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Resistance to motion,$F$, of a moving vehicle is given by,$F=\large\frac{5}{x}$$+100x.$ Determine the minimum value of resistance.

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  • Second derivative test: Suppose $f$ is continuous on an open interval that contains
  • (i)If $f'(c)=0\; and\; f''(c) >0$ then $f$ has a local minimum at $c$.
  • (ii)If $f'(c)=0\;and \; f''(c)<0$ then $f$ has a local maximum at $c$
Step 1:
$F=\large\frac{5}{x}$$+100x$ is the resistance function
$ \large\frac{dF}{dx}=\frac{-5}{x^2}$$+100$ and $\large\frac{d^2F}{dx^2}=\frac{10}{x^3}$
Step 2:
When the resistance is minimum $\large\frac{dF}{dx}$$=0$ and $\large\frac{d^2F}{dx^2} $$ > 0$
Now $\large\frac{dF}{dx}=$$0=>\large\frac{-5}{x^2}$$+100=0$
Step 3:
Now $\large\frac{dF}{dx}=$$0=>\large\frac{-5}{x^2}$$+100=0$
$\therefore x^2=-\large\frac{1}{20} $$\;or\; x=\large\frac{1}{2 \sqrt 5} $
When $x=\large\frac{1}{2 \sqrt 5 }$$ ( > 0) , \large\frac{d^2F}{dx^2}=\frac{10}{(2 \sqrt 5)^3} $$ > 0$
$\therefore$ F is minimum
Step 4:
The minimum value of F is
$ F \bigg(\large\frac{1}{2 \sqrt 5}\bigg)$$=5 \times 2 \sqrt 5 +\large\frac{100}{2 \sqrt 5}$
$\qquad=\large \frac{50+50}{\sqrt 5}$
$\qquad=\large \frac{100}{\sqrt 5}$
$\qquad=\large \frac{100 \times \sqrt 5}{5}$
$\qquad=20 \sqrt 5 $
answered Aug 5, 2013 by meena.p

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