1. Find the triangle vertices

The sides of the triangle are: $x+y=11,$ $x+2y=16,$ $2x+3y=29.$

Let us find their pairwise intersections.

• From $x+y=11$ and $x+2y=16$: $(x+2y)-(x+y)=16-11 \Rightarrow y=5,$ so $x=11-5=6.$ Vertex: $(6,5)$.

• From $x+y=11$ and $2x+3y=29$: From $x=11-y$, $2(11-y)+3y=29 \Rightarrow 22-2y+3y=29 \Rightarrow y=7,$ hence $x=4.$ Vertex: $(4,7)$.

• From $x+2y=16$ and $2x+3y=29$: From $x=16-2y$, $2(16-2y)+3y=29 \Rightarrow 32-4y+3y=29 \Rightarrow y=3,$ hence $x=10.$ Vertex: $(10,3)$.

So the triangle has vertices: $(6,5),\ (4,7),\ (10,3).$

2. Consider the vertical line $x=\dfrac{11}{2}$

We need points of the form: $\left(\frac{11}{2},\alpha\right)$ that lie on or inside the triangle.

So we intersect the line $x=\dfrac{11}{2}$ with the sides of the triangle.

3. Intersection with each side

• On $x+y=11$: $\frac{11}{2}+y=11 \Rightarrow y=\frac{11}{2}.$

• On $x+2y=16$: $\frac{11}{2}+2y=16 \Rightarrow 2y=\frac{21}{2} \Rightarrow y=\frac{21}{4}.$

• On $2x+3y=29$: $2\cdot \frac{11}{2}+3y=29 \Rightarrow 11+3y=29 \Rightarrow 3y=18 \Rightarrow y=6.$

So the line $x=\dfrac{11}{2}$ meets the three side-lines at: $\left(\frac{11}{2},\frac{11}{2}\right),\quad \left(\frac{11}{2},\frac{21}{4}\right),\quad \left(\frac{11}{2},6\right).$

4. Determine the segment inside the triangle

The triangle vertices have $x$-coordinates $4,6,10$, so the line $x=\dfrac{11}{2}=5.5$ passes through the interior of the triangle.

At $x=5.5$, the relevant boundary points of the triangle are on the two sides meeting the left vertex region, namely the segment joining $(6,5)$ to $(4,7)$ and the segment joining $(6,5)$ to $(10,3)$.

These correspond to:

• $x+y=11 \Rightarrow y=\dfrac{11}{2}$,
• $2x+3y=29 \Rightarrow y=6$.

Thus the allowed values of $\alpha$ are: $\frac{11}{2} \le \alpha \le 6.$

So,

• smallest value of $\alpha$ is $\dfrac{11}{2}$,
• largest value of $\alpha$ is $6$.

5. Compute the product

$\frac{11}{2}\times 6=33.$

So the required product is $\boxed{33}.$

This corresponds to Option B.