A rectangle has one corner in quadrant I on the graph of �another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis. Find the maximum area of the rectangle.

To find the maximum area of the rectangle, we need to determine the dimensions that would result in the largest area possible.

Let's label the corner in quadrant I as (x, y). Since the other corner is at the origin, its coordinates are (0, 0). The third corner is on the positive y-axis, so it has coordinates (0, y). Similarly, the fourth corner is on the positive x-axis, so its coordinates are (x, 0).

The length of the rectangle is the distance between the points (0, 0) and (x, 0), which is simply x. The width of the rectangle is the distance between the points (0, 0) and (0, y), which is y.

The area of the rectangle is given by the formula: Area = length * width.

Substituting the values, we have: Area = x * y.

Since we are trying to find the maximum area, we need to express the area in terms of one variable. In this case, we can express y in terms of x using the equation of the line passing through the points (0, y), and (x, 0). The equation of a line passing through two points is given by: y = mx + b, where m is the slope and b is the y-intercept.

The slope (m) of the line passing through (0, y) and (x, 0) can be calculated as the change in y divided by the change in x: m = (0 - y) / (x - 0) = -y / x.

Substituting the values, we have the equation: y = (-y / x) * x + b.

Simplifying the equation, we get: y = -y + b.

To find the value of b, we can substitute the coordinates of either point (0, y) or (x, 0) into the equation. Let's use the point (0, y):

y = -y + b
y = -y + b
2y = b

So, b = 2y.

Substituting the value of b back into the equation, we now have: y = -y + 2y.

Simplifying further, we get: y = 2y.

Dividing both sides by y, we have: 1 = 2.

Since this is not possible, it means that a line passing through these points does not exist. Therefore, the area of the rectangle can be maximized when the rectangle becomes a square.

In a square, all sides are equal. Since the corner in quadrant I has coordinates (x, y), the side length of the square would be the minimum of the two values, which is min(x, y).

The area of a square is calculated by squaring the side length, so the maximum area of the rectangle is (min(x, y))^2.