Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola. (Round your answers to the nearest hundredth.)

Complete question is;

Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola. (Round your answers to the nearest hundredth.) y = 6 - x²

height = 4 and base = 2√2

Step-by-step explanation:

Area of a rectangle is given by;

A = base(b) × height(h)

To start off, we will first have to consider half of the rectangle.

Due to the fact that it is bounded by a parabola which is symmetric over the y-axis, it means that the rectangle will have the same area on the right and left hand sides.

So if we maximize the area of the first quadrant, it means that we are also maximizing the area of the entire rectangle.

Now, since we are dealing with the base and other 2 vertices on the x-axis, it means that the base(b) of this half rectangle will be x.

The height of this rectangle is the y coordinate of the corner that is directly above the x coordinate, which is also on the rectangle.

Therefore, the coordinates of the upper corner will be (x, y), or (x, 6 - x²). Thus our half base is x and full height is 6 - x². Area will be;

A = x(6 - x²)

A = 6x - x³

We will maximize this area by finding the derivative and equating to zero.

Thus;

A' = 6 - 3x²

At A' = 0,we have;

6 - 3x² = 0

3x² = 6

x² = 6/3

x² = 2

x = √2

Thus, from y = 6 - x², we have;

y = 6 - (√2)²

y = 6 - 2

y = 4

Since x is half base, it means our full width is 2x = 2 × √2 = 2√2

So, height = 4 and base = 2√2