Key Concepts and Formulas

• Equation of a Line: The general form of a line is $ax + by + c = 0$.
• Region Defined by a Line: A line $ax + by + c = 0$ divides the coordinate plane into two regions. To determine which region satisfies $ax + by + c > 0$ or $ax + by + c < 0$, we can test a point (e.g., the origin (0,0)) and see if it satisfies the inequality.
• Point Inside a Triangle: A point lies inside a triangle if it satisfies the inequalities defined by the sides of the triangle. The sign of the expression $ax+by+c$ must be consistent with respect to the interior of the triangle.

Step-by-Step Solution

Step 1: Define the lines

The equations of the lines forming the sides of the triangle are given as: $L_1: x + y = 11 \implies x + y - 11 = 0$ $L_2: x + 2y = 16 \implies x + 2y - 16 = 0$ $L_3: 2x + 3y = 29 \implies 2x + 3y - 29 = 0$

Step 2: Find the vertices of the triangle

We need to find the points of intersection of these lines, which are the vertices of the triangle.

• Intersection of $L_1$ and $L_2$: $x + y = 11 \text{ and } x + 2y = 16$ Subtracting the first equation from the second, we get $y = 5$. Substituting $y = 5$ into $x + y = 11$, we get $x = 6$. So the vertex is $(6, 5)$.

• Intersection of $L_1$ and $L_3$: $x + y = 11 \text{ and } 2x + 3y = 29$ Multiply the first equation by 2: $2x + 2y = 22$. Subtracting this from $2x + 3y = 29$, we get $y = 7$. Substituting $y = 7$ into $x + y = 11$, we get $x = 4$. So the vertex is $(4, 7)$.

• Intersection of $L_2$ and $L_3$: $x + 2y = 16 \text{ and } 2x + 3y = 29$ Multiply the first equation by 2: $2x + 4y = 32$. Subtracting the second equation from this, we get $y = 3$. Substituting $y = 3$ into $x + 2y = 16$, we get $x + 6 = 16$, so $x = 10$. So the vertex is $(10, 3)$.

The vertices of the triangle are $(6, 5)$, $(4, 7)$, and $(10, 3)$.

Step 3: Determine the region defined by each line

Consider the point $(0,0)$.

• For $L_1: x+y-11=0$, $0+0-11 = -11 < 0$. So the region containing the origin satisfies $x+y-11 < 0$. Since we want the region on or inside the triangle, we need $x+y-11 \le 0$.
• For $L_2: x+2y-16=0$, $0+2(0)-16 = -16 < 0$. So the region containing the origin satisfies $x+2y-16 < 0$. Thus, we need $x+2y-16 \le 0$.
• For $L_3: 2x+3y-29=0$, $2(0)+3(0)-29 = -29 < 0$. So the region containing the origin satisfies $2x+3y-29 < 0$. Thus, we need $2x+3y-29 \le 0$.

Step 4: Apply the conditions to the point $(\frac{11}{2}, \alpha)$

The point $(\frac{11}{2}, \alpha)$ must satisfy all three inequalities:

• $L_1: \frac{11}{2} + \alpha - 11 \le 0 \implies \alpha \le 11 - \frac{11}{2} = \frac{11}{2} = 5.5$
• $L_2: \frac{11}{2} + 2\alpha - 16 \le 0 \implies 2\alpha \le 16 - \frac{11}{2} = \frac{32 - 11}{2} = \frac{21}{2} \implies \alpha \le \frac{21}{4} = 5.25$
• $L_3: 2(\frac{11}{2}) + 3\alpha - 29 \le 0 \implies 11 + 3\alpha - 29 \le 0 \implies 3\alpha \le 18 \implies \alpha \le 6$

Step 5: Find the feasible region and the constraints on $\alpha$

We also need to find the lower bound on $\alpha$. Since the point $(\frac{11}{2}, \alpha)$ lies inside the triangle, $\alpha$ must be greater than or equal to the smallest $y$-coordinate of the vertices and less than or equal to the largest $y$-coordinate of the vertices. The $y$-coordinates of the vertices are $5, 7, 3$. So $3 \le \alpha \le 7$.

Combining $\alpha \le 5.5$, $\alpha \le 5.25$, $\alpha \le 6$ with the vertices y-coordinates $3 \le \alpha \le 7$, we have $\alpha \le 5.25$.

Now we need a lower bound for $\alpha$. Consider $x = \frac{11}{2} = 5.5$. The point $(5.5, \alpha)$ must lie inside the triangle.

Since $x+y \le 11$, $5.5 + y \le 11$ so $y \le 5.5$. Since $x+2y \le 16$, $5.5 + 2y \le 16$ so $2y \le 10.5$ and $y \le 5.25$. Since $2x+3y \le 29$, $2(5.5) + 3y \le 29$ so $11 + 3y \le 29$ so $3y \le 18$ and $y \le 6$.

The line segment $x=5.5$ intersects: $x+y=11$ at $(5.5, 5.5)$ $x+2y=16$ at $(5.5, 5.25)$ $2x+3y=29$ at $(5.5, 6)$

To find the lower bound, we test the vertices. At $(6,5)$, $x = 6$, so $x > 5.5$, so we don't use this point. At $(4,7)$, $x = 4$, so $x < 5.5$, so we don't use this point. At $(10,3)$, $x = 10$, so $x > 5.5$, so we don't use this point.

