Question

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:

• A. 22
• B. 55
• C. 33
• D. 44

Answer 1

Answer: (C)

33

Answer 2

The vertices of the triangle are A(6,5), B(4,7), and C(10,3). A point (x,y) lies on or inside the triangle if it satisfies:

1. $x+y \ge 11$
2. $x+2y \ge 16$
3. $2x+3y \le 29$

For the point $(\frac{11}{2}, \alpha)$, we have $x = 5.5$. Substituting into the inequalities:

1. $5.5 + \alpha \ge 11 \implies \alpha \ge 5.5$
2. $5.5 + 2\alpha \ge 16 \implies 2\alpha \ge 10.5 \implies \alpha \ge 5.25$
3. $2(5.5) + 3\alpha \le 29 \implies 11 + 3\alpha \le 29 \implies 3\alpha \le 18 \implies \alpha \le 6$

Combining these, we get $5.5 \le \alpha \le 6$. The smallest value of $\alpha$ is $5.5$ and the largest value is $6$. The product is $5.5 \times 6 = \frac{11}{2} \times 6 = 33$.