1. **State the problem:** We are given the marginal revenue function $R'(q) = 100 q^{-\frac{1}{2}}$ and the marginal cost function $C'(q) = 0.4q$. We know the total profit at $q=16$ pastries is 520 dollars. We need to find the total profit when $q=25$ pastries. 2. **Recall formulas:** Total profit $P(q)$ is total revenue minus total cost: $$P(q) = R(q) - C(q)$$ where $$R(q) = \int R'(q) \, dq$$ $$C(q) = \int C'(q) \, dq$$ 3. **Integrate marginal revenue:** $$R(q) = \int 100 q^{-\frac{1}{2}} \, dq = 100 \int q^{-\frac{1}{2}} \, dq = 100 \cdot 2 q^{\frac{1}{2}} + K_R = 200 \sqrt{q} + K_R$$ 4. **Integrate marginal cost:** $$C(q) = \int 0.4 q \, dq = 0.4 \cdot \frac{q^2}{2} + K_C = 0.2 q^2 + K_C$$ 5. **Express total profit:** $$P(q) = R(q) - C(q) = (200 \sqrt{q} + K_R) - (0.2 q^2 + K_C) = 200 \sqrt{q} - 0.2 q^2 + (K_R - K_C)$$ Let $K = K_R - K_C$ be the combined constant. 6. **Use initial condition at $q=16$:** $$P(16) = 200 \sqrt{16} - 0.2 \cdot 16^2 + K = 200 \cdot 4 - 0.2 \cdot 256 + K = 800 - 51.2 + K = 748.8 + K$$ Given $P(16) = 520$, solve for $K$: $$520 = 748.8 + K \implies K = 520 - 748.8 = -228.8$$ 7. **Find total profit at $q=25$:** $$P(25) = 200 \sqrt{25} - 0.2 \cdot 25^2 - 228.8 = 200 \cdot 5 - 0.2 \cdot 625 - 228.8 = 1000 - 125 - 228.8 = 646.2$$ **Final answer:** The total profit when production reaches 25 pastries is **646.2 dollars**.