### Related Formula For a vertical line x = x_0$x = x_0$ crossing a bounded region, the valid coordinates of y$y$ sit between the boundary lines intersecting that specific line vertical plane. ### Core Logic The point given is fixed at x = frac112 = 5.5$x = \frac{11}{2} = 5.5$[cite: 3285, 3925]. We evaluate the values of y$y$ along this vertical line segment across each boundary edge.

### Step 1: Evaluate Intersections Substitute x = frac112$x = \frac{11}{2}$ into the three linear constraints: 1. From x + y = 11$x + y = 11$: frac112 + y = 11 Rightarrow y = 11 - 5.5 = 5.5$\frac{11}{2} + y = 11 \Rightarrow y = 11 - 5.5 = 5.5$ 2. From x + 2y = 16$x + 2y = 16$: frac112 + 2y = 16 Rightarrow 2y = 16 - 5.5 = 10.5 Rightarrow y = 5.25$\frac{11}{2} + 2y = 16 \Rightarrow 2y = 16 - 5.5 = 10.5 \Rightarrow y = 5.25$ 3. From 2x + 3y = 29$2x + 3y = 29$: 2left(frac112right) + 3y = 29 Rightarrow 11 + 3y = 29 Rightarrow 3y = 18 Rightarrow y = 6$2\left(\frac{11}{2}\right) + 3y = 29 \Rightarrow 11 + 3y = 29 \Rightarrow 3y = 18 \Rightarrow y = 6$ ### Step 2: Define Extrema and Multiply Checking the internal region of the \triangle bounded by these lines [cite: 3286, 3287], the valid range for alpha$\alpha$ along the section line is delimited by y = 5.5$y = 5.5$ and y = 6$y = 6$: alpha_min = frac112 = 5.5$\alpha_{\min} = \frac{11}{2} = 5.5$ alpha_max = 6$\alpha_{\max} = 6$ The product of the limits is : alpha_min cdot alpha_max = frac112 times 6 = 33$\alpha_{\min} \cdot \alpha_{\max} = \frac{11}{2} \times 6 = 33$ ### Pattern Recognition Instead of drawing full coordinate diagrams or computing all three vertex points, evaluating values directly at the fixed coordinate constraint x = 5.5$x = 5.5$ saves time during multi-line area problems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Linear Inequalities Class 11 Mathematics: Straight Lines