Suppose the length of the rectangle = 2x i..e. from -x to + x; since the upper corners are on the parabola, the breadth is given by b = 3 - x^2 so the area is A = 2x(3 - x^2) = 6x - 2x^3; dA/dx = 6 - 6x^2 = 6(x^2 - 1) so dA/dx = 0 for a Stationary Point i..e. x^2 = 1 so x = 1 or x = -1 - these obviously give the same rectangle, since the length was fro -x to + x. I'll leave it up to you to use a Table of Signs (or the Second Derivative) to verify that this is indeed a maximum.

y = 3 - x²

Area = A

A(x,y) = xy

Subbing 3 - x² for y,

A(x) = x(3 - x²)

A(x) = 3x - x³

Max. Area: A'(x) = 0:

A'(x) = 3 - 3x²

3 - 3x² = 0

3x² = 3

x² = 3 / 3

x² = 1

x = √1

x = ± 1

and

y = 3 - (1)²

y = 3 - 1

y = 2

Since x crosses the y-axis,

x = 2x:

x = 2(1)

x = 2

Dimensions:

Length = 2

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Width = 2

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