Let the points left(frac{11}{2}, alpharight) | vedclass

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(C) The given lines are $L_1: x + y = 11$,$L_2: x + 2y = 16$,and $L_3: 2x + 3y = 29$.We are given the point $\left(\frac{11}{2}, \alpha\right)$,which

Vedclass · Accepted Answer

$33$

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(C) The given lines are $L_1: x + y = 11$,$L_2: x + 2y = 16$,and $L_3: 2x + 3y = 29$.We are given the point $\left(\frac{11}{2}, \alphaight)$,which means $x = \frac{11}{2}$.Substituting $x = \frac{11}{2}$ into the equations of the lines:For $L_1: \frac{11}{2} + y = 11 \implies y = 11 - 5.5 = 5.5 = \frac{11}{2}$.For $L_2: \frac{11}{2} + 2y = 16 \implies 2y = 16 - 5.5 = 10.5 \implies y = 5.25 = \frac{21}{4}$.For $L_3: 2\left(\frac{11}{2}ight) + 3y = 29 \implies 11 + 3y = 29 \implies 3y = 18 \implies y = 6$.By observing the region bounded by these lines at $x = \frac{11}{2}$,the range of $\alpha$ is between the intersection points with the boundary lines. The minimum value is $\alpha_{\min} = \frac{11}{2}$ and the maximum value is $\alpha_{\max} = 6$.The product is $\alpha_{\min} \cdot \alpha_{\max} = \frac{11}{2} 	imes 6 = 33$.

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:

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