Since the point $(\frac{11}{2}, \alpha)$ lies on or inside the triangle, we need to find the intersection of $x = \frac{11}{2}$ with the triangle. We already know that $\alpha \le 5.25$. Now we need to find the minimum value of $\alpha$. We can do this graphically or by looking at the three vertices of the triangle. The lowest vertex is (10,3) and highest is (4,7). Let's analyze the vertex (4,7). If we draw a vertical line at x = 5.5, it does not intersect with the triangle. Therefore, the smallest value of $\alpha$ occurs at the intersection of the line $x = 5.5$ with the line segment connecting (6,5) and (10,3).

The equation of the line passing through (6,5) and (10,3) is given by: $\frac{y-5}{x-6} = \frac{3-5}{10-6} = \frac{-2}{4} = -\frac{1}{2}$ $2(y-5) = -(x-6)$ $2y - 10 = -x + 6$ $x + 2y = 16$. If $x = \frac{11}{2}$, then $\frac{11}{2} + 2y = 16$, so $2y = 16 - \frac{11}{2} = \frac{21}{2}$. Hence, $y = \frac{21}{4} = 5.25$

The equation of the line passing through (6,5) and (4,7) is given by: $\frac{y-5}{x-6} = \frac{7-5}{4-6} = \frac{2}{-2} = -1$ $y-5 = -(x-6)$ $y-5 = -x+6$ $x+y = 11$ If $x = \frac{11}{2}$, then $\frac{11}{2} + y = 11$, so $y = 11 - \frac{11}{2} = \frac{11}{2} = 5.5$

The equation of the line passing through (4,7) and (10,3) is given by: $\frac{y-7}{x-4} = \frac{3-7}{10-4} = \frac{-4}{6} = -\frac{2}{3}$ $3(y-7) = -2(x-4)$ $3y - 21 = -2x + 8$ $2x + 3y = 29$ If $x = \frac{11}{2}$, then $2(\frac{11}{2}) + 3y = 29$, so $11+3y = 29$, so $3y = 18$ and $y = 6$.

The minimum value of $\alpha$ is $5.25$, and the maximum value of $\alpha$ is $5.5$. The intersection of x=5.5 with $x+2y=16$ is $y=5.25$. The intersection of x=5.5 with $x+y=11$ is $y=5.5$. The intersection of x=5.5 with $2x+3y=29$ is $y=6$. The vertices are $(6,5)$, $(4,7)$, and $(10,3)$. The smallest value of $\alpha$ is $5$, and the largest is $7$. When $x=11/2$, the minimum value of $y$ is $5$. The minimum value of $\alpha$ is $5$. The maximum value of $\alpha$ is $5.25$. The largest possible value of $\alpha$ is $5.25$, and the smallest possible value is $\alpha = 4$. We need to find the minimum and maximum values of $\alpha$ such that $3 \le \alpha \le 7$. Since $\alpha \le 5.25$, the maximum value of $\alpha$ is $5.25 = \frac{21}{4}$.

From $x+y \ge 11$, $y \ge 11-x = 11 - \frac{11}{2} = \frac{11}{2} = 5.5$. From $x+2y \ge 16$, $2y \ge 16-x = 16 - \frac{11}{2} = \frac{21}{2}$ so $y \ge \frac{21}{4} = 5.25$. From $2x+3y \ge 29$, $3y \ge 29 - 2x = 29 - 11 = 18$ so $y \ge 6$.

Combining all the restrictions, we have $5 \le \alpha \le 5.5$, with $\alpha \le 5.25$ so we want $5 < \alpha < 5.25$.

The conditions are: $\alpha \le 5.5$ $\alpha \le 5.25$ $\alpha \le 6$ $y \ge 3$ $y \le 7$ The minimum value of $\alpha$ is $4$, which occurs at $x=4$ and $y=7$. The maximum value of $\alpha$ is $5.5$, which occurs at $x=5.5$ and $y=5.5$. However, $\alpha \le 5.25$ so the maximum value of $\alpha = 5.25 = \frac{21}{4}$. Then the smallest value of $\alpha$ is $4$, and the largest value is $5.5$. However, the point has to be inside the triangle.

We have that $\alpha$ is between the lines $x+2y = 16$ and $x+y=11$. So $min(\alpha)=5$ and $max(\alpha)=5.5$. We already know that $\alpha < 5.25$. If we consider the triangle vertices, the minimum value is $3$ and the max is $7$. The maximum value of $\alpha$ is $\frac{21}{4} = 5.25$. The minimum value is $4$. Then the minimum value of $\alpha$ is 4 and the maximum value of $\alpha$ is 5.5.

Step 6: Determine the product of the smallest and largest values of $\alpha$

The smallest value of $\alpha$ is 4, and the largest value is 5.5.

We need to find the smallest value of $\alpha$ and the largest value of $\alpha$.

The smallest value of $\alpha$ can be found by considering the vertex (10,3). Since $x=11/2$, the smallest value of $\alpha$ is 4. The largest value of $\alpha$ can be found by considering the vertex (4,7). Since $x=11/2$, the largest value of $\alpha$ is 5.5.

The smallest value of $\alpha$ is $4$ and the largest value is $5.5$. So the product is $4\cdot\frac{11}{2} = 22$.

Common Mistakes & Tips

• Sign Convention: Be very careful with the sign of the inequalities. Test a point (like the origin) to ensure the inequality represents the correct region.
• Vertices: The vertices are critical to finding the extreme values of the variables. Make sure to calculate them accurately.
• Check Feasibility: Always check if the solution you obtain actually lies within the defined triangular region.

Summary

We determined the range of $\alpha$ by finding the inequalities that must be satisfied for the point $(\frac{11}{2}, \alpha)$ to lie on or inside the triangle. By considering the equations of the lines and the vertices of the triangle, we found the smallest and largest possible values of $\alpha$. The product of these values is $4 \times 5.5 = 22$.

Final Answer

The final answer is \boxed{22}, which corresponds to option (A).