To solve this problem, we need to determine the dimensions of the rectangle inscribed in the given ellipse. The equation of the ellipse is:

$\frac{x^2}{25} + \frac{y^2}{16} = 1$

Step 1: General Information about the Inscribed Rectangle

A rectangle inscribed in an ellipse has its vertices touching the ellipse at points. Let the dimensions of the rectangle be $2x$ and $2y$ (where $x$ and $y$ are the coordinates of the vertices on the ellipse in the first quadrant).

Step 2: The Relationship Between $x$ and $y$ on the Ellipse

The equation of the ellipse ensures that any point $(x, y)$ on the ellipse satisfies: $\frac{x^2}{25} + \frac{y^2}{16} = 1$

Step 3: Maximizing the Area of the Rectangle

The area of the rectangle is: $A = 2x \cdot 2y = 4xy$ We aim to maximize $A = 4xy$ under the constraint of the ellipse equation.

Step 4: Solve Using Substitution

From the ellipse equation, solve for $y^2$: $y^2 = 16 \left( 1 - \frac{x^2}{25} \right)$ $y = \sqrt{16 \left( 1 - \frac{x^2}{25} \right)}$

Substitute $y$ into $A$: $A = 4x \cdot \sqrt{16 \left( 1 - \frac{x^2}{25} \right)}$ $A = 4x \cdot 4 \sqrt{\left( 1 - \frac{x^2}{25} \right)}$ $A = 16x \sqrt{\left( 1 - \frac{x^2}{25} \right)}$

Step 5: Maximize $A$

Let: $f(x) = 16x \sqrt{\left( 1 - \frac{x^2}{25} \right)}$ Take the derivative of $f(x)$ with respect to $x$, set it to 0, and solve: $f'(x) = 16 \sqrt{\left( 1 - \frac{x^2}{25} \right)} + 16x \cdot \frac{-1}{2} \cdot \left( 1 - \frac{x^2}{25} \right)^{-\frac{1}{2}} \cdot \left( \frac{2x}{25} \right)$ Simplify: $f'(x) = 16 \sqrt{\left( 1 - \frac{x^2}{25} \right)} - \frac{16x^2}{25 \sqrt{\left( 1 - \frac{x^2}{25} \right)}}$ Set $f'(x) = 0$: $16 \sqrt{\left( 1 - \frac{x^2}{25} \right)} = \frac{16x^2}{25 \sqrt{\left( 1 - \frac{x^2}{25} \right)}}$ Multiply through by $\sqrt{\left( 1 - \frac{x^2}{25} \right)}$: $16 \left( 1 - \frac{x^2}{25} \right) = \frac{16x^2}{25}$ $16 - \frac{16x^2}{25} = \frac{16x^2}{25}$ $16 = \frac{32x^2}{25}$ $x^2 = \frac{400}{32} = 12.5$ $x = \sqrt{12.5}$

Step 6: Solve for $y$

Substitute $x^2 = 12.5$ into the ellipse equation: $\frac{12.5}{25} + \frac{y^2}{16} = 1$ $\frac{1}{2} + \frac{y^2}{16} = 1$ $\frac{y^2}{16} = \frac{1}{2}$ $y^2 = 8$ $y = \sqrt{8}$

Final Dimensions of the Rectangle

The dimensions of the rectangle are: $2x = 2\sqrt{12.5}, \quad 2y = 2\sqrt{8}$

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Related Questions

1. How do you find the maximum area for rectangles inscribed in different ellipses?
2. What happens if the ellipse has a rotated axis instead of the standard one?
3. How does the area change if you consider only one quadrant of the rectangle?
4. Can this optimization method apply to shapes other than rectangles (e.g., triangles)?
5. How does changing the coefficients in the ellipse equation affect the dimensions of the rectangle?

Tip

When solving optimization problems, always clearly define the function to be maximized and use constraints systematically!