Example 36 - Let AP and BQ be two vertical poles. If AP = 16 m

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) S﷐𝑥﷯=﷐𝑥﷮2﷯+﷐﷐16﷯﷮2﷯+﷐﷐20−𝑥﷯﷮2﷯+﷐﷐22﷯﷮2﷯ S﷐𝑥﷯=﷐𝑥﷮2﷯+﷐﷐16﷯﷮2﷯+﷐﷐20−𝑥﷯﷮2﷯+﷐﷐22﷯﷮2﷯ = ﷐𝑥﷮2﷯+256+﷐﷐﷐20﷯﷮2﷯+﷐﷐𝑥﷯﷮2﷯−2﷐20﷯﷐𝑥﷯﷯+484 = ﷐𝑥﷮2﷯+256+﷐400+﷐𝑥﷮2﷯−40𝑥 ﷯+484 = ﷐2𝑥﷮2﷯−40𝑥+1140 Finding s’﷐𝑥﷯ S’﷐𝑥﷯=﷐𝑑﷐2﷐𝑥﷮2﷯ − 40𝑥 + 1140﷯﷮𝑑𝑥﷯ S’﷐𝑥﷯=4𝑥−40+0 S’﷐𝑥﷯=4﷐𝑥−10﷯ Putting S’﷐x﷯=0 4﷐𝑥−10﷯=0 𝑥−10=0 𝑥=10 Finding S’’﷐𝑥﷯ S’’﷐𝑥﷯=4﷐𝑑﷐𝑥 − 10﷯﷮𝑑𝑥﷯ S’’﷐𝑥﷯=4﷐1−0﷯ S’’﷐𝑥﷯=4 > 0 At 𝑥=10, S’’﷐𝑥﷯>0 ⇒ 𝑥=10 is point of minima ⇒ S﷐𝑥﷯ is minimum when 𝑥=10 Thus, the distance of R from A on AB is AR = x = 10 𝒎.

Example 36 - Chapter 6 Class 12 Application of Derivatives