Question

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

• A. $12$
• B. $9$
• C. $8$
• D. $6$

Answer 1

The correct option is A

$12$

Explanation for the correct option

Step 1: Solve for the critical points

Given equation is, $2x+3y$ such that $xy=6$

$\begin{array}{cc}& f\left(x\right)=2x+3y\\ \Rightarrow & f\left(x\right)=2x+3·\frac{6}{x}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{xy}\mathbf{=}\mathbf{6}\mathbf{]}\\ \Rightarrow & f\left(x\right)=2x+\frac{18}{x}\end{array}$

We know that a function has minima at $a$ if $f\text{'}\left(a\right)=0$ and $f\text{'}\text{'}\left(a\right)>0$

So we have,
$f\text{'}\left(x\right)=2-\frac{18}{x^2}$

To find the critical point, set $f\text{'}\left(x\right)=0$, i.e.,
$\begin{array}{cc}\Rightarrow & 2-\frac{18}{x^2}=0\\ \Rightarrow & x^2=\frac{18}{2}\\ \Rightarrow & x=\pm 3\end{array}$

Step 2: Solve for the minimum value
$f\text{'}\text{'}\left(x\right)=\frac{36}{x^3}$

We see that,
$\begin{array}{rcl}f\text{'}\text{'}\left(-3\right)& =& \frac{36}{{\left(-3\right)}^3}\\ & =& \frac{36}{-27}\\ & =& -\frac{4}{3}<0\\ f\text{'}\text{'}\left(3\right)& =& \frac{36}{3^3}\\ & =& \frac{36}{27}\\ & =& \frac{4}{3}>0\end{array}$
So, it has a minima at $x=3$
At that point, $y=\frac{6}{3}=2$ $\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{xy}\mathbf{=}\mathbf{6}\mathbf{]}$

So the minimum value of $2x+3y$ is $2\left(3\right)+3\left(2\right)=12$

Hence, option(A) is correct.