1. **Problem statement:** We have 17 pieces of paper shaped as circular sectors with radius 12 cm, used to form a right circular cone with height $h$. We want to find the maximum volume of this cone. 2. **Formula for volume of a cone:** $$V = \frac{1}{3} \pi r^2 h$$ where $r$ is the base radius and $h$ is the height. 3. **Relating the sector to the cone:** The sector's arc length forms the cone's base circumference: $$\text{arc length} = 2 \pi r$$ The arc length of one sector is: $$L = \theta \times 12$$ where $\theta$ is the central angle in radians. 4. **Total arc length for 17 sectors:** $$L_{total} = 17 \times L = 17 \theta \times 12$$ This equals the circumference of the cone base: $$17 \theta \times 12 = 2 \pi r \Rightarrow r = \frac{17 \theta \times 12}{2 \pi} = \frac{102 \theta}{2 \pi} = \frac{51 \theta}{\pi}$$ 5. **Height and radius relation:** The slant height of the cone equals the radius of the sector, which is 12 cm: $$l = 12 = \sqrt{r^2 + h^2} \Rightarrow h = \sqrt{12^2 - r^2} = \sqrt{144 - r^2}$$ 6. **Volume as a function of $r$:** $$V = \frac{1}{3} \pi r^2 \sqrt{144 - r^2}$$ 7. **Maximizing volume:** Set derivative $\frac{dV}{dr} = 0$ and solve for $r$: Let $$V = \frac{1}{3} \pi r^2 (144 - r^2)^{1/2}$$ Using the product and chain rules: $$\frac{dV}{dr} = \frac{1}{3} \pi \left(2r \sqrt{144 - r^2} + r^2 \frac{1}{2} (144 - r^2)^{-1/2} (-2r) \right)$$ Simplify: $$= \frac{1}{3} \pi \left(2r \sqrt{144 - r^2} - r^3 (144 - r^2)^{-1/2} \right)$$ Set equal to zero: $$2r \sqrt{144 - r^2} = r^3 (144 - r^2)^{-1/2}$$ Multiply both sides by $(144 - r^2)^{1/2}$: $$2r (144 - r^2) = r^3$$ Divide both sides by $r$ (assuming $r \neq 0$): $$2 (144 - r^2) = r^2 \Rightarrow 288 - 2r^2 = r^2 \Rightarrow 288 = 3r^2 \Rightarrow r^2 = 96 \Rightarrow r = \sqrt{96} = 4 \sqrt{6}$$ 8. **Calculate height $h$:** $$h = \sqrt{144 - 96} = \sqrt{48} = 4 \sqrt{3}$$ 9. **Calculate maximum volume:** $$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (96)(4 \sqrt{3}) = \frac{1}{3} \pi \times 384 \sqrt{3} = 128 \pi \sqrt{3}$$ 10. **Compare with options:** Option د) $\pi 3 \sqrt{128}$ can be rewritten: $$3 \sqrt{128} = 3 \times 8 \sqrt{2} = 24 \sqrt{2}$$ Our volume is $128 \pi \sqrt{3}$, which is not equal to any option exactly, but the closest is option د) if we consider a typo or simplification. **Final answer:** $$\boxed{128 \pi \sqrt{3} \text{ cm}^3}$$ --- **Note:** The second problem about the function $d$ and its graph is not solved as per instructions to solve only the first problem.