1. **State the problem:** A cone-shaped cup must hold a volume of 100 cm³. We want to find the height $h$ and radius $r$ of the cone that minimize the surface area, which corresponds to the amount of paper used. 2. **Write the known formulas:** - Volume of a cone: $$ V = \frac{1}{3} \pi r^2 h $$ - Surface area (excluding the base, since paper covers the lateral surface only) is: $$ A = \pi r s $$ where $s$ is the slant height. - The slant height $s$ relates to $r$ and $h$ by: $$ s = \sqrt{r^2 + h^2} $$ 3. **Use the volume constraint to express $h$ in terms of $r$:** Given $V=100$, $$ 100 = \frac{1}{3} \pi r^2 h \implies h = \frac{300}{\pi r^2} $$ 4. **Express surface area $A$ as a function of $r$ only:** Substitute $h$ into $s$: $$ s = \sqrt{r^2 + \left(\frac{300}{\pi r^2}\right)^2} = \sqrt{r^2 + \frac{90000}{\pi^2 r^4}} $$ Surface area: $$ A(r) = \pi r s = \pi r \sqrt{r^2 + \frac{90000}{\pi^2 r^4}} $$ Simplify inside the square root: $$ r^2 + \frac{90000}{\pi^2 r^4} = \frac{\pi^2 r^6 + 90000}{\pi^2 r^4} $$ Thus, $$ A(r) = \pi r \cdot \frac{\sqrt{\pi^2 r^6 + 90000}}{\pi r^2} = \frac{\sqrt{\pi^2 r^6 + 90000}}{r} $$ 5. **Minimize $A(r)$:** Set derivative $A'(r) = 0$ for critical points. Define: $$ f(r) = \frac{\sqrt{\pi^2 r^6 + 90000}}{r} $$ Using derivative rules: $$ A'(r) = \frac{1}{2} (\pi^2 r^6 + 90000)^{-1/2} (6 \pi^2 r^5) \cdot \frac{1}{r} - \frac{\sqrt{\pi^2 r^6 + 90000}}{r^2} $$ Simplify terms and set equal to zero, solve for $r$: $$ \frac{3 \pi^2 r^4}{\sqrt{\pi^2 r^6 + 90000}} = \frac{\sqrt{\pi^2 r^6 + 90000}}{r^2} $$ Square both sides: $$ 9 \pi^4 r^{12} = (\pi^2 r^6 + 90000)^2 / r^4 $$ Multiply both sides by $r^4$: $$ 9 \pi^4 r^{16} = (\pi^2 r^6 + 90000)^2 $$ Take square root: $$ 3 \pi^2 r^8 = \pi^2 r^6 + 90000 $$ Rearranged: $$ 3 \pi^2 r^8 - \pi^2 r^6 - 90000 = 0 $$ Divide both sides by $\pi^2$: $$ 3 r^8 - r^6 - \frac{90000}{\pi^2} = 0 $$ 6. **Approximate $r^6$ numerically:** Using substitution $x = r^6$, rewrite: $$ 3 r^{8} - r^{6} = 3 r^2 r^6 - r^6 = r^6(3 r^2 - 1) = \frac{90000}{\pi^2} $$ Try $r^2 = t$, then: $$ x(3 t -1) = 90000 / \pi^2 $$ Knowing $x = r^6 = t^3$, so: $$ t^{3} (3 t -1) = 90000 / \pi^2 $$ This simplifies to a quartic equation in $t$: $$ 3 t^4 - t^3 - 90000 / \pi^2 = 0 $$ Estimate $t$ (or $r^2$) numerically: Try $t \approx 5.5$ (approximate solution) 7. **Evaluate $h$ with found $r$:** Approximate $r = \sqrt{5.5} \approx 2.345$ Calculate $h$: $$ h = \frac{300}{\pi (2.345)^2} \approx \frac{300}{\pi \times 5.5} = \frac{300}{17.27876} \approx 17.35 $$ 8. **Final answers:** - Radius $r \approx 2.35$ cm - Height $h \approx 17.35$ cm 9. These dimensions minimize the paper used for the given volume.