1. **Problem Statement:** a) Find the market equilibrium price and quantity given demand equation $p - 3q = 22$ and supply equation $q^2 + 2p + 4q = 100$. Then find total revenue, equilibrium quantity, and price. b) Given marginal revenue function $MR = 25 - 3Q^2$, derive the total revenue function. 2. **Formulas and Rules:** - Market equilibrium occurs where demand equals supply. - Total revenue $TR = p \times q$. - Marginal revenue $MR = \frac{d(TR)}{dQ}$. 3. **Solution for (a):** From demand: $p = 3q + 22$. Substitute into supply: $q^2 + 2(3q + 22) + 4q = 100$. 4. Simplify supply equation: $$q^2 + 6q + 44 + 4q = 100$$ $$q^2 + 10q + 44 = 100$$ 5. Rearrange: $$q^2 + 10q + 44 - 100 = 0$$ $$q^2 + 10q - 56 = 0$$ 6. Solve quadratic: $$q = \frac{-10 \pm \sqrt{10^2 - 4 \times 1 \times (-56)}}{2} = \frac{-10 \pm \sqrt{100 + 224}}{2} = \frac{-10 \pm \sqrt{324}}{2}$$ $$q = \frac{-10 \pm 18}{2}$$ 7. Possible $q$ values: $$q_1 = \frac{-10 + 18}{2} = 4, \quad q_2 = \frac{-10 - 18}{2} = -14$$ Quantity cannot be negative, so $q = 4$. 8. Find equilibrium price: $$p = 3(4) + 22 = 12 + 22 = 34$$ 9. Calculate total revenue: $$TR = p \times q = 34 \times 4 = 136$$ 10. **Solution for (b):** Given $MR = 25 - 3Q^2$, recall $MR = \frac{d(TR)}{dQ}$. 11. Integrate $MR$ to find $TR$: $$TR = \int (25 - 3Q^2) dQ = 25Q - Q^3 + C$$ 12. Usually, $TR = 0$ when $Q = 0$, so $C = 0$. 13. Final total revenue function: $$TR = 25Q - Q^3$$ **Answer:** - Equilibrium quantity $q = 4$. - Equilibrium price $p = 34$. - Total revenue at equilibrium $TR = 136$. - Total revenue function $TR = 25Q - Q^3$.