1. **Problem statement:** Find the market equilibrium price $p^*$ and quantity $t^*$ by solving the system where supply price equals demand price, then calculate the producer and consumer surplus using integral calculus. 2. **Equilibrium condition:** At equilibrium, supply price equals demand price: $$P_S(t) = P_D(t)$$ Given: $$P_S(t) = 2t - 3$$ $$P_D(t) = 18 - 2t - t^2$$ 3. **Set equal and solve for $t$:** $$2t - 3 = 18 - 2t - t^2$$ Rearranged: $$2t - 3 - 18 + 2t + t^2 = 0$$ $$t^2 + 4t - 21 = 0$$ 4. **Solve quadratic equation:** Using quadratic formula: $$t = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-21)}}{2} = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2}$$ $$t = \frac{-4 \pm 10}{2}$$ Two solutions: $$t_1 = \frac{-4 + 10}{2} = 3$$ $$t_2 = \frac{-4 - 10}{2} = -7$$ Since quantity cannot be negative, take $t^* = 3$. 5. **Find equilibrium price $p^*$:** Substitute $t^* = 3$ into supply or demand function: $$p^* = P_S(3) = 2(3) - 3 = 6 - 3 = 3$$ 6. **Producer surplus (PS):** PS is area above supply curve and below equilibrium price from $0$ to $t^*$: $$PS = \int_0^{3} (p^* - P_S(t)) dt = \int_0^{3} (3 - (2t - 3)) dt = \int_0^{3} (6 - 2t) dt$$ Calculate integral: $$= \left[6t - t^2\right]_0^{3} = (6 \times 3 - 3^2) - 0 = (18 - 9) = 9$$ 7. **Consumer surplus (CS):** CS is area below demand curve and above equilibrium price from $0$ to $t^*$: $$CS = \int_0^{3} (P_D(t) - p^*) dt = \int_0^{3} (18 - 2t - t^2 - 3) dt = \int_0^{3} (15 - 2t - t^2) dt$$ Calculate integral: $$= \left[15t - t^2 - \frac{t^3}{3}\right]_0^{3} = (15 \times 3 - 3^2 - \frac{27}{3}) - 0 = (45 - 9 - 9) = 27$$ **Final answers:** - Equilibrium quantity $t^* = 3$ - Equilibrium price $p^* = 3$ - Producer surplus = 9 - Consumer surplus = 27