A rectangle in the first quadrant has one vertex as the origin, and other vertex on the curve y= -x^2 -2x +4. If the rectangle has one side lying on the x-axis, find the largest possible area of the rectangle.
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To find the largest possible area of the rectangle, we need to maximize the product of its length and width. Let's start by visualizing the problem.
We know that one vertex of the rectangle is at the origin (0,0). The other vertex lies on the curve y = -x^2 - 2x + 4. The x-coordinate of this vertex determines the length of the rectangle, and the y-coordinate determines the width.
Let's consider a general point (x, y) on the curve. The length of the rectangle is x units, and since one side lies on the x-axis, the width is y units.
The area of the rectangle is given by A = length * width = x * y.
To find the largest possible area, we need to find the maximum value of A.
Now, let's express y in terms of x using the given curve equation:
y = -x^2 - 2x + 4
To find the x-coordinate of the other vertex, we need to take the derivative of the curve equation and set it equal to zero to find any potential maximum or minimum points. We can do this by finding the critical points of the curve. Differentiating the curve equation with respect to x:
dy/dx = -2x - 2
Setting dy/dx = 0 and solving for x:
-2x - 2 = 0
-2x = 2
x = -1
So, the x-coordinate of the other vertex is -1. To find the y-coordinate, substitute x = -1 into the curve equation:
y = -(-1)^2 - 2(-1) + 4
y = -1 - (-2) + 4
y = 1 + 2 + 4
y = 7
Therefore, the other vertex of the rectangle is at (-1, 7).
Now, we know that one side of the rectangle lies on the x-axis, so the length of the rectangle is (-1) - 0 = 1 unit. The width is given by the y-coordinate of the other vertex, which is 7 units.
Finally, we can find the area of the rectangle:
A = length * width = 1 * 7 = 7 square units
So, the largest possible area of the rectangle is 7 square units.