1. **Problem statement:** Ju Shen designs a plastic cover shaped as a cylinder with a hemisphere on top, open at the bottom. The cylinder has height $h$ cm and radius $r$ cm. 2. **(a) Total internal surface area $A$:** - The curved surface area of the cylinder is $2\pi rh$. - The curved surface area of the hemisphere is half the surface area of a sphere: $2\pi r^2$. - Since the cover is open at the bottom, the total internal surface area is: $$A = 2\pi rh + 2\pi r^2$$ 3. **(b) Total volume $V$ enclosed:** - Volume of the cylinder: $\pi r^2 h$. - Volume of the hemisphere: half the volume of a sphere: $\frac{2}{3}\pi r^3$. - Total volume: $$V = \pi r^2 h + \frac{2}{3} \pi r^3 = \frac{3\pi r^2 h + 2\pi r^3}{3} = \frac{2\pi r^3 + 3\pi r^2 h}{3}$$ 4. **(c) Express $h$ in terms of $r$ given $V=10000$ cm³:** $$10000 = \frac{2\pi r^3 + 3\pi r^2 h}{3}$$ Multiply both sides by 3: $$30000 = 2\pi r^3 + 3\pi r^2 h$$ Isolate $h$: $$3\pi r^2 h = 30000 - 2\pi r^3$$ Divide both sides by $3\pi r^2$: $$h = \frac{30000 - 2\pi r^3}{3\pi r^2}$$ 5. **(d) Expression for $A$ in terms of $r$ only:** Substitute $h$ from (c) into $A$ from (a): $$A = 2\pi r h + 2\pi r^2$$ $$= 2\pi r \times \frac{30000 - 2\pi r^3}{3\pi r^2} + 2\pi r^2$$ Simplify numerator and denominator: $$= \frac{2\pi r (30000 - 2\pi r^3)}{3\pi r^2} + 2\pi r^2$$ Cancel $\pi$ and $r$ terms: $$= \frac{2 \cancel{\pi} \cancel{r} (30000 - 2\pi r^3)}{3 \cancel{\pi} r^2} + 2\pi r^2 = \frac{2 (30000 - 2\pi r^3)}{3 r} + 2\pi r^2$$ Distribute numerator: $$= \frac{60000 - 4\pi r^3}{3 r} + 2\pi r^2 = \frac{60000}{3 r} - \frac{4\pi r^3}{3 r} + 2\pi r^2$$ Simplify terms: $$= \frac{20000}{r} - \frac{4\pi r^2}{3} + 2\pi r^2 = \frac{20000}{r} + \left(2\pi r^2 - \frac{4\pi r^2}{3}\right)$$ $$= \frac{20000}{r} + \frac{6\pi r^2 - 4\pi r^2}{3} = \frac{20000}{r} + \frac{2\pi r^2}{3}$$ 6. **(e) Find $\frac{dA}{dr}$:** $$A = \frac{20000}{r} + \frac{2\pi r^2}{3}$$ Differentiate term by term: $$\frac{dA}{dr} = -\frac{20000}{r^2} + \frac{4\pi r}{3}$$ 7. **(f) Find $r$ and $h$ minimizing $A$:** Set derivative to zero for minimum: $$-\frac{20000}{r^2} + \frac{4\pi r}{3} = 0$$ Rearranged: $$\frac{4\pi r}{3} = \frac{20000}{r^2}$$ Multiply both sides by $r^2$: $$\frac{4\pi r^3}{3} = 20000$$ Solve for $r^3$: $$r^3 = \frac{20000 \times 3}{4\pi} = \frac{60000}{4\pi} = \frac{15000}{\pi}$$ So: $$r = \sqrt[3]{\frac{15000}{\pi}}$$ Calculate $h$ using (c): $$h = \frac{30000 - 2\pi r^3}{3\pi r^2} = \frac{30000 - 2\pi \times \frac{15000}{\pi}}{3\pi r^2} = \frac{30000 - 30000}{3\pi r^2} = 0$$ 8. **(g) Interpretation:** The minimum plastic use occurs when $h=0$, meaning the cover is just a hemisphere without a cylindrical part.