1. **Problem statement:** A balloon is being filled with helium at the rate of 4 ft³/min. We need to find the rate at which the surface area is increasing (in ft²/min) when the volume is $32\pi$ ft³. 2. **Relevant formulas:** - Volume of a sphere: $$V = \frac{4}{3}\pi r^3$$ - Surface area of a sphere: $$S = 4\pi r^2$$ 3. **Given:** - $$\frac{dV}{dt} = 4$$ ft³/min - $$V = 32\pi$$ ft³ at the instant of interest 4. **Find:** $$\frac{dS}{dt}$$ when $$V = 32\pi$$. 5. **Step 1: Find radius $$r$$ when $$V = 32\pi$$** $$32\pi = \frac{4}{3}\pi r^3 \implies 32 = \frac{4}{3} r^3 \implies r^3 = 32 \times \frac{3}{4} = 24$$ $$r = \sqrt[3]{24}$$ 6. **Step 2: Differentiate volume formula with respect to time $$t$$:** $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$ 7. **Step 3: Solve for $$\frac{dr}{dt}$$:** $$4 = 4\pi r^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{4}{4\pi r^2} = \frac{1}{\pi r^2}$$ 8. **Step 4: Differentiate surface area formula with respect to time:** $$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$$ 9. **Step 5: Substitute $$\frac{dr}{dt}$$ and $$r$$ into $$\frac{dS}{dt}$$:** $$\frac{dS}{dt} = 8\pi r \times \frac{1}{\pi r^2} = \frac{8\pi r}{\pi r^2} = \frac{8}{r}$$ 10. **Step 6: Calculate $$\frac{dS}{dt}$$ using $$r = \sqrt[3]{24}$$:** $$\frac{dS}{dt} = \frac{8}{\sqrt[3]{24}}$$ 11. **Step 7: Simplify $$\sqrt[3]{24}$$:** $$24 = 8 \times 3 \implies \sqrt[3]{24} = \sqrt[3]{8} \times \sqrt[3]{3} = 2 \times \sqrt[3]{3}$$ 12. **Step 8: Final expression:** $$\frac{dS}{dt} = \frac{8}{2 \times \sqrt[3]{3}} = \frac{4}{\sqrt[3]{3}}$$ 13. **Step 9: Approximate $$\sqrt[3]{3} \approx 1.442$$:** $$\frac{dS}{dt} \approx \frac{4}{1.442} \approx 2.77$$ 14. **Step 10: Match closest answer choice:** The closest choice is (C) 4, but our exact answer is approximately 2.77, which is closest to (B) 2. **Answer:** (B) 2