Question

The minimum value of $\frac{(6+x)(11+x)}{(2+x)},x\ge 0$ is

• A. $45$
• B. $25$
• C. $5$
• D. $15$

Answer 1

The correct option is B $25$

Let $y=\frac{(6+x)(11+x)}{2+x}=\frac{x^2+17x+66}{2+x}$
$y^{\prime }=\frac{(2+x)(2x+17)-(x^2+17x+66)}{(2+x{)}^2}$
$\Rightarrow y^{\prime }=\frac{2x^2+21x+34-x^2-17x-66}{(2+x{)}^2}$
$\Rightarrow y^{\prime }=\frac{x^2+4x-32}{(2+x{)}^2}$
$\Rightarrow y^{\prime }=\frac{(x+8)(x-4)}{(2+x{)}^2}$
Critical points are $x=-8,-2,4$


But given that $x\ge 0$
Our critical point is $x=4$ only
$f\left(x\right)$ changes sign from negative to positive as $x$ crosses $4$ from left to right. So $x=4$ is the point of local minima.
Now compare value of function at $x=4$ with value of function at the boundary points. So,
$f\left(4\right)=\frac{10\times 15}{6}=25$
$f\left(0\right)=33$
when $x\to \infty ,f\to \infty$
Hence, the minimum value of $f\left(x\right)$ is $25$