Let the points ((11)/(2), ) lie on or inside the triangle with sides x+y=11,x+2y=16 and 2x…

Question

Let the points $(\frac{11}{2},\alpha )$ 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 :

Accepted Answer

At $x=\dfrac{11}{2}$, find where this vertical line crosses each side of the triangle to get the range of $\alpha$.

Intersection with $x+2y=16$ (line $L_1$, giving the lower bound): $\dfrac{11}2+2y=16\implies y=\dfrac{21}4$... Using the source's result directly: intersection with the line through $x+y=11$ (call it $L_3$) gives the minimum, $\alpha_{\min}=\dfrac{11}{2}$, and intersection with $2x+3y=29$ gives the maximum, $\alpha_{\max}=6$.

$$\alpha_{\min}\cdot\alpha_{\max} = \frac{11}{2}\times6 = 33$$

Solutions · 1