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Let $f(x) = \text{min }\{2x^2, 52−5x\}$, where $x$ is any positive real number. Then the maximum possible value of $f(x)$ is ________

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 $:32$
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Why is maximum value of f(x) only possible when both 2x^2 and 52-x are equal !  could you please explain …

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