1. **Problem statement:** Find the value of $x \geq 4$ that maximizes the area of the rectangle with vertices at $(4,0)$, $(x,0)$, $(4,-\frac{8}{4^3})$, and $(x,-\frac{8}{x^3})$ where $y=f(x)=-\frac{8}{x^3}$. 2. **Formula for area:** The area $A$ of the rectangle is given by the base times the height. The base is the horizontal distance between $x$ and 4, i.e., $|x-4|=x-4$ since $x\geq4$. The height is the vertical distance between $0$ and $f(x)=-\frac{8}{x^3}$, which is $\left| -\frac{8}{x^3} \right|=\frac{8}{x^3}$. So, $$A(x) = (x-4) \times \frac{8}{x^3} = \frac{8(x-4)}{x^3}.$$ 3. **Maximizing area:** To find the maximum, differentiate $A(x)$ with respect to $x$ and set to zero. $$A(x) = \frac{8(x-4)}{x^3} = 8(x-4)x^{-3}.$$ Using the product rule: $$A'(x) = 8 \left[ 1 \cdot x^{-3} + (x-4)(-3x^{-4}) \right] = 8 \left( x^{-3} - 3(x-4)x^{-4} \right).$$ Simplify inside the bracket: $$x^{-3} - 3(x-4)x^{-4} = \frac{1}{x^3} - \frac{3(x-4)}{x^4} = \frac{x - 3(x-4)}{x^4} = \frac{x - 3x + 12}{x^4} = \frac{-2x + 12}{x^4}.$$ So, $$A'(x) = 8 \times \frac{-2x + 12}{x^4} = \frac{8(-2x + 12)}{x^4} = \frac{8(12 - 2x)}{x^4}.$$ 4. **Set derivative to zero:** $$A'(x) = 0 \implies 8(12 - 2x) = 0 \implies 12 - 2x = 0 \implies 2x = 12 \implies x = 6.$$ 5. **Check endpoints and critical points:** - At $x=4$, $A(4) = \frac{8(4-4)}{4^3} = 0$. - At $x=6$, $$A(6) = \frac{8(6-4)}{6^3} = \frac{8 \times 2}{216} = \frac{16}{216} = \frac{2}{27} \approx 0.0741.$$ - As $x \to \infty$, $A(x) \to 0$ because $\frac{8(x-4)}{x^3} \approx \frac{8x}{x^3} = \frac{8}{x^2} \to 0$. Thus, $x=6$ gives the maximum area. **Final answer:** $\boxed{6}$ (Option B).