Philippine Mathematical Olympiad Area Stage

Solution:
Let us simplify the expression inside the square root:

Let $s = \sin^2 x$, $c = \cos^2 x$. Then $s + c = 1$.

$4 \sin^4 x + 4 \cos^4 x = 4(s^2 + c^2)$

But $s^2 + c^2 = (s + c)^2 - 2sc = 1 - 2sc$

So $4(s^2 + c^2) = 4(1 - 2sc) = 4 - 8sc$

Also, $-\sin^2 x \cos^2 x = -sc$

So the expression inside the square root is:

$4 - 8sc - sc = 4 - 9sc$

Therefore,

$f(x) = \sqrt{4 - 9 \sin^2 x \cos^2 x}$

Recall that $\sin^2 x \cos^2 x = (\sin x \cos x)^2 = \left(\frac{\sin 2x}{2}\right)^2 = \frac{\sin^2 2x}{4}$

So $9 \sin^2 x \cos^2 x = 9 \cdot \frac{\sin^2 2x}{4} = \frac{9}{4} \sin^2 2x$

Thus,

$f(x) = \sqrt{4 - \frac{9}{4} \sin^2 2x} = \sqrt{\frac{16 - 9 \sin^2 2x}{4}} = \frac{\sqrt{16 - 9 \sin^2 2x}}{2}$

Now, $\sin^2 2x$ ranges from $0$ to $1$.

So $16 - 9 \sin^2 2x$ ranges from $16$ (when $\sin^2 2x = 0$) to $7$ (when $\sin^2 2x = 1$).

Therefore, the maximum value of $f(x)$ is when $\sin^2 2x = 0$:

$M = \frac{\sqrt{16}}{2} = \frac{4}{2} = 2$

The minimum value is when $\sin^2 2x = 1$:

$m = \frac{\sqrt{7}}{2}$

Therefore, the product $M \cdot m = 2 \cdot \frac{\sqrt{7}}{2} = \sqrt{7}$

Answer: $\boxed{\sqrt{7}}$