Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.3, 23 Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.Cone of largest volume inscribed in the sphere of radius R Let OC = x Radius of cone = BC Height of cone = h = OC + OA Finding OC Ξ BOC is a right angled triangle Using Pythagoras theorem in βBOC γππ΅γ^2=γπ΅πΆγ^2+γππΆγ^2 R2 =γπ΅πΆγ^2+π₯^2 BC2 = π^2 β x2 BC = β(π^2βπ₯^2 ) Thus, Radius of cone = BC = β(π^2βπ₯^2 ) Height of cone = OC + OA = R + x We need to show Maximum volume of cone = 8/27 Γ Volume of sphere = 8/27 Γ 4/3 ππ^3 = 32/81 ππ^3 Let V be the volume of a cone We know that Volume of a cone = 1/3 π(πππππ’π  )^2 (βπππβπ‘ ) V = 1/3 π(β(π^2βπ₯^2 ))^2 (π+π₯) V = 1/3 π(π^2βπ₯^2 )(π+π₯) V = 1/3 π(π^2 (π+π₯)βπ₯^2 (π+π₯)) V = 1/3 π(π^(3 )+π^2 π₯βπ₯^2 πβπ₯^3 ) Diff w.r.t π ππ/ππ₯=π/ππ₯ [1/3 π(π^(3 )+π^2 π₯βπ₯^2 πβπ₯^3 )] ππ/ππ₯=π/3 [π/ππ₯ (π^(3 )+π^2 π₯βπ₯^2 πβπ₯^3 )] ππ/ππ₯=π/3 (0+π^2.1βπ Γ2π₯β3π₯^2 ) ππ/ππ₯=π/3 (π^2β2ππ₯β3π₯^2 ) Putting ππ½/ππ=π 1/3 π(π^2β2ππ₯β3π₯^2 )=0 π^2β2ππ₯β3π₯^2=0 π^2β3ππ₯+ππ₯β3π₯^2=0 R(πβ3π₯)+π₯(πβ3π₯)=0 (π+π₯)(πβ3π₯)=0 So, x = βR & π₯=π/3 Since x cannot be negative π₯=π/3 Finding (π^π π)/(ππ^π ) ππ£/ππ₯=1/3 π(π^2β2ππ₯β3π₯^2 ) (π^2 π£)/(ππ₯^2 )=π/3 π/ππ₯ (π^2β2ππ₯β3π₯^2 ) (π^2 π£)/(ππ₯^2 )=π/3 (0β2πβ6π₯) (π^2 π£)/(ππ₯^2 )=(βπ)/3 (2π+6π₯) Putting π=πΉ/π γ(π^2 π£)/(ππ₯^2 )βγ_(π₯ = π/3) =(βπ)/3 (2π+6(π/3)) =(βπ)/3 (2π+2π) =(βπ)/3 (4π) =(β4ππ)/3 < 0 Thus (π^2 π£)/(ππ₯^2 )<0 when π₯=π/3 β΄ Volume is Maximum when π₯=π/3 Finding maximum volume From (1) Volume of cone = 1/3 π(π^2βπ₯^2 )(π+π₯) Putting π₯ = π/3 = 1/3 π(π^2β(π/3)^2 )(π+π/3) = 1/3 π(π^2βπ^2/9)((3π + π)/3) = 1/3 π((9π^2β π^2)/9)(4π/3) = ππ/ππ ππΉ^π Hence proved