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Let $f\left (x \right ) = \max\left \{5x, 52 – 2x^{2}\right \}$ , where $x$ is any positive real numbers. Then the minimum possible value of $f(x)$ is ________

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 $: 20$

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