Let AP and BQ be two vertical poles at points | vedclass

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(B) Let $R$ be a point on $AB$ such that $AR=x \, m.$Then $RB=(20-x) \, m$ (as $AB=20 \, m$).From the geometry of the problem,we have right-angled tri

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$10$

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(B) Let $R$ be a point on $AB$ such that $AR=x \, m.$Then $RB=(20-x) \, m$ (as $AB=20 \, m$).From the geometry of the problem,we have right-angled triangles $	riangle PAR$ and $	riangle QBR$.$RP^2 = AR^2 + AP^2 = x^2 + 16^2 = x^2 + 256$$RQ^2 = RB^2 + BQ^2 = (20-x)^2 + 22^2 = (400 - 40x + x^2) + 484 = x^2 - 40x + 884$Let $S(x) = RP^2 + RQ^2 = (x^2 + 256) + (x^2 - 40x + 884) = 2x^2 - 40x + 1140.$To find the minimum,we differentiate $S(x)$ with respect to $x$:$S'(x) = 4x - 40.$Setting $S'(x) = 0$ gives $4x - 40 = 0 \implies x = 10.$Now,we check the second derivative: $S''(x) = 4.$Since $S''(10) = 4 > 0,$ the function $S(x)$ has a local minimum at $x = 10.$Thus,the distance of point $R$ from $A$ is $10 \, m.$

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 $RP^2 + RQ^2$ is minimum. (in $, m$)

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