Arectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=6-x62. what are the dimensions of such a rectangle with the greatest possible area?

So we have a rectangle, whose area is A=bh. We don't know any of those, so we've got a bit of thinking to do.  (Wish I could draw a picture) First of all, we only need to consider half the rectangle. Since it is bounded by a parabola that is symmetric over the y-axis (ie not moved left or right), the rectangle has the same area on the right and left. So let's just consider the first quadrant. Note that if we maximize this area, we maximize the entire rectangle's area. The width of this half rectangle is x (it's the x coordinate). The height of this rectangle is whatever the y coordinate of the corner is. That corner is directly above the x coordinate, and is on the rectangle. So the upper corner must have coordinates of (x, y), or (x, 8-x2). Those will serve as our base and height. A = x (8 - x2)A = 8x - x3 Maximize implies derivative = 0, so... dA = 8 - 3x20  = 8 - 3x23x2 = 8x2 = 8/3x = ±1.633 Notice that this is giving us two answers -- one of them is on the left (second quadrant), while the other is on the right. The y coordinates will be the same for both, due to the nature of this parabola. y = 8 - (1.633)2y = 8 - (8/3)y = 5.333 Our upper corners are (±1.633, 5.333). The dimensions are: Height: 5.333Width = (2 · 1.633) = 3.266 :)

The answer is idk yo