Question

The minimum value of $2x+3y$, when $xy=6$ is

• A. $9$
• B. $12$
• C. $8$
• D. $6$

Answer 1

The correct option is B

$12$

Explanation for the correct option:

Step 1: Find the critical points of the given function.

A function $f\left(x\right)=2x+3y$ is given.

Also, It is given that $xy=6$.

$\Rightarrow y=\frac{6}{x}$

Therefore, the given function becomes $f\left(x\right)=2x+\frac{18}{x}$

Differentiate both sides with respect to $x$.

$f\text{'}\left(x\right)=2-\frac{18}{x^2}...1$

Put $f\text{'}\left(x\right)$ equal to zero to find the critical points.

$$\begin{gathered} 2-\frac{18}{x^2}=0 \\ \Rightarrow 2=\frac{18}{x^2} \\ \Rightarrow x^2=\frac{18}{2} \\ \Rightarrow x^2=9 \\ \Rightarrow x=\pm 3 \end{gathered}$$

So, the critical points are $x=\pm 3$.

Step 2: Find the minimum value of the given function.

Since, the critical point is $x=-3,3$.

Differentiate equation $1$ with respect to $x$.

$f"\left(x\right)=\frac{36}{x^3}$

Since, $f"\left(3\right)>0$. Therefore, $f\left(x\right)$ is minimum at $x=3$.

Evaluate $f\left(x\right)$ for $x=3$ as follows:

$$\begin{gathered} f\left(x\right)=6+6 \\ \Rightarrow f\left(x\right)=12 \end{gathered}$$

So, the value of $f\left(x\right)$ for $x=3$ is $12$.

Therefore, the minimum value of the given function is $12$.

Hence, option $B$ is the correct answer.