Question

The least value of $\alpha ϵR$ for which $4\alpha x^2+\frac{1}{x}\ge 1$, for all x > 0, is

• A. $\frac{1}{64}$
• B. $\frac{1}{32}$
• C. $\frac{1}{27}$
• D. $\frac{1}{25}$

Answer 1

The correct option is C $\frac{1}{27}$

Let $f\left(x\right)=4\alpha x^2+\frac{1}{x}$
For x > 0, $f_{min}=1$
$f^{\prime }\left(x\right)=8\alpha x-\frac{1}{x^2}=0\Rightarrow x=\frac{1}{2{\alpha }^{1/3}}$
$f_{min}=1\Rightarrow 4\alpha {\left(\frac{1}{2{\alpha }^{1/3}}\right)}^2+2{\alpha }^{1/3}=1$
$\Rightarrow 3{\alpha }^{1/3}=1\text{or}\alpha =\frac{1}{27}$