Let the points $\left(\frac{11}{2}, \alpha\right)$ lie on or inside the triangle with sides $x + y = 11$, $x + 2y = 16$, and $2x + 3y = 29$. Then the product of the smallest and the largest values of $\alpha$ is equal to:

Step 1: Find the vertices of the triangle

First, we need to find the vertices of the triangle formed by the given lines. We will find the intersection points of each pair of lines. Let the lines be $L_1: x + y = 11$, $L_2: x + 2y = 16$, and $L_3: 2x + 3y = 29$.

Step 2: Calculate intersection points

Solving $L_1$ and $L_2$: Subtracting $L_1$ from $L_2$ gives $y = 5$. Substituting into $L_1$ gives $x = 6$. So, vertex $A = (6, 5)$. Solving $L_2$ and $L_3$: Multiply $L_2$ by 2 to get $2x + 4y = 32$. Subtract $L_3$ from this to get $y = 3$. Substituting into $L_2$ gives $x + 2(3) = 16 \implies x = 10$. So, vertex $B = (10, 3)$. Solving $L_3$ and $L_1$: Multiply $L_1$ by 2 to get $2x + 2y = 22$. Subtract this from $L_3$ to get $y = 7$. Substituting into $L_1$ gives $x + 7 = 11 \implies x = 4$. So, vertex $C = (4, 7)$. Wait, let's recheck the intersection of $L_2$ and $L_3$. $L_2: x + 2y = 16 \implies x = 16 - 2y$ Substitute into $L_3: 2(16 - 2y) + 3y = 29 \implies 32 - 4y + 3y = 29 \implies 32 - y = 29 \implies y = 3$. Then $x = 16 - 2(3) = 16 - 6 = 10$. So $B = (10, 3)$. Let's recheck the intersection of $L_1$ and $L_2$: $L_1: x + y = 11$ $L_2: x + 2y = 16$ Subtracting $L_1$ from $L_2$: $(x + 2y) - (x + y) = 16 - 11 \implies y = 5$. Substitute $y=5$ into $L_1$: $x + 5 = 11 \implies x = 6$. So $A = (6, 5)$. Let's recheck the intersection of $L_1$ and $L_3$: $L_1: x + y = 11 \implies x = 11 - y$ $L_3: 2x + 3y = 29$ Substitute $x = 11 - y$ into $L_3$: $2(11 - y) + 3y = 29 \implies 22 - 2y + 3y = 29 \implies 22 + y = 29 \implies y = 7$. Substitute $y=7$ into $L_1$: $x + 7 = 11 \implies x = 4$. So $C = (4, 7)$. The vertices are $A(6, 5)$, $B(10, 3)$, and $C(4, 7)$.

Step 3: Substitute the point into the line equations

The point $P\left(\frac{11}{2}, \alpha\right)$ lies on or inside the triangle. This means that when we substitute the coordinates of $P$ into the equations of the lines, the resulting values must satisfy the inequalities that define the interior of the triangle. To determine the correct inequality direction, we can test a point known to be inside the triangle, for example, the centroid. The centroid is $\left(\frac{6+10+4}{3}, \frac{5+3+7}{3}\right) = \left(\frac{20}{3}, 5\right)$. For $L_1: x + y - 11 = 0$, $P_c = \frac{20}{3} + 5 - 11 = \frac{20+15-33}{3} = \frac{2}{3} > 0$. So for points inside, $x+y-11 \ge 0$. For $L_2: x + 2y - 16 = 0$, $P_c = \frac{20}{3} + 2(5) - 16 = \frac{20}{3} + 10 - 16 = \frac{20}{3} - 6 = \frac{20-18}{3} = \frac{2}{3} > 0$. So for points inside, $x+2y-16 \ge 0$. For $L_3: 2x + 3y - 29 = 0$, $P_c = 2\left(\frac{20}{3}\right) + 3(5) - 29 = \frac{40}{3} + 15 - 29 = \frac{40}{3} - 14 = \frac{40-42}{3} = -\frac{2}{3} < 0$. So for points inside, $2x+3y-29 \le 0$. Now, substitute $x = \frac{11}{2}$ into these inequalities: 1. $x + y - 11 \ge 0 \implies \frac{11}{2} + \alpha - 11 \ge 0 \implies \alpha - \frac{11}{2} \ge 0 \implies \alpha \ge \frac{11}{2}$ 2. $x + 2y - 16 \ge 0 \implies \frac{11}{2} + 2\alpha - 16 \ge 0 \implies 2\alpha \ge 16 - \frac{11}{2} \implies 2\alpha \ge \frac{32 - 11}{2} \implies 2\alpha \ge \frac{21}{2} \implies \alpha \ge \frac{21}{4}$ 3. $2x + 3y - 29 \le 0 \implies 2\left(\frac{11}{2}\right) + 3\alpha - 29 \le 0 \implies 11 + 3\alpha - 29 \le 0 \implies 3\alpha - 18 \le 0 \implies 3\alpha \le 18 \implies \alpha \le 6$

Step 4: Determine the range of $\alpha$

We have the following conditions for $\alpha$: $\alpha \ge \frac{11}{2} = 5.5$ $\alpha \ge \frac{21}{4} = 5.25$ $\alpha \le 6$ Combining these inequalities, we get: $\alpha \ge \max(5.5, 5.25) \implies \alpha \ge 5.5$ So, the range for $\alpha$ is $5.5 \le \alpha \le 6$.

Step 5: Calculate the product of smallest and largest values

The smallest value of $\alpha$ is $\frac{11}{2}$ and the largest value of $\alpha$ is $6$. The product of these values is $\frac{11}{2} \times 6 = 11 \times 3 = 33$.

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