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 $\frac{32\pi}{3}$ 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 - $V = \frac{32\pi}{3}$ ft³ at the instant of interest 4. **Find:** $\frac{dS}{dt}$ when $V = \frac{32\pi}{3}$. 5. **Step 1: Find radius $r$ when $V = \frac{32\pi}{3}$** $$\frac{32\pi}{3} = \frac{4}{3}\pi r^3 \implies \frac{32}{3} = \frac{4}{3} r^3 \implies 32 = 4 r^3 \implies r^3 = 8$$ $$r = \sqrt[3]{8} = 2$$ 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 $\frac{dr}{dt}$:** $$4 = 4\pi (2)^2 \frac{dr}{dt} \implies 4 = 16\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{4}{16\pi} = \frac{1}{4\pi}$$ 8. **Step 4: Differentiate surface area formula with respect to time:** $$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$$ 9. **Step 5: Substitute $r=2$ and $\frac{dr}{dt} = \frac{1}{4\pi}$ into $\frac{dS}{dt}$:** $$\frac{dS}{dt} = 8\pi \times 2 \times \frac{1}{4\pi} = \frac{16\pi}{4\pi} = 4$$ 10. **Final answer:** The surface area is increasing at a rate of 4 ft²/min when the volume is $\frac{32\pi}{3}$ ft³. **Answer:** 4