Given that, $ f(x) = \max \{ 5x, 52-2x^{2} \} \quad \longrightarrow (1) $
And, $x$ is a positive real number.
The minimum value occurs when both the graphs intersect.
- Let $ y_{1} = 5x \quad \longrightarrow (2) $
- And, $ y_{2} = 52 – 2x^{2} \quad \longrightarrow (3) $
From equation $(2),$
$ y_{1} = 5x$ is an equation of line.
$ y = mx+c\,,$ where $m =$ slope of the line , $c =$ intercept on the $y$ – axis.
From equation $(3),$
$ y_{2} = 52 – 2x^{2}$ is a quadractic equation.
$ y_{2} = 0 \Rightarrow x = \sqrt{26} = \pm 5 \cdot 09 $
$ x = 0 \Rightarrow y_{2} = 52 $
Now, combine both graphs, we get
From the above graph,
$ 5x = 52 – 2x^{2} $
$ \Rightarrow – 2x^{2} + 52 – 5x = 0 $
$ \Rightarrow 2x^{2} + 5x – 52 = 0 $
$ \Rightarrow 2x^{2} + 13x – 8x – 52 = 0 $
$ \Rightarrow x( 2x+13) – 4(2x+13) = 0 $
$ \Rightarrow (2x+13) (x-4) = 0 $
$ \Rightarrow 2x+13 = o , x-4 = 0 $
$ \Rightarrow \boxed{x = \frac{-13}{4}} , \boxed{x=4}$
Since, $x$ is positive real number, so we take $x = 4.$
Now, from equation $(1),$
$ f(x) = \max \{ 5(4), 52 – 2(4)^{2} \}$
$ \Rightarrow f(x) = \max \{ 20, 52 – 32 \} $
$ \Rightarrow f(x) = \max \{ 20, 20 \} $
$ \Rightarrow \boxed{f(x) = 20} $
$\therefore$ The minimum possible value of $f(x)$ is $20.$
Correct Answer $: 20 $