A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions?

Steps to Solve

1. Determine the relationship between the dimensions of the rectangle and the semicircle

   Let the rectangle have a width $w$ and a height $h$. The semicircle's equation is $x^2 + y^2 = r^2$ with $r = 2$. The upper part of the semicircle can be described as $y = \sqrt{4 - x^2}$. The width of the rectangle will be $2x$ (since it extends equally in both directions from the center), and the height will be $h = \sqrt{4 - x^2}$.

2. Express the area of the rectangle

   The area $A$ of the rectangle can be expressed as: $$ A = w \cdot h = 2x \cdot \sqrt{4 - x^2} $$

3. Differentiate the area to find critical points

   First, rewrite the area: $$ A = 2x\sqrt{4 - x^2} $$

   To differentiate, use the product rule: $$ \frac{dA}{dx} = 2 \sqrt{4 - x^2} + 2x \cdot \frac{d}{dx}(\sqrt{4 - x^2}) $$

   The derivative of $\sqrt{4 - x^2}$ is: $$ \frac{d}{dx}(\sqrt{4 - x^2}) = \frac{-x}{\sqrt{4 - x^2}} $$

   Thus, $$ \frac{dA}{dx} = 2\sqrt{4 - x^2} - \frac{2x^2}{\sqrt{4 - x^2}} $$

   Set $\frac{dA}{dx} = 0$ to find critical points.

4. Solve the equation for $x$

   Multiply both sides by $\sqrt{4 - x^2}$: $$ 2(4 - x^2) - 2x^2 = 0 $$ Simplifying gives: $$ 8 - 4x^2 = 0 \implies x^2 = 2 \implies x = \sqrt{2} $$

5. Calculate the dimensions and area

   Using $x = \sqrt{2}$, find $h$: $$ h = \sqrt{4 - (\sqrt{2})^2} = \sqrt{4 - 2} = \sqrt{2} $$

   Now, find the width of the rectangle: $$ w = 2x = 2\sqrt{2} $$

   Therefore, the maximum area is: $$ A = 2x \cdot h = 2\sqrt{2} \cdot \sqrt{2} = 4 $$