Let $x$ be a real number between 0 and $\tfrac{\pi}{2}$ for which the function  $3\sin^2 x + 8\sin x \cos x + 9\cos^2 x$ obtains its maximum value, $M$. Find the value of $M + 100\cos^2x$.
1. **Rewrite the given function:**
   \[
   3\sin^2 x + 8\sin x \cos x + 9\cos^2 x
   \]
   We can split \(9\cos^2 x\) into \(6\cos^2 x + 3\cos^2 x\) and factor the function as:
   \[
   3(\sin^2 x + \cos^2 x) + 6\cos^2 x + 8\sin x \cos x
   \]
   Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), we simplify this to:
   \[
   3(1) + 6\cos^2 x + 8\sin x \cos x = 3 + 6\cos^2 x + 8\sin x \cos x
   \]

2. **Use double angle formulas:**
   Recall the double angle formulas:
   \[
   \sin 2x = 2\sin x \cos x \quad \text{and} \quad \cos 2x = \cos^2 x - \sin^2 x
   \]
   We can rewrite \(8\sin x \cos x\) as \(4 \cdot 2\sin x \cos x = 4\sin 2x\). Also, note that:
   \[
   \cos^2 x = \frac{1 + \cos 2x}{2}
   \]
   Substituting these into the expression, we get:
   \[
   3 + 6\left(\frac{1 + \cos 2x}{2}\right) + 4\sin 2x = 3 + 3(1 + \cos 2x) + 4\sin 2x = 3 + 3 + 3\cos 2x + 4\sin 2x = 6 + 3\cos 2x + 4\sin 2x
   \]

3. **Maximize the expression \(6 + 3\cos 2x + 4\sin 2x\):**
   To find the maximum value of \(3\cos 2x + 4\sin 2x\), we use the Cauchy-Schwarz inequality in the form:
   \[
   a\cos \theta + b\sin \theta \leq \sqrt{a^2 + b^2}
   \]
   Here, \(a = 3\) and \(b = 4\), so:
   \[
   3\cos 2x + 4\sin 2x \leq \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
   \]
   Therefore, the maximum value of \(3\cos 2x + 4\sin 2x\) is 5, and the maximum value of the entire expression is:
   \[
   6 + 5 = 11
   \]

4. **Find \(M + 100\cos^2 x\):**
   Since the maximum value \(M\) is 11, we need to find \(100\cos^2 x\) when this maximum is achieved. From the Cauchy-Schwarz equality condition, the maximum is achieved when:
   \[
   \frac{3}{5} = \cos 2x \quad \text{and} \quad \frac{4}{5} = \sin 2x
   \]
   Using \(\cos 2x = \frac{3}{5}\), we find:
   \[
   \cos^2 x = \frac{1 + \cos 2x}{2} = \frac{1 + \frac{3}{5}}{2} = \frac{\frac{8}{5}}{2} = \frac{4}{5}
   \]
   Therefore:
   \[
   100\cos^2 x = 100 \cdot \frac{4}{5} = 80
   \]
   Finally:
   \[
   M + 100\cos^2 x = 11 + 80 = 91
   \]

The final answer is \(\boxed{91}\)