Question

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

(a) 12

(b) 9

(c) 8

(d) 6

Answer 1

SOLUTION —  Given that, $f(x)=2 x+3 y$ and $x y=6$

$\begin{array}{ll}\Rightarrow & f(x)=2 x+\frac{18}{x} \\\Rightarrow & f^{\prime}(x)=2-\frac{18}{x^2}\end{array}$

Put $f^{\prime}(x)=0$ for maxima or minima

$\Rightarrow \quad 0=2-\frac{18}{x^2} \Rightarrow x= \pm 3$

and

$\begin{array}{l}f^{\prime \prime}(x)=\frac{36}{x^3} \\f^{\prime \prime}(3)=\frac{36}{3^3}>0\end{array}$

$\Rightarrow \quad f^{\prime \prime}(3)=\frac{36}{3^3}>0$

$\therefore$ At $x=3, f(x)$ is minimum.

The minimum value is

$f(3)=2(3)+3(2)=12$

So, The correct option of this question will be (A).