Example 36 - Chapter 6 Class 12 Application of Derivatives
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.