Let AD and BC be two vertical poles at A and | vedclass

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(B) Let the distance of point $M$ from $A$ be $x$. Then $AM = x$ and $MB = 10 - x$.In right-angled triangles $	riangle DAM$ and $	riangle CBM$:$MD^2

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$5$

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(B) Let the distance of point $M$ from $A$ be $x$. Then $AM = x$ and $MB = 10 - x$.In right-angled triangles $	riangle DAM$ and $	riangle CBM$:$MD^2 = AM^2 + AD^2 = x^2 + 8^2 = x^2 + 64$$MC^2 = MB^2 + BC^2 = (10 - x)^2 + 11^2 = (10 - x)^2 + 121$Let $f(x) = MD^2 + MC^2 = x^2 + 64 + (10 - x)^2 + 121$$f(x) = x^2 + 64 + 100 - 20x + x^2 + 121$$f(x) = 2x^2 - 20x + 285$To find the minimum,differentiate $f(x)$ with respect to $x$:$f'(x) = 4x - 20$Set $f'(x) = 0$ for critical points:$4x - 20 = 0 \Rightarrow x = 5$Since $f''(x) = 4 > 0$,the function $f(x)$ is minimum at $x = 5$ meters.

Let $AD$ and $BC$ be two vertical poles at $A$ and $B$ respectively on a horizontal ground. If $AD = 8 \ m$,$BC = 11 \ m$ and $AB = 10 \ m$,then the distance (in meters) of point $M$ on $AB$ from the point $A$ such that $MD^2 + MC^2$ is minimum,is

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