Ex 6.3, 20 - Chapter 6 Class 12 Application of Derivatives (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 6 Class 12 Application of Derivatives
Question 7 Important Deleted for CBSE Board 2025 Exams
Question 12 Deleted for CBSE Board 2025 Exams
Question 15 Important Deleted for CBSE Board 2025 Exams
Question 26 (MCQ) Important Deleted for CBSE Board 2025 Exams
Example 23 Important
Example 25 Important
Example 26 Important
Example 28 Important
Ex 6.3, 1 (i) Important
Ex 6.3, 5 (i)
Ex 6.3,7 Important
Ex 6.3,11 Important
Ex 6.3,18 Important
Ex 6.3, 20 Important You are here
Ex 6.3,23 Important
Ex 6.3, 26 Important
Ex 6.3,28 (MCQ) Important
Question 14 Important Deleted for CBSE Board 2025 Exams
Example 33 Important
Misc 3 Important
Misc 8 Important
Misc 10 Important
Misc 14 Important
Question 6 (MCQ) Deleted for CBSE Board 2025 Exams
Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 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 𝑟, ℎ be the Radius & Height of Cylinder respectively & 𝑉, 𝑆 be the Volume & Surface area of Cylinder respectively Given Surface Area of Cylinder = 2𝜋𝑟^2+ 2𝜋𝑟ℎ S = 2𝜋𝑟^2+ 2𝜋𝑟ℎ S – 2𝜋𝑟^2= 2𝜋𝑟ℎ (𝑆 − 2𝜋𝑟^2)/2𝜋𝑟=ℎ ℎ=(𝑆 − 2𝜋𝑟^2)/2𝜋𝑟 Volume of Cylinder = 𝜋𝑟2ℎ V = 𝜋𝑟2ℎ We need to maximum volume Now, V = πr2h V = πr2 ((𝑆 − 2𝜋𝑟^2)/2𝜋𝑟) V = (𝜋𝑟^2)/2𝜋𝑟 (𝑆 −2𝜋𝑟^2 ) V = 𝑟/2 (𝑆 −2𝜋𝑟^2 ) V = 1/2 (𝑆𝑟 −2𝜋𝑟^3 ) Diff w.r.t 𝒓 𝑑𝑉/𝑑𝑟=1/2 𝑑(𝑆𝑟−2𝜋𝑟^3 )/𝑑𝑟 𝑑𝑉/𝑑𝑟=1/2 (𝑆−6𝜋𝑟^2 ) Putting 𝒅𝑽/𝒅𝒓=𝟎 1/2 (𝑆−6𝜋𝑟^2 )=0 𝑆−6𝜋𝑟^2=0 Putting value of 𝑆 = 2𝜋𝑟2+ 2𝜋𝑟ℎ (2𝜋𝑟^2+2𝜋𝑟ℎ)−6𝜋𝑟^2=0 −4𝜋 𝑟2 + 2𝜋𝑟ℎ = 0 2𝜋𝑟ℎ (−2𝑟+ℎ)=0 2𝜋𝑟ℎ(ℎ−2𝑟)=0 ℎ−2𝑟=0 ℎ=2𝑟 Finding (𝒅^𝟐 𝒗)/(𝒅𝒓^𝟐 ) 𝑑𝑉/𝑑𝑟=1/2 (𝑠−6𝜋𝑟^2 ) (𝑑^2 𝑉)/(𝑑𝑟^2 )=1/2 𝑑(𝑆 − 6𝜋𝑟^2 )/𝑑𝑟 \ (𝑑^2 𝑣)/(𝑑𝑟^2 )=1/2 (0−12𝜋𝑟) (𝑑^2 𝑣)/(𝑑𝑟^2 )=−6𝜋𝑟 ∴ (𝑑^2 𝑣)/(𝑑𝑟^2 )<0 for ℎ=2𝑟 Hence, Volume of a cylinder is Maximum when 𝒉=𝟐𝒓