1. **State the problem:** We are given the marginal revenue function $$\frac{dR}{dq} = 100 - \frac{3}{2}\sqrt{2q}$$ and need to find the corresponding demand equation. 2. **Recall the relationship:** Marginal revenue $$\frac{dR}{dq}$$ is the derivative of the revenue function $$R(q)$$ with respect to quantity $$q$$. Revenue $$R(q)$$ is price $$p(q)$$ times quantity $$q$$, i.e., $$R(q) = p(q) \times q$$. 3. **Integrate marginal revenue to find revenue:** $$R(q) = \int \left(100 - \frac{3}{2}\sqrt{2q}\right) dq = \int 100 dq - \int \frac{3}{2} \sqrt{2q} dq$$ 4. **Calculate each integral:** - $$\int 100 dq = 100q$$ - For $$\int \frac{3}{2} \sqrt{2q} dq$$, rewrite $$\sqrt{2q} = \sqrt{2} \sqrt{q} = \sqrt{2} q^{1/2}$$: $$\int \frac{3}{2} \sqrt{2} q^{1/2} dq = \frac{3\sqrt{2}}{2} \int q^{1/2} dq = \frac{3\sqrt{2}}{2} \times \frac{2}{3} q^{3/2} = \sqrt{2} q^{3/2}$$ 5. **Combine results:** $$R(q) = 100q - \sqrt{2} q^{3/2} + C$$ where $$C$$ is the constant of integration. 6. **Express revenue in terms of price:** Since $$R(q) = p(q) q$$, we have $$p(q) = \frac{R(q)}{q} = \frac{100q - \sqrt{2} q^{3/2} + C}{q} = 100 - \sqrt{2} q^{1/2} + \frac{C}{q}$$ 7. **Interpretation:** Usually, for demand functions, the constant $$C$$ is zero or determined by boundary conditions. Assuming $$C=0$$, **Demand equation:** $$p(q) = 100 - \sqrt{2} \sqrt{q}$$ This is the demand function corresponding to the given marginal revenue. **Final answer:** $$p(q) = 100 - \sqrt{2} \sqrt{q}$$