      1. Chapter 6 Class 12 Application of Derivatives (Term 1)
2. Serial order wise
3. Miscellaneous

Transcript

Misc 17 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2𝑅/√3 . Also find the maximum volume.Given Radius of sphere = R Let h be the height & 𝒙 be the diameter of cylinder In ∆ 𝑨𝑩𝑪 Using Pythagoras theorem (𝐶𝐵)^2+(𝐴𝐵)^2=(𝐴𝐶)^2 h2 + 𝑥^2=(𝑅+𝑅)^2 h2 + 𝑥2 =(2𝑅)^2 h2 + 𝑥2 = 4R2 𝒙2 = 4R2 – h2 We need to find maximum volume of cylinder Let V be the volume of cylinder V = π (𝑟𝑎𝑑𝑖𝑢𝑠)^2×(ℎ𝑒𝑖𝑔ℎ𝑡) V = π (𝑥/2)^2× ℎ V = π × 𝑥^2/4× ℎ V = π ((4𝑅^2 − ℎ^2 ))/4 × ℎ V = (4𝑅^2 𝜋ℎ)/4−(𝜋ℎ^3)/4 V = πhR2 – (𝝅𝒉^𝟑)/𝟒 Differentiating w.r.t 𝒉 𝑑𝑉/𝑑ℎ=𝑑(𝜋ℎ𝑅^2 − 𝜋ℎ^3/4)/𝑑ℎ 𝑑𝑉/𝑑ℎ= πR2 𝑑(ℎ)/𝑑ℎ−𝜋/4 𝑑(ℎ^3 )/𝑑ℎ 𝑑𝐵/𝑑ℎ= πR2 – 𝜋/4 (3ℎ^2 ) 𝑑𝑉/𝑑ℎ= πR2 – 3𝜋/4 h2 Putting 𝒅𝑽/𝒅𝒉=𝟎 π R2 – 3/4 𝜋 ℎ^2=0 3/4 𝜋ℎ^2=𝜋𝑅^2 h2 = (𝜋𝑅^2)/(3/4 𝜋) h2 = (4𝑅^2)/3 h =√((4𝑅^2)/3) h = 𝟐𝑹/√𝟑 Finding (𝒅^𝟐 𝑽)/(𝒅𝒉^𝟐 ) 𝑑𝑉/𝑑ℎ=𝜋𝑅^2−3/(4 ) 𝜋 ℎ^2 Differentiating w.r.t. h (𝑑^2 𝑉)/(𝑑ℎ^2 )= 𝑑(𝜋𝑅^2 − 3/4 𝜋ℎ^2 )/𝑑ℎ (𝑑^2 𝑉)/(𝑑ℎ^2 )= 0 – 3𝜋/4 ×2ℎ (𝒅^𝟐 𝑽)/(𝒅𝒉^𝟐 )=(−𝟑𝝅𝒉)/𝟐 Since (𝒅^𝟐 𝑽)/(𝒅𝒉^𝟐 )<𝟎 for h = 2𝑅/√3 ∴ Volume is maximum for h = 2𝑅/√3 We also need to find Maximum Volume V = πhR2 – (𝜋ℎ^3)/4 V = πR2 × 2𝑅/√3 – 𝜋/4 × (2𝑅/√3)^3 V = (2𝜋𝑅^3)/√3 – 𝜋/4 ×(8𝑅^3)/(3√3) V = (2𝜋𝑅^3)/√3 – (2𝜋𝑅^3)/(3√3) V = (2𝜋𝑅^3)/√3 (1−1/3) V = (2𝜋𝑅^3)/√3 ×2/3 V = (𝟒𝝅𝑹^𝟑)/(𝟑√𝟑) cubic unit 