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Ex 6.5, 20 - Show that cylinder of given surface and maximum volume

Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5, 20 - Chapter 6 Class 12 Application of Derivatives - Part 4


Transcript

Ex 6.5, 20 Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base. Let π‘Ÿ, β„Ž be the Radius & Height of Cylinder respectively & 𝑉, 𝑆 be the Volume & Surface area of Cylinder respectively Given Surface Area of Cylinder = 2πœ‹π‘Ÿ^2+ 2πœ‹π‘Ÿβ„Ž S = 2πœ‹π‘Ÿ^2+ 2πœ‹π‘Ÿβ„Ž S – 2πœ‹π‘Ÿ^2= 2πœ‹π‘Ÿβ„Ž (𝑆 βˆ’ 2πœ‹π‘Ÿ^2)/2πœ‹π‘Ÿ=β„Ž β„Ž=(𝑆 βˆ’ 2πœ‹π‘Ÿ^2)/2πœ‹π‘Ÿ Volume of Cylinder = πœ‹π‘Ÿ2β„Ž V = πœ‹π‘Ÿ2β„Ž We need to maximum volume Now, V = Ο€r2h V = Ο€r2 ((𝑆 βˆ’ 2πœ‹π‘Ÿ^2)/2πœ‹π‘Ÿ) V = (πœ‹π‘Ÿ^2)/2πœ‹π‘Ÿ (𝑆 βˆ’2πœ‹π‘Ÿ^2 ) V = π‘Ÿ/2 (𝑆 βˆ’2πœ‹π‘Ÿ^2 ) V = 1/2 (π‘†π‘Ÿ βˆ’2πœ‹π‘Ÿ^3 ) Diff w.r.t 𝒓 𝑑𝑉/π‘‘π‘Ÿ=1/2 𝑑(π‘†π‘Ÿβˆ’2πœ‹π‘Ÿ^3 )/π‘‘π‘Ÿ 𝑑𝑉/π‘‘π‘Ÿ=1/2 (π‘†βˆ’6πœ‹π‘Ÿ^2 ) Putting 𝒅𝑽/𝒅𝒓=𝟎 1/2 (π‘†βˆ’6πœ‹π‘Ÿ^2 )=0 π‘†βˆ’6πœ‹π‘Ÿ^2=0 Putting value of 𝑆 = 2πœ‹π‘Ÿ2+ 2πœ‹π‘Ÿβ„Ž (2πœ‹π‘Ÿ^2+2πœ‹π‘Ÿβ„Ž)βˆ’6πœ‹π‘Ÿ^2=0 βˆ’4πœ‹ π‘Ÿ2 + 2πœ‹π‘Ÿβ„Ž = 0 2πœ‹π‘Ÿβ„Ž (βˆ’2π‘Ÿ+β„Ž)=0 2πœ‹π‘Ÿβ„Ž(β„Žβˆ’2π‘Ÿ)=0 β„Žβˆ’2π‘Ÿ=0 β„Ž=2π‘Ÿ Finding (𝒅^𝟐 𝒗)/(𝒅𝒓^𝟐 ) 𝑑𝑉/π‘‘π‘Ÿ=1/2 (π‘ βˆ’6πœ‹π‘Ÿ^2 ) (𝑑^2 𝑉)/(π‘‘π‘Ÿ^2 )=1/2 𝑑(𝑆 βˆ’ 6πœ‹π‘Ÿ^2 )/π‘‘π‘Ÿ \ (𝑑^2 𝑣)/(π‘‘π‘Ÿ^2 )=1/2 (0βˆ’12πœ‹π‘Ÿ) (𝑑^2 𝑣)/(π‘‘π‘Ÿ^2 )=βˆ’6πœ‹π‘Ÿ ∴ (𝑑^2 𝑣)/(π‘‘π‘Ÿ^2 )<0 for β„Ž=2π‘Ÿ Hence, Volume of a cylinder is Maximum when 𝒉=πŸπ’“

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.