Let $f(x)=$ max{$5x$, $52-2x^2$}, where $x$ is any positive real number. Then the minimum possible value of $f(x)$ is

We have $f(x) = \max\{5x, 52-2x^2\}$ where $x > 0$.

The notation $\max\{a, b\}$ means "take the larger of the two values $a$ and $b$."

So at any point $x$, we calculate both $5x$ and $52-2x^2$, then pick whichever is bigger.

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The minimum value of $f(x)$ occurs exactly when the two expressions inside the max function are equal.

When $5x < 52-2x^2$, we have $f(x) = 52-2x^2$ (the larger value)

When $5x > 52-2x^2$, we have $f(x) = 5x$ (the larger value)

When $5x = 52-2x^2$, we have $f(x) = 5x = 52-2x^2$ (both are equal)

The minimum happens at the "switching point" where neither function dominates the other.

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We set the two expressions equal:

$5x = 52 - 2x^2$

$2x^2 + 5x - 52 = 0$

Using the quadratic formula where $a = 2$, $b = 5$, $c = -52$:

$x = \dfrac{-5 \pm \sqrt{5^2 - 4(2)(-52)}}{2(2)}$

$= \dfrac{-5 \pm \sqrt{25 + 416}}{4}$

$= \dfrac{-5 \pm \sqrt{441}}{4}$

Since $\sqrt{441} = 21$:

$x = \dfrac{-5 \pm 21}{4}$

This gives us $x = \dfrac{16}{4} = 4$ or $x = \dfrac{-26}{4} = -6.5$

Since we need $x > 0$, we take $x = 4$.

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At $x = 4$:

$5x = 5(4) = 20$

$52 - 2x^2 = 52 - 2(16) = 52 - 32 = 20$

Both expressions equal 20, confirming our intersection point.

Therefore, the minimum value of $f(x)$ is 20.