Question

Any Genius Here please answer I challenge to Geniuses

Question = Let AP and BQ two vertical pole at point A and B respectively if AP = 16 m and BQ = 22 m and AB = 20 m then find the distance of a point R on AB from the point such that RP^2+RQ^2 is minimum.

Answer 1

$\boxed{\boxed{\mathtt{ANSWER = 10\:metres. }}}$

$Let \: R \: be \: the \: point \: on \: AB \: such \: that \: AR = \:x \: and \: RB = (20-x)\\\\ </p><p>Then \: \: RP^2 = AR^2 +AP^2</p><p>= x^2+(16)^{2} = x^2 + 256\\\\ and\:\:\:\:RQ^2 = RB^2 + BQ^2 </p><p> = (20-x)^2+(22)^2\\\\ </p><p> \implies \: 400 + x^2- 40x+ 484 </p><p> = x^2 -40x +884\\\\Let\:\:\:\: f(x) \:\:RP^2 +RQ^2 = (x^2 + 256)+(x^2 -40x +884)\\\\ \implies 2x^2-40x+1140\\\\then\:\:f'(x)= 4x-40\:\:and\:\:f"(x) = 4 > 0\\\\ Also, f'(x) = 0 \: \implies = 4x-40 = 0 \implies x = 10\\\\ \therefore \:f(x)\:is\:minimum \:when \:x =10.\\\\Hence,\:\:f(x) \:is \: minimum \:when \: AR=10 \: metres.$