Last updated at April 19, 2021 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 minimumGiven AP & BQ be two poles where AP = 16m, BQ = 22m & AB = 20m Let R be a point on AB Let AR = π m. RB = AB β AR RB = 20 β π Here, β A = 90Β° & β B = 90Β° as they are vertical poles In βARP , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RP2 = AR2 + AP2 RP2 = x2 + 162 In βBRQ , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RQ2 = RB2 + BQ2 RQ2 = (20 β 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 π such that RP2 + RQ2 is minimum In βARP , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RP2 = AR2 + AP2 RP2 = x2 + 162 In βBRQ , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RQ2 = RB2 + BQ2 RQ2 = (20 β 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 π₯ such that RP2 + RQ2 is maximum Let S(x) = RP2 + RQ2 From (1) & (2) S(x) = π₯^2+(16)^2+(20βπ₯)^2+(22)^2 = π₯^2+(16)^2+(20βπ₯)^2+(22)^2 = π₯^2+256+((20)^2+(π₯)^2β2(20)(π₯))+484 = π₯^2+256+(400+π₯^2β40π₯ )+484 = γππγ^πβπππ+ππππ Finding Sβ(π) Sβ(π₯)=π(2π₯^2 β 40π₯ + 1140)/ππ₯ Sβ(π₯)=4π₯β40+0 Sβ(π₯)=4(π₯β10) Putting Sβ(π)=π 4(π₯β10)=0 π₯β10=0 π=ππ Finding Sββ(π₯) Sββ(π₯)=4 π(π₯ β 10)/ππ₯ Sββ(π₯)=4(1β0) Sββ(π₯)=4 Since Sββ(π₯)>0 at π₯=10 β΄ π=ππ is point of minima Hence, S(π₯) is minimum when π₯=10 Thus, the distance of R from A on AB is AR = x = 10 π

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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 10 years. He provides courses for Maths and Science at Teachoo.