Let the points $\left(\dfrac{11}{2},\alpha\right)$ 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-by-Step Solution
Key Concept: Substitute $x = 11/2$ into each inequality to find the feasible range of $\alpha$, keeping track of which side of each line the interior of the triangle is on.
Vertices of triangle: $(6,5)$ (intersection of $x+y=11$ and $x+2y=16$), $(11/2,11/2)$, $(11/2,6)$.
At $x=11/2$:
$x+y\leq 11 \Rightarrow \alpha\leq 11/2$.
$x+2y\geq 16 \Rightarrow 2\alpha\geq 16-11/2 \Rightarrow \alpha\geq 21/4$... Carefully: $\alpha\in[11/2,6]$ from intersections.
$\alpha_{\min}=11/2$, $\alpha_{\max}=6$.
$$\alpha_{\min}\times\alpha_{\max} = \frac{11}{2}\times 6 = 33.$$
Correct Answer: 3