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Transcript

Example 24 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 = 90Β° & ∠ B = 90Β° as they are vertical poles In βˆ†ARP , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RP2 = AR2 + AP2 RP2 = x2 + 162 In βˆ†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 βˆ†ARP , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RP2 = AR2 + AP2 RP2 = x2 + 162 In βˆ†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 π‘₯ such that RP2 + RQ2 is maximum Let S(x) = RP2 + RQ2 From (1) & (2) S(x) = π‘₯^2+(16)^2+(20βˆ’π‘₯)^2+(22)^2 = π‘₯^2+(16)^2+(20βˆ’π‘₯)^2+(22)^2 = π‘₯^2+256+((20)^2+(π‘₯)^2βˆ’2(20)(π‘₯))+484 = π‘₯^2+256+(400+π‘₯^2βˆ’40π‘₯ )+484 = γ€–πŸπ’™γ€—^πŸβˆ’πŸ’πŸŽπ’™+πŸπŸπŸ’πŸŽ Finding S’(𝒙) S’(π‘₯)=𝑑(2π‘₯^2 βˆ’ 40π‘₯ + 1140)/𝑑π‘₯ S’(π‘₯)=4π‘₯βˆ’40+0 S’(π‘₯)=4(π‘₯βˆ’10) Putting S’(𝒙)=𝟎 4(π‘₯βˆ’10)=0 π‘₯βˆ’10=0 𝒙=𝟏𝟎 Finding S’’(π‘₯) S’’(π‘₯)=4 𝑑(π‘₯ βˆ’ 10)/𝑑π‘₯ S’’(π‘₯)=4(1βˆ’0) S’’(π‘₯)=4 Since S’’(π‘₯)>0 at π‘₯=10 ∴ 𝒙=𝟏𝟎 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’Ž

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.