1. **State the problem:** Find the maximum area of a rectangle inscribed in the region bounded by the curve $y = 8 - x^3$, the positive $x$-axis, and the positive $y$-axis. 2. **Set up the problem:** The rectangle's vertices are at $(0,0)$, $(x,0)$, $(x,y)$, and $(0,y)$ where $y = 8 - x^3$. 3. **Area formula:** The area $A$ of the rectangle is given by $$A = x \times y = x(8 - x^3) = 8x - x^4$$ 4. **Find critical points:** To maximize $A$, differentiate with respect to $x$ and set to zero: $$\frac{dA}{dx} = 8 - 4x^3 = 0$$ 5. **Solve for $x$:** $$8 = 4x^3$$ $$\Rightarrow \cancel{4} \times 2 = \cancel{4} x^3$$ $$2 = x^3$$ $$\Rightarrow x = \sqrt[3]{2}$$ 6. **Find corresponding $y$:** $$y = 8 - (\sqrt[3]{2})^3 = 8 - 2 = 6$$ 7. **Calculate maximum area:** $$A_{max} = x \times y = \sqrt[3]{2} \times 6 = 6 \sqrt[3]{2}$$ 8. **Check endpoints:** At $x=0$, $A=0$; at $x=2$, $A=8(2) - 2^4 = 16 - 16 = 0$. Thus, the maximum area is $6 \sqrt[3]{2}$. **Final answer:** $$\boxed{6 \sqrt[3]{2}}$$