Determine the maximum area of a rectangle inscribed in the ellipse x2/16+y2/36=1. Enter only the maximum area and do not include any units.

To find the maximum area of a rectangle inscribed in the ellipse \( \frac{x^2}{16} + \frac{y^2}{36} = 1 \), we can use the following approach.

The area of the rectangle is given by \( A = 2x \cdot 2y = 4xy \).

Given the ellipse equation \( \frac{x^2}{16} + \frac{y^2}{36} = 1 \), we can express \( y \) in terms of \( x \) as:

\[
y = \sqrt{36\left(1 - \frac{x^2}{16}\right)}
\]

The area becomes:

\[
A = 4x \sqrt{36\left(1 - \frac{x^2}{16}\right)} = 24x \sqrt{1 - \frac{x^2}{16}}
\]

To maximize the area, we take the derivative of \( A \) with respect to \( x \) and set it equal to zero:

\[
\frac{dA}{dx} = 24\sqrt{1 - \frac{x^2}{16}} - \frac{24x^2}{16\sqrt{1 - \frac{x^2}{16}}}
\]

Setting \( \frac{dA}{dx} = 0 \), we solve for \( x \):

\[
\sqrt{1 - \frac{x^2}{16}} = \frac{x^2}{16\sqrt{1 - \frac{x^2}{16}}}
\]

Squaring both sides and simplifying:

\[
1 - \frac{x^2}{16} = \frac{x^4}{256(1 - \frac{x^2}{16})}
\]

After solving the resulting equation, the maximum value of \( x \) is \( x = \sqrt{8} \).

Substituting \( x = \sqrt{8} \) back into the ellipse equation to find \( y \):

\[
y = \sqrt{36\left(1 - \frac{8}{16}\right)} = \sqrt{18}
\]

So, the maximum area is:

\[
A_{\text{max}} = 4xy = 4(\sqrt{8})(\sqrt{18}) = 24
\]

The maximum area of the rectangle is \( \boxed{24} \).

Discover how to maximize the area of a rectangle inscribed in the ellipse x^2/16 + y^2/36 = 1. Learn the mathematical approach involving the ellipse equation, area calculation, and derivative optimization. Suitable for college-level mathematics enthusiasts.

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