1. **Problem Statement:** Find the market equilibrium price and quantity given demand and supply equations, then find total revenue. 2. **Given:** Demand: $p = 3q - 22$ Supply: $q^2 + 2p + 4q = 100$ 3. **Step 1: Find equilibrium where demand equals supply price:** Set demand price equal to supply price. 4. From demand: $p = 3q - 22$ From supply: rearranged to express $p$: $$q^2 + 2p + 4q = 100 \implies 2p = 100 - q^2 - 4q \implies p = \frac{100 - q^2 - 4q}{2}$$ 5. Set equal: $$3q - 22 = \frac{100 - q^2 - 4q}{2}$$ Multiply both sides by 2: $$2(3q - 22) = 100 - q^2 - 4q$$ $$6q - 44 = 100 - q^2 - 4q$$ 6. Rearrange: $$6q - 44 - 100 + q^2 + 4q = 0$$ $$q^2 + 10q - 144 = 0$$ 7. Solve quadratic: $$q = \frac{-10 \pm \sqrt{10^2 - 4 \times 1 \times (-144)}}{2} = \frac{-10 \pm \sqrt{100 + 576}}{2} = \frac{-10 \pm \sqrt{676}}{2} = \frac{-10 \pm 26}{2}$$ 8. Possible $q$ values: $$q = \frac{-10 + 26}{2} = 8$$ $$q = \frac{-10 - 26}{2} = -18$$ (discard negative quantity) 9. Equilibrium quantity $q = 8$ 10. Find equilibrium price $p$: $$p = 3(8) - 22 = 24 - 22 = 2$$ 11. Total revenue $TR = p \times q = 2 \times 8 = 16$ --- 12. **Problem:** Given marginal revenue $MR = 25 - 3Q^2$, find total revenue $TR$. 13. Recall $MR = \frac{dTR}{dQ}$, so integrate: $$TR = \int MR \, dQ = \int (25 - 3Q^2) dQ = 25Q - Q^3 + C$$ 14. Usually $TR = 0$ when $Q=0$, so $C=0$. 15. Final total revenue function: $$TR = 25Q - Q^3$$ --- 16. **Problem:** Find total amount paid to firms A, B, C using matrix method. 17. Stones loads: A=40, B=35, C=25; Sand loads: A=10, B=5, C=8 Cost per load: Stones=1200, Sand=500 18. Define matrices: $$Q = \begin{bmatrix}40 & 10 \\ 35 & 5 \\ 25 & 8\end{bmatrix}, \quad C = \begin{bmatrix}1200 \\ 500\end{bmatrix}$$ 19. Total cost per firm: $$T = Q \times C = \begin{bmatrix}40 & 10 \\ 35 & 5 \\ 25 & 8\end{bmatrix} \times \begin{bmatrix}1200 \\ 500\end{bmatrix} = \begin{bmatrix}40 \times 1200 + 10 \times 500 \\ 35 \times 1200 + 5 \times 500 \\ 25 \times 1200 + 8 \times 500\end{bmatrix} = \begin{bmatrix}53000 \\ 44500 \\ 34000\end{bmatrix}$$ --- 20. **Problem:** Find number of students in courses A, B, C given total 18000 and enrollment relations. 21. Let highest enrollment = $x$ Then: - Lowest = $x - 7000$ - Second highest = $x - 4000$ 22. Sum: $$x + (x - 4000) + (x - 7000) = 18000$$ $$3x - 11000 = 18000$$ $$3x = 29000$$ $$x = \frac{29000}{3} \approx 9666.67$$ 23. Students: - Highest: 9666.67 - Second highest: 9666.67 - 4000 = 5666.67 - Lowest: 9666.67 - 7000 = 2666.67 24. **Number of ways to arrange 4 boxes from 9:** Permutation $P(9,4) = \frac{9!}{(9-4)!} = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3024$ --- 25. **Problem:** Radio listeners data 26. Given: - Total citizens = 600 - Listened 7 o'clock news = 310 - Listened late night news = 370 - Listened both = 120 27. Number who listened late night but not 7 o'clock: $$370 - 120 = 250$$ 28. Number who listened at least one: $$310 + 370 - 120 = 560$$ 29. Number who did not listen either: $$600 - 560 = 40$$ --- 30. **Problem:** Calculate future value of quarterly deposits 31. Given: - Deposit $R = 450$ - Interest rate per annum $i = 12\%$ compounded quarterly $\Rightarrow i_q = \frac{12\%}{4} = 3\% = 0.03$ - Number of years $t = 10$ - Number of quarters $n = 4 \times 10 = 40$ 32. Future value of annuity formula: $$FV = R \times \frac{(1 + i_q)^n - 1}{i_q}$$ 33. Calculate: $$FV = 450 \times \frac{(1 + 0.03)^{40} - 1}{0.03}$$ 34. Compute $(1.03)^{40} \approx 3.262$ (approximate) 35. So: $$FV = 450 \times \frac{3.262 - 1}{0.03} = 450 \times \frac{2.262}{0.03} = 450 \times 75.4 = 33930$$ --- 36. **Problem:** Calculate amount received by child with annual deposits 37. Given: - Annual deposit $R = 2500$ - Interest rate $i = 4.5\% = 0.045$ - Number of years $n = 18 - 3 = 15$ 38. Future value of annuity: $$FV = R \times \frac{(1 + i)^n - 1}{i}$$ 39. Calculate: $$(1.045)^{15} \approx 1.938$$ 40. So: $$FV = 2500 \times \frac{1.938 - 1}{0.045} = 2500 \times \frac{0.938}{0.045} = 2500 \times 20.84 = 52100$$ --- 41. **Problem:** Calculate equal installments for loan repayment 42. Given: - Loan amount $P = 20000$ - Interest rate $i = 8\% = 0.08$ - Number of installments $n = 12$ 43. Installment formula (annuity payment): $$A = P \times \frac{i(1+i)^n}{(1+i)^n - 1}$$ 44. Calculate: $$(1.08)^{12} \approx 2.518$$ 45. So: $$A = 20000 \times \frac{0.08 \times 2.518}{2.518 - 1} = 20000 \times \frac{0.2014}{1.518} = 20000 \times 0.1327 = 2654$$ --- **Final answers:** - Equilibrium price: $2$, quantity: $8$, total revenue: $16$ - Total revenue function: $TR = 25Q - Q^3$ - Total paid to firms: A=53000, B=44500, C=34000 - Students: Highest=9667, Second=5667, Lowest=2667 - Ways to arrange boxes: 3024 - Radio listeners: Late night only=250, At least one=560, Neither=40 - Future value quarterly deposits: 33930 - Child education trust amount: 52100 - Loan installment: 2654