1. **State the problem:** We want to find the largest area of a rectangle inscribed in a semicircle of radius 2. The rectangle's base lies along the diameter from $-x$ to $x$, and its height is the $y$-value on the semicircle. 2. **Set up the function for the area:** The semicircle is described by $$y = \sqrt{4 - x^2}$$ where $x$ ranges from 0 to 2. The rectangle's width is $2x$ and height is $y$, so the area $A$ is: $$A = \text{width} \times \text{height} = 2x \times \sqrt{4 - x^2}$$ 3. **Express the area function:** $$A(x) = 2x \sqrt{4 - x^2}$$ 4. **Find the critical points by differentiating:** Use the product rule and chain rule: $$A'(x) = 2 \sqrt{4 - x^2} + 2x \cdot \frac{d}{dx} \sqrt{4 - x^2}$$ Calculate derivative inside: $$\frac{d}{dx} \sqrt{4 - x^2} = \frac{1}{2\sqrt{4 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{4 - x^2}}$$ So, $$A'(x) = 2 \sqrt{4 - x^2} + 2x \left( \frac{-x}{\sqrt{4 - x^2}} \right) = 2 \sqrt{4 - x^2} - \frac{2x^2}{\sqrt{4 - x^2}}$$ 5. **Combine terms over common denominator:** $$A'(x) = \frac{2(4 - x^2) - 2x^2}{\sqrt{4 - x^2}} = \frac{8 - 2x^2 - 2x^2}{\sqrt{4 - x^2}} = \frac{8 - 4x^2}{\sqrt{4 - x^2}}$$ 6. **Set derivative equal to zero to find critical points:** $$\frac{8 - 4x^2}{\sqrt{4 - x^2}} = 0$$ Since denominator $\sqrt{4 - x^2} > 0$ for $0 < x < 2$, numerator must be zero: $$8 - 4x^2 = 0$$ 7. **Solve for $x$:** $$8 = 4x^2$$ $$\cancel{4} \times 2 = \cancel{4} x^2$$ $$2 = x^2$$ $$x = \sqrt{2}$$ (only positive root since $x > 0$) 8. **Find the corresponding height $y$:** $$y = \sqrt{4 - x^2} = \sqrt{4 - 2} = \sqrt{2}$$ 9. **Calculate the maximum area:** $$A = 2x \times y = 2 \times \sqrt{2} \times \sqrt{2} = 2 \times 2 = 4$$ 10. **Answer:** The largest area is $4$ and the rectangle's dimensions are width $2\sqrt{2}$ and height $\sqrt{2}$.