Given that, $ f(x) = \min \{2x^{2}, 52 – 5x\} \quad \longrightarrow (1)$
For maximum value of $f(x):$
$ 2x^{2} = 52 – 5x $
$ \Rightarrow 2x^{2} + 5x – 52 = 0 $
$ \Rightarrow 2x^{2} + 13x – 8x – 52 = 0 $
$ \Rightarrow x(2x+13) – 4(2x+13) = 0 $
$ \Rightarrow (x-4) (2x+13) = 0 $
$ \Rightarrow x-4=0, 2x+13=0 $
$ \Rightarrow x=4, x= \frac{-13}{2} (\text {rejected, because $x$ is not positive real number})$
$ \Rightarrow \boxed{x = 4}$
Now, $f(x) = \min \{2x^{2}, 52 – 5x\}$
$ \Rightarrow f(x) = \min \{2(4)^{2}, 52 – 20\}$
$ \Rightarrow f(x) = \min \{32,32\}$
$ \Rightarrow \boxed{f(x) = 32}$
$\therefore$ The maximum possible valve of $ f(x) = 32 $
Correct Answer $:32 $