# Example 36

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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 minimum Given 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 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) S𝑥=𝑥2+162+20−𝑥2+222 S𝑥=𝑥2+162+20−𝑥2+222 = 𝑥2+256+202+𝑥2−220𝑥+484 = 𝑥2+256+400+𝑥2−40𝑥 +484 = 2𝑥2−40𝑥+1140 Finding s’𝑥 S’𝑥=𝑑2𝑥2 − 40𝑥 + 1140𝑑𝑥 S’𝑥=4𝑥−40+0 S’𝑥=4𝑥−10 Putting S’x=0 4𝑥−10=0 𝑥−10=0 𝑥=10 Finding S’’𝑥 S’’𝑥=4𝑑𝑥 − 10𝑑𝑥 S’’𝑥=41−0 S’’𝑥=4 > 0 At 𝑥=10, S’’𝑥>0 ⇒ 𝑥=10 is point of minima ⇒ S𝑥 is minimum when 𝑥=10 Thus, the distance of R from A on AB is AR = x = 10 𝒎.

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18

Example 19

Example 20

Example 21

Example 22

Example 23

Example 24

Example 25

Example 26

Example 27

Example 28

Example 29

Example 30

Example 31

Example 32

Example 33

Example 34

Example 35 Important

Example 36 You are here

Example 37 Important

Example 38 Important

Example 39

Example 40 Important

Example 41

Example 42

Example 43

Example 44

Example 45

Example 46 Important

Example 47 Important

Example 48

Example 49

Example 50

Example 51

Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .