﷮5﷯ ﷮ 2 𝑟﷮6﷯﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷮ 2 𝑟﷮6﷯﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯ − 4 2 𝑟﷮6﷯﷯ 𝑟﷮3﷯ + 10 𝑟﷮9﷯﷯ 2 𝑟﷮6﷯ − 2 𝑟﷮6﷯﷯﷯﷮2 ﷮ 2 𝑟﷮6﷯﷯ 𝑟﷮4﷯ + 𝑟﷮10﷯﷯﷯﷯ = 𝜋 12 𝑟﷮5﷯ ﷮2 𝑟﷮10﷯ + 𝑟﷮10﷯﷯﷮2 2 𝑟﷮10﷯ + 𝑟﷮10﷯﷯﷯ −0﷯ = 𝜋 12 𝑟﷮5﷯ ﷮3 𝑟﷮10﷯﷯﷮2 3 𝑟﷮10﷯﷯﷯ −0﷯ = 6𝜋𝑟﷮5﷯﷮ ﷮3 𝑟﷮10﷯﷯﷯ > 0 ⇒ 𝑑﷮2﷯𝑠﷮𝑑 𝑠﷮2﷯﷯>0 when 𝑘﷮2﷯=2 𝑟﷮6﷯ ⇒ S is minimum when 𝑘﷮2﷯=2 𝑟﷮6﷯ Now, ℎ= 𝑘﷮ 𝑟﷮2﷯﷯ ℎ 𝑟﷮2﷯=𝑘 ℎ 𝑟﷮2﷯﷯﷮2﷯= 𝑘﷮2﷯ ℎ﷮2﷯ 𝑟﷮4﷯= 𝑘﷮2﷯ Putting value of 𝑘﷮2﷯=2 𝑟﷮6﷯ ℎ﷮2﷯ 𝑟﷮4﷯=2 𝑟﷮6﷯ ℎ﷮2﷯= 2 𝑟﷮6﷯﷮ 𝑟﷮4﷯﷯ ℎ﷮2﷯=2 𝑟﷮2﷯ ℎ= ﷮2 𝑟﷮2﷯﷯ 𝒉=𝒓 ﷮𝟐﷯ Hence altitude equal to ﷮2﷯ times the radius of the base Hence proved