Question

Minimum value of $\frac{(6+x)(11+x)}{2+x}$ is

• A. 5
• B. 15
• C. 45
• D. 25

Answer 1

The correct option is D 25

$f\left(x\right)=\frac{(6+x)(11+x)}{2+x}$
$f^{\prime }\left(x\right)=\frac{x^2+4x-32}{(2+x{)}^2}$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow x^2+4x-32=0$
$\Rightarrow x=-8,4$
Now $f^{\prime \prime }\left(x\right)=\frac{36(2x+4)}{(2+x{)}^4}$
$f^{\prime \prime }(-8)<0$ and $f^{\prime \prime }\left(4\right)>0$
$\Rightarrow f\left(x\right)$ has a minimum value at $x=4$
$f\left(4\right)=25$