# Example 36 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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.

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Chapter 6 Class 12 Application of Derivatives

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.