1. **Problem statement:** We have a rectangle ABCD with vertices A and B on the x-axis at points (-x,0) and (x,0), and vertices C and D on the curve $y=3e^{-x^2}$. We want to find the value of $x$ that maximizes the area of ABCD and then find the maximum area. 2. **Formula for area:** The width of the rectangle is the distance between A and B, which is $2x$. The height of the rectangle is the y-coordinate of points C and D, which is $3e^{-x^2}$. So, the area $A$ as a function of $x$ is: $$A(x) = \text{width} \times \text{height} = 2x \times 3e^{-x^2} = 6xe^{-x^2}$$ 3. **Maximizing the area:** To find the maximum area, we take the derivative of $A(x)$ with respect to $x$ and set it to zero. Using the product rule: $$A'(x) = 6 \left(e^{-x^2} + x \cdot \frac{d}{dx} e^{-x^2} \right)$$ The derivative of $e^{-x^2}$ is: $$\frac{d}{dx} e^{-x^2} = e^{-x^2} \cdot (-2x) = -2xe^{-x^2}$$ So, $$A'(x) = 6 \left(e^{-x^2} + x(-2xe^{-x^2})\right) = 6 \left(e^{-x^2} - 2x^2 e^{-x^2}\right) = 6e^{-x^2}(1 - 2x^2)$$ 4. **Setting the derivative to zero:** $$6e^{-x^2}(1 - 2x^2) = 0$$ Since $6e^{-x^2} \neq 0$ for all real $x$, we solve: $$1 - 2x^2 = 0 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$$ Since $x$ represents a positive distance, we take: $$x = \frac{1}{\sqrt{2}}$$ 5. **Second derivative test or interval chart:** - For $x < \frac{1}{\sqrt{2}}$, $1 - 2x^2 > 0$ so $A'(x) > 0$ (area increasing). - For $x > \frac{1}{\sqrt{2}}$, $1 - 2x^2 < 0$ so $A'(x) < 0$ (area decreasing). Thus, $x = \frac{1}{\sqrt{2}}$ is a maximum. 6. **Maximum area:** Substitute $x = \frac{1}{\sqrt{2}}$ into the area formula: $$A\left(\frac{1}{\sqrt{2}}\right) = 6 \cdot \frac{1}{\sqrt{2}} \cdot e^{-\left(\frac{1}{\sqrt{2}}\right)^2} = 6 \cdot \frac{1}{\sqrt{2}} \cdot e^{-\frac{1}{2}} = \frac{6}{\sqrt{2}} e^{-\frac{1}{2}}$$ This is the exact simplified form of the maximum area. --- **Summary:** - The value of $x$ that maximizes the area is $x = \frac{1}{\sqrt{2}}$. - The maximum area is $$A_{max} = \frac{6}{\sqrt{2}} e^{-\frac{1}{2}}$$. --- **Interval chart for $A'(x)$:** - $(-\infty, -\frac{1}{\sqrt{2}})$: $A'(x) < 0$ (decreasing) - $(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$: $A'(x) > 0$ (increasing) - $(\frac{1}{\sqrt{2}}, \infty)$: $A'(x) < 0$ (decreasing)