Get live Maths 1-on-1 Classs - Class 6 to 12

Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8 Important

Example 9 Important

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14 Deleted for CBSE Board 2023 Exams

Example 15 Deleted for CBSE Board 2023 Exams

Example 16 Deleted for CBSE Board 2023 Exams

Example 17 Important Deleted for CBSE Board 2023 Exams

Example 18 Deleted for CBSE Board 2023 Exams

Example 19 Deleted for CBSE Board 2023 Exams

Example 20 Deleted for CBSE Board 2023 Exams

Example 21 Deleted for CBSE Board 2023 Exams

Example 22 Deleted for CBSE Board 2023 Exams

Example 23 Deleted for CBSE Board 2023 Exams

Example 24

Example 25

Example 26

Example 27

Example 28 Important

Example 29

Example 30 Important

Example 31

Example 32 Important

Example 33 Important

Example 34

Example 35 Important

Example 36

Example 37 Important

Example 38 Important You are here

Example 39

Example 40 Important

Example 41 Important

Example 42 Important

Example 43 Important

Example 44 Important

Example 45 Important Deleted for CBSE Board 2023 Exams

Example 46 Important Deleted for CBSE Board 2023 Exams

Example 47 Important

Example 48 Important

Example 49

Example 50 Important

Example 51

Chapter 6 Class 12 Application of Derivatives

Serial order wise

Last updated at March 30, 2023 by Teachoo

Example 38 Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that cone. Let OC = r be the radius of cone & OA = h be height of cone & ∠ OAQ = α be the semi-vertical angle of cone And, Let OE = x be the radius of cylinder Height of cylinder = OO’ Since cone is given, radius (r) and height (h) of cone are constant And, radius (x) and height (OO’) of cylinder is variable Finding Height OO’ in terms of h, r and x From (1) & (2) 𝒙/(𝒉 − 𝑶^′ 𝑶)=𝒓/𝒉 ℎ𝑥/𝑟=ℎ−𝑂^′ 𝑂 𝑂^′ 𝑂=ℎ−ℎ𝑥/𝑟 𝑂^′ 𝑂= (ℎ𝑟 − ℎ𝑥)/𝑟 In Δ AO’Q tan α = (𝑂^′ 𝑄)/𝐴𝑂′ tan α = 𝑂𝐸/(𝑂𝐴 − 𝑂^′ 𝑂) tan 𝜶 = 𝑥/(ℎ − 𝑂^′ 𝑂) In ∆𝑨𝑶𝑪 tan α = 𝑂𝐶/𝑂𝐴 tan 𝜶 = 𝑟/ℎ 𝑶^′ 𝑶= 𝒉(𝒓 − 𝒙)/𝒓 Now, Curved Surface Area of Cylinder = 2𝜋 × Radius × Height S = 2𝜋" × " 𝑥" × " 𝑂^′ 𝑂 S = 2𝜋𝑥 ℎ(𝑟 − 𝑥)/𝑟 S = 2𝜋ℎ/𝑟 (𝑟𝑥−𝑥^2 ) S = 𝒌(𝒓𝒙−𝒙^𝟐 ) We need to minimize S (Where k = 2𝜋ℎ/𝑟 is a constant) Finding S’(x) S’ (x) = 𝑑(𝑘(𝑟𝑥 − 𝑥^2 ))/𝑑𝑥 S’ (x) = 𝑘 𝑑(𝑟𝑥 − 𝑥^2 )/𝑑𝑥 S’ (x) = 𝑘(𝑟−2𝑥) Putting S’ = 0 0 = 𝑘(𝑟−2𝑥) 𝑟−2𝑥 = 0 𝒙 = 𝒓/𝟐 Finding S’’(x) at x = 𝒓/𝟐 S’’ (x) = 𝑑(𝑘(𝑟 − 2𝑥))/𝑑𝑥 S’’ (x) = 𝑘 𝑑(𝑟 − 2𝑥)/𝑑𝑥 S’’ = 𝑘 (0−2) S’’ = −2𝑘 Therefore, S’’ (x) < 0 for 𝑥=𝑟/2 Thus, 𝒙=𝒓/𝟐 is maxima of S. Hence, radius of cylinder with greatest curved surface area which can be inscribed in a given cone is half of that cone