Question

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

• A. $9$
• B. $12$
• C. $8$
• D. $6$

Answer 1

Answer: (B)

$12$

Let $f\left(x\right)=2x+3y$
$f\left(x\right)=2x+\frac{18}{x}$ $(\because xy=6given)$
On differentiating, we get
$f^{\prime }x=2-\frac{18}{x^2}$
Put $f^{\prime }\left(x\right)=0$ for maximum or minima.
$\Rightarrow 0=2-\frac{18}{x^2}$
$\Rightarrow x=\pm 3$
And $f^{\prime \prime }x=\frac{36}{x^3}$
$\Rightarrow f^{\prime \prime }3=\frac{36}{3^3}>0$
$\therefore$ At $x=3$, If $x$ is minimum.
The minimum value is
$f\left(3\right)=2\left(3\right)+3\left(2\right)=12$