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 You are here

Example 37 Important

Example 38 Important

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 April 19, 2021 by Teachoo

Example 36 Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m, BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A such that RP2 + RQ2 is minimumGiven AP & BQ be two poles where AP = 16m, BQ = 22m & AB = 20m Let R be a point on AB Let AR = 𝒙 m. RB = AB – AR RB = 20 – 𝒙 Here, ∠ A = 90° & ∠ B = 90° as they are vertical poles In ∆ARP , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RP2 = AR2 + AP2 RP2 = x2 + 162 In ∆BRQ , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RQ2 = RB2 + BQ2 RQ2 = (20 – x)2 + 222 We need to find distance of R on AB from point A such that RP2 + RQ2 is minimum i.e. we need to find 𝒙 such that RP2 + RQ2 is minimum In ∆ARP , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RP2 = AR2 + AP2 RP2 = x2 + 162 In ∆BRQ , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RQ2 = RB2 + BQ2 RQ2 = (20 – x)2 + 222 We need to find distance of R on AB from point A such that RP2 + RQ2 is maximum i.e. we need to find 𝑥 such that RP2 + RQ2 is maximum Let S(x) = RP2 + RQ2 From (1) & (2) S(x) = 𝑥^2+(16)^2+(20−𝑥)^2+(22)^2 = 𝑥^2+(16)^2+(20−𝑥)^2+(22)^2 = 𝑥^2+256+((20)^2+(𝑥)^2−2(20)(𝑥))+484 = 𝑥^2+256+(400+𝑥^2−40𝑥 )+484 = 〖𝟐𝒙〗^𝟐−𝟒𝟎𝒙+𝟏𝟏𝟒𝟎 Finding S’(𝒙) S’(𝑥)=𝑑(2𝑥^2 − 40𝑥 + 1140)/𝑑𝑥 S’(𝑥)=4𝑥−40+0 S’(𝑥)=4(𝑥−10) Putting S’(𝒙)=𝟎 4(𝑥−10)=0 𝑥−10=0 𝒙=𝟏𝟎 Finding S’’(𝑥) S’’(𝑥)=4 𝑑(𝑥 − 10)/𝑑𝑥 S’’(𝑥)=4(1−0) S’’(𝑥)=4 Since S’’(𝑥)>0 at 𝑥=10 ∴ 𝒙=𝟏𝟎 is point of minima Hence, S(𝑥) is minimum when 𝑥=10 Thus, the distance of R from A on AB is AR = x = 10 𝒎