Let $f ( x )=2 x +3 y$ $f ( x )=2 x +\frac{18}{ x } $ $(\because xy =6$ given $)$ On differentiating, we get $f' x=2-\frac{18}{x^{2}}$ Put $f '( x )=0$ for maximum or minima. $\Rightarrow 0=2-\frac{18}{x^{2}} $ $\Rightarrow x=\pm 3$ And $f '' x =\frac{36}{ x ^{3}}$ $\Rightarrow f '' 3=\frac{36}{3^{3}}>0$ $\therefore$ At $x =3$, If $x$ is minimum. The minimum value is $f(3)=2(3)+3(2)=12$
The extrema of a function are very well known as Maxima and minima. Maxima is the maximum and minima is the minimum value of a function within the given set of ranges.
There are two types of maxima and minima that exist in a function, such as: