# Resistance to motion,$F$, of a moving vehicle is given by,$F=\large\frac{5}{x}$$+100x. Determine the minimum value of resistance. Can you answer this question? ## 1 Answer 0 votes Toolbox: • Second derivative test: Suppose f is continuous on an open interval that contains • (i)If f'(c)=0\; and\; f''(c) >0 then f has a local minimum at c. • (ii)If f'(c)=0\;and \; f''(c)<0 then f has a local maximum at c Step 1: F=\large\frac{5}{x}$$+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