Question: Problem 6: Find the area of the largest rectangle that can be inscribed in the ellipse $4x^2 + 16y^2 = 16$
1. **State the problem:** We want to find the area of the largest rectangle that can fit inside the ellipse given by the equation $$4x^2 + 16y^2 = 16$$.
2. **Rewrite the ellipse equation in standard form:** Divide both sides by 16:
$$\frac{4x^2}{16} + \frac{16y^2}{16} = \frac{16}{16} \implies \frac{x^2}{4} + y^2 = 1$$
3. **Set up the rectangle:** Consider a rectangle centered at the origin with vertices at $(x,y)$, $(-x,y)$, $(-x,-y)$, and $(x,-y)$. The area of this rectangle is:
$$A = 2x \times 2y = 4xy$$
4. **Express $y$ in terms of $x$ using the ellipse equation:** From $$\frac{x^2}{4} + y^2 = 1$$ we get
$$y^2 = 1 - \frac{x^2}{4}$$
$$y = \sqrt{1 - \frac{x^2}{4}}$$
5. **Write the area as a function of $x$ only:**
$$A(x) = 4x \times \sqrt{1 - \frac{x^2}{4}}$$
6. **Maximize $A(x)$:** To find the maximum area, differentiate $A(x)$ with respect to $x$ and set the derivative equal to zero.
First, write $A(x)$ as:
$$A(x) = 4x \left(1 - \frac{x^2}{4}\right)^{1/2}$$
7. **Differentiate using the product rule:**
$$A'(x) = 4 \left(1 - \frac{x^2}{4}\right)^{1/2} + 4x \times \frac{1}{2} \left(1 - \frac{x^2}{4}\right)^{-1/2} \times \left(-\frac{x}{2}\right)$$
Simplify:
$$A'(x) = 4 \sqrt{1 - \frac{x^2}{4}} - \frac{4x^2}{4 \sqrt{1 - \frac{x^2}{4}}} = 4 \sqrt{1 - \frac{x^2}{4}} - \frac{x^2}{\sqrt{1 - \frac{x^2}{4}}}$$
8. **Set $A'(x) = 0$ and solve:**
$$4 \sqrt{1 - \frac{x^2}{4}} = \frac{x^2}{\sqrt{1 - \frac{x^2}{4}}}$$
Multiply both sides by $\sqrt{1 - \frac{x^2}{4}}$:
$$4 \left(1 - \frac{x^2}{4}\right) = x^2$$
Expand:
$$4 - x^2 = x^2$$
Combine like terms:
$$4 = 2x^2$$
$$x^2 = 2$$
9. **Find $x$ and $y$ values:**
$$x = \sqrt{2}$$
From step 4:
$$y = \sqrt{1 - \frac{x^2}{4}} = \sqrt{1 - \frac{2}{4}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}$$
10. **Calculate the maximum area:**
$$A_{max} = 4xy = 4 \times \sqrt{2} \times \frac{\sqrt{2}}{2} = 4 \times \sqrt{2} \times \frac{\sqrt{2}}{2} = 4 \times \frac{2}{2} = 4$$
**Final answer:** The area of the largest rectangle inscribed in the ellipse is $$4$$.