








Ex 6.5
Ex 6.5, 1 (ii)
Ex 6.5, 1 (iii) Important
Ex 6.5, 1 (iv)
Ex 6.5, 2 (i)
Ex 6.5, 2 (ii) Important
Ex 6.5, 2 (iii)
Ex 6.5, 2 (iv) Important
Ex 6.5, 2 (v) Important
Ex 6.5, 3 (i)
Ex 6.5, 3 (ii)
Ex 6.5, 3 (iii)
Ex 6.5, 3 (iv) Important
Ex 6.5, 3 (v)
Ex 6.5, 3 (vi)
Ex 6.5, 3 (vii) Important
Ex 6.5, 3 (viii)
Ex 6.5, 4 (i)
Ex 6.5, 4 (ii) Important
Ex 6.5, 4 (iii)
Ex 6.5, 5 (i)
Ex 6.5, 5 (ii)
Ex 6.5, 5 (iii) Important
Ex 6.5, 5 (iv)
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important
Ex 6.5,15 Important
Ex 6.5,16
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
Ex 6.5,23 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important You are here
Ex 6.5, 27 (MCQ)
Ex 6.5,28 (MCQ) Important
Ex 6.5,29 (MCQ)
Last updated at April 15, 2021 by Teachoo
Ex 6.5, 26 Show that semi-vertical angle of right circular cone of given surface area and maximum volume is tan β1 (1/3) Let π , h & l be the radius, height & slant height of a cone respectively And Let V & S be the volume & surface area & ΞΈ be a semi vertical angle of a cone Given surface Area of a cone is constant Surface Area of a cone = Οπ^2+πππ S = Οπ^2+πππ S β Οπ^2=πππ (π β ππ^2)/(\ ππ)=π π = (π β ππ^2)/(\ ππ) We need to find minimize volume of a cone & show that semi vertical angle is sin (β1)/3 i.e. ΞΈ =π ππ (β1)/3 sin ΞΈ =1/3 We know that sin ΞΈ =π/π Volume of a cone = 1/3 ππ^2 β V = 1/3 ππ^2 β(π^2βπ^2 ) V = 1/3 ππ^2 β(((π β ππ^2)/ππ)^2βπ^2 ) We need to find minimize volume of a cone & show that semi vertical angle is sin (β1)/3 i.e. ΞΈ =π ππ (β1)/3 sin ΞΈ =1/3 We know that sin ΞΈ =π/π Volume of a cone = 1/3 ππ^2 β V = 1/3 ππ^2 β(π^2βπ^2 ) V = 1/3 ππ^2 β(((π β ππ^2)/ππ)^2βπ^2 ) We need to find minimize volume of a cone & show that semi vertical angle is sin (β1)/3 i.e. ΞΈ =π ππ (β1)/3 sin ΞΈ =1/3 We know that sin ΞΈ =π/π Volume of a cone = 1/3 ππ^2 β V = 1/3 ππ^2 β(π^2βπ^2 ) V = 1/3 ππ^2 β(((π β ππ^2)/ππ)^2βπ^2 ) V = 1/3 ππ^2 β((π β ππ^2 )^2/(π^2 π^2 )βπ^2 ) V = 1/3 ππ^2 β(((π β ππ^2 )^2 β ππ^2 (π^2 ))/(π^2 π^2 )) V = 1/3 ππ^2 β(((π β ππ^2 )^2 β π^2 π^4)/(π^2 π^2 π)) V = (ππ^2)/3ππ β((π βππ^2 )^2βπ^2 π^4 ) V = ((π))/3 β(γ(π )^2+(ππ^2 )γ^2β2π γππγ^2βπ^2 π^4 ) V = π/3 β(π ^2+π^2 π^4β2πππ^2βπ^2 π^4 ) V = π/3 β(π ^2β2 πππ^2 ) V = 1/3 β(π^2 (π ^2β2 π ππ^2 ) ) V = 1/3 β(π^2 π ^2β2 π ππ^4 ) Since V has square root It will be difficult to differentiate So, we take Z = V2 Z = 1/3 (π^2 π ^2β2 π ππ^4 ) Since V is positive, Z is maximum if V2 is maximum So, we maximize Z = V2 Diff. Z w.r.t π πZ/ππ=π(1/3 (π^2 π ^2 β 2π ππ^4 ))/ππ πZ/ππ=1/3 [π ^2 (2π)β2π π (4π^3 )] πZ/ππ=1/3 [2ππ ^2β8π ππ^3 ] Putting π π/π π = 0 1/3 [2ππ ^2β8π ππ^3 ]=0 2ππ ^2β8π ππ^3=0 2ππ ^2=8π ππ^3 (2π ^2)/(4π π )=π^3/π π /(4π )=π^2 π =4ππ^2 Finding (π ^π π)/(ππ^π ) πZ/ππ=1/3 [2ππ ^2β8π ππ^3 ] Diff w.r.t π₯ (π^2 Z)/(ππ^2 ) = π/ππ [1/3 [2ππ ^2β8π ππ^3 ] " " ] (π^2 Z)/(πr^2 ) = 1/3 [2π ^2β8π π(3π^2) ] (π^2 Z)/(πr^2 ) = 1/3 [2π ^2β24π ππ^2 ] Putting π =4ππ^2 (π^2 Z)/(πr^2 ) = 1/3 [2γ(4ππ^2)γ^2β24(4ππ^2)ππ^2 ] (π^2 Z)/(πr^2 ) = 1/3 [32π^2 π^4β96π^2 π^4 ] (π^2 Z)/(πr^2 ) = 1/3 [β64π^2 π^4 ] Since (π^2 Z)/(ππ^2 ) < 0 for π =4ππ^2 Volume is maximum for π =4ππ^2 Now, Surface area of cone = ππ^2+πππ π=ππ^2+πππ Putting S = 4ππ^2 4ππ^2=ππ^2+πππ ππ^2+πππ=4ππ^2 Dividing both sides by ππ (ππ^2+ πππ)/ππ=(4ππ^2)/ππ π+π=4π π=4πβπ π=3π π/π=3 π/π=1/3 But we know that sin ΞΈ =π/π Putting value of π/π sin ΞΈ =1/3 ΞΈ =γπππγ^(βπ)β‘ π/π Hence proved