A rectangle with sides parallel to the axes is inscribed in the region bounded by the axes and the line x+2y = 2. How do you find the maximum area of this triangle?
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To find the maximum area of the rectangle inscribed in the region bounded by the axes and the line (x+2y=2), you first need to find the coordinates of the vertices of the rectangle. Since the rectangle's sides are parallel to the axes, the vertices will be where the line intersects the axes.
The x-intercept is found by setting (y=0) in the equation (x+2y=2), which gives (x=2). The y-intercept is found by setting (x=0), yielding (2y=2) or (y=1).
So, the vertices of the rectangle are at (0, 1), (2, 0), (0, 0), and (2, 1).
The area of the rectangle is given by (A = \text{length} \times \text{width}). The length of the rectangle is the difference in x-coordinates of the vertices, which is (2-0=2), and the width is the difference in y-coordinates, which is (1-0=1).
So, the area (A = 2 \times 1 = 2). Therefore, the maximum area of the rectangle is (2).
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