Let $f(x) =$ min{$2x², 52-5x$}, where x is any positive real number. Then the maximum possible value of $f(x)$ is

We have $f(x) = \min\{2x^2, 52-5x\}$ where $x$ is any positive real number.

The function $f(x)$ takes the smaller value between $2x^2$ and $52-5x$ for any given $x$.

$2x^2$ is a parabola opening upward (gets larger as $x$ increases)

$52-5x$ is a decreasing straight line (gets smaller as $x$ increases)

The maximum value of $f(x)$ occurs at the intersection point of these two functions.

Before the intersection: $2x^2 < 52-5x$, so $f(x) = 2x^2$ (increasing)

After the intersection: $2x^2 > 52-5x$, so $f(x) = 52-5x$ (decreasing)

At the intersection: Both functions are equal, giving us the peak of the minimum function. This is true for any min, max function.

---

We set the two functions equal:

$2x^2 = 52-5x$

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

$2x^2 + 13x - 8x - 52 = 0$

$x(2x + 13) - 4(2x + 13) = 0$

$(2x + 13)(x - 4) = 0$

---

From $(2x + 13)(x - 4) = 0$:

$2x + 13 = 0 \rightarrow x = -\dfrac{13}{2}$

$x - 4 = 0 \rightarrow x = 4$

Since $x$ must be positive, we take $x = 4$.

---

At $x = 4$:

$f(4) = 2(4)^2 = 2 \times 16 = 32$

Therefore, the maximum possible value of $f(x)$ is $32$.