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

The maximum area is obtained when the vertices are the points

(-√2, 0), (√2, 0), (√2, 4), and (-√2, 4).

This rectangle has base 2√2 and height 4, giving a maximum area of 8√2.

The area of a rectangle is expressed as shown:

A = xy

x is the length of the rectangle

y is the width

Given that y = 6 - x²

If its other two vertices above the x-axis and lying on the parabola, hence

the length of the rectangle will be 2x

The area of the rectangle will be A = 2x(6-x²)

If the dimension of the rectangle is at the maximum, hence dA/dx=0

A = 12x - 2x³

dA/dx = 12 - 6x²

0 = 12 - 6x²

6x² = 12

x² = 12/6

x² = 2

x = ±√2

Hence the maximum area is obtained when the vertices are the points

(-√2, 0), (√2, 0), (√2, 4), and (-√2, 4).

This rectangle has base 2√2 and height 4, giving a maximum area of 8√2.

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If one of the vertices on the x-axis is (x, 0), then the other vertex on the same axis is (-x, 0), so that the rectangle has base 2x. The other two vertices on the parabola are the points (x, 6 - x²) and (-x, 6 - x²), so the height of the rectangle is 6 - x².

Then the area of the rectangle is given by the function

Compute the critical points of A:

So the maximum area is obtained when the vertices are the points (-√2, 0), (√2, 0), (√2, 4), and (-√2, 4). This rectangle has base 2√2 and height 4, giving a maximum area of 8√2.

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