Question

if the height of tower is given by the minimum value x+1/2x, x>0 then it's distance
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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

The height of tower is given by the minimum value of

$\displaystyle \sf{x + \frac{1}{2x} \: } \: \: \: \: where \: x > 0$

TO DETERMINE

The height of the tower

CALCULATION

$\displaystyle \sf{ Let \: \: h\: = \: x + \frac{1}{2x} \: } \: \: \:$

Differentiating both sides with respect to x two times we get

$\displaystyle \sf{ \frac{dh}{dx} = 1 - \frac{1}{2 {x}^{2} } \: \: }$

$\displaystyle \sf{ \frac{{d}^{2} h}{d {x}^{2} } = \frac{1}{ {x}^{3} } \: \: }$

For minimum value of h we have

$\displaystyle \sf{ \frac{dh}{dx} = 0} \: \:$

$\implies \: \displaystyle \sf{ 1 - \frac{1}{2 {x}^{2} } \: = 0 \: }$

$\implies \: \displaystyle \sf{ 2 {x}^{2} = 1\: }$

$\implies \: \displaystyle \sf{ {x}^{2} = \frac{1}{2} \: }$

$\implies \: \displaystyle \sf{ {x} = \pm \: \frac{1}{ \sqrt{2} } \: }$

Since x > 0

$\displaystyle \sf{ So \: \: \: {x} = \frac{1}{ \sqrt{2} } \: }$

$\displaystyle \sf{ For \: \: \: \: {x} = \frac{1}{ \sqrt{2} } \: }$

$\displaystyle \sf{ \frac{{d}^{2} h}{d {x}^{2} } = {( \sqrt{2} )}^{3} = 2 \sqrt{2} \: > 0 \: }$

$\displaystyle \sf{ So \: h \: \: has \: a \: minimum \: value \: at \: \: {x} = \frac{1}{ \sqrt{2} } \: }$

Minimum value of h

$= \displaystyle \sf{ \frac{1}{ \sqrt{2} } + \frac{ \sqrt{2} }{2} \: } \:$

$= \displaystyle \sf{ \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } \: } \:$

$= \displaystyle \sf{ \frac{2}{ \sqrt{2} } } \:$

$= \displaystyle \sf{ \sqrt{2} } \: \: \: \: unit$

RESULT

$\boxed{ \sf{ The \: height \: of \: tower \: is \: \: \: \sqrt{2} \: \: \: unit\: }}$

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