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

Question

Aicharya · Accepted Answer

Clearly, $x=\frac{11}{2}$ intersect $x+y-11=0$ at $\left(\frac{11}{2}, \frac{11}{2}\right)$ and $2 x+3 y-29=0$ at $\left(\frac{11}{2}, 6\right) \Rightarrow \alpha=\left[\frac{11}{2}, 6\right]$
$\alpha_{\min } \cdot \alpha_{\max }=\frac{11}{2} \cdot 6=33$ *

Let the points $\left(\frac{11}{2}, \alpha\right)$ lie on or inside the triangle with sides $x+y=11, x+2…

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