Let m, n be real numbers such that 0 leq m le | vedclass

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(C) The region is bounded by the lines $y = mx + 3$,$y = nx + 3$,and the $x$-axis $(y = 0)$.For $y = 0$,the $x$-intercepts are $x_1 = -\frac{3}{m}$ (f

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$3 \sqrt{3}$

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(C) The region is bounded by the lines $y = mx + 3$,$y = nx + 3$,and the $x$-axis $(y = 0)$.For $y = 0$,the $x$-intercepts are $x_1 = -\frac{3}{m}$ (for $m > 0$) and $x_2 = -\frac{3}{n}$ (for $n The intersection point of the two lines $y = mx + 3$ and $y = nx + 3$ is $(0, 3)$.The base of the triangle on the $x$-axis has length $b = |x_2 - x_1| = |-\frac{3}{n} - (-\frac{3}{m})| = 3 |\frac{1}{m} - \frac{1}{n}|$.The height of the triangle is $h = 3$.Area $A = \frac{1}{2} 	imes b 	imes h = \frac{1}{2} 	imes 3 |\frac{1}{m} - \frac{1}{n}| 	imes 3 = \frac{9}{2} (\frac{1}{m} - \frac{1}{n})$.To minimize the area,we maximize $m$ and $n$. Given $m \leq \sqrt{3}$ and $n \geq -\sqrt{3}$,we take $m = \sqrt{3}$ and $n = -\sqrt{3}$.Area $= \frac{9}{2} (\frac{1}{\sqrt{3}} - \frac{1}{-\sqrt{3}}) = \frac{9}{2} (\frac{2}{\sqrt{3}}) = \frac{9}{\sqrt{3}} = 3 \sqrt{3}$.

Let $m, n$ be real numbers such that $0 \leq m \leq \sqrt{3}$ and $-\sqrt{3} \leq n \leq 0$. The minimum possible area of the region of the plane consisting of points $(x, y)$ satisfying the inequalities $y \geq 0$,$y - 3 \leq mx$,and $y - 3 \leq nx$ is

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