1. Problem: Compute the net present value of the project which costs 12000, yields annual savings 3600 and additional revenue 2000 for 3 years, uses straight-line depreciation, has tax rate 40, and the firm uses after-tax WACC 10. 2. Formula and rules: NPV = -Initial + \sum_{t=1}^{n} CashFlow_t/(1+r)^t. 3. Important rule: After-tax operating cash flow with depreciation is given by $\text{CF}=(\text{Benefit})(1-T)+\text{Depreciation}\times T$. 4. Work: Annual benefit $=3600+2000=5600$. 5. Work: Straight-line depreciation $=12000/3=4000$. 6. Work: Annual after-tax cash flow $=5600(1-0.4)+4000(0.4)=5600\times0.6+1600=3360+1600=4960$. 7. Work: Present value of three-year annuity at $r=0.10$ is $\displaystyle \frac{1-(1+0.10)^{-3}}{0.10}=2.486852$. 8. Work: PV of inflows $=4960\times2.486852\approx12342.71$. 9. Work: NPV $= -12000+12342.71\approx342.71$. 10. Final answer: NPV $\approx342.71$ (positive so project is acceptable). 11. Problem: Find the first derivative of $y=x^{3}-12x+13$ using the definition of the derivative. 12. Formula and rules: Definition $f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. 13. Work: Compute $\displaystyle\frac{(x+h)^{3}-12(x+h)+13-(x^{3}-12x+13)}{h}=\frac{3x^{2}h+3xh^{2}+h^{3}-12h}{h}$. 14. Work: Simplify to $3x^{2}+3xh+h^{2}-12$ and take $h\to0$ to get $3x^{2}-12$. 15. Final answer: $y'=3x^{2}-12$. 16. Problem: Find relative maxima/minima and inflection of $y=\dfrac{x}{\sqrt{1-x^{2}}}$ and describe the curve. 17. Domain and rule: Domain is $|x|<1$ because of $\sqrt{1-x^{2}}$. 18. Work: Rewrite $y=x(1-x^{2})^{-1/2}$ and differentiate. 19. Work: $y'= (1-x^{2})^{-1/2}+x\cdot(-\tfrac{1}{2})(1-x^{2})^{-3/2}(-2x)$. 20. Work: Simplify to $y'=\dfrac{1}{(1-x^{2})^{3/2}}$. 21. Interpretation: $y'>0$ for all $-10$ (local minimum at $x=-1$) and $y''(1)=-6<0$ (local maximum at $x=1$). 30. Work: Values $y(-1)=4-3+1=2$ and $y(1)=4+3-1=6$. 31. Curve shape: Cubic with negative leading coefficient, local max at $(1,6)$ and local min at $(-1,2)$, inflection at $x=0$ with $y(0)=4$. 32. Final: Key points for sketch are $(-1,2),(0,4),(1,6)$ and end behavior down to the right and up to the left. 33. Problem 5a: For MMC revenue $R=24x-3x^{2}$ find maximum revenue. 34. Formula: Max where $R'=0$ and $R''<0$. 35. Work: $R'=24-6x$, set $0\Rightarrow x=4$. 36. Work: $R''=-6<0$ so maximum, and $R(4)=24\times4-3\times16=96-48=48$. 37. Final: Maximum revenue $48$ at $x=4$. 38. Problem 5b: What is the average revenue function. 39. Work: Average revenue $AR=R/x=24-3x$ for $x\neq0$. 40. Final: $AR=24-3x$. 41. Problem 5c: What is the marginal revenue function. 42. Work: Marginal revenue $MR=R'=24-6x$. 43. Final: $MR=24-6x$. 44. Problem 5d: Plot total, average, and marginal revenue on one graph (description and desmos provided). 45. Description: $R(x)=24x-3x^{2}$ is a downward parabola with vertex at $(4,48)$, $AR(x)=24-3x$ is a line with slope -3, $MR(x)=24-6x$ is a steeper line with slope -6 that crosses the $x$-axis at $x=4$. 46. Note: On a graph the TR parabola, AR line and MR line intersect consistent with the calculus results. 47. Problem 6: Demand $y_{d}=30-2x^{2}$ and supply $y_{s}=3+x^{2}$, find the tax per unit $t$ that maximizes tax revenue and the maximum tax revenue. 48. Model and rule: With a per-unit tax $t$ imposed on sellers, buyer price $p_{b}$ satisfies $p_{b}=30-2x^{2}$ and seller receives $p_{s}=p_{b}-t=3+x^{2}$. 49. Work: Equate $30-2x^{2}=3+x^{2}+t$ giving $3x^{2}=27-t$ and $x^{2}=9-\tfrac{t}{3}$. 50. Work: Quantity $x=\sqrt{9-\tfrac{t}{3}}$ and tax revenue $TR(t)=t\,x=t\sqrt{9-\tfrac{t}{3}}$ for $0\le t\le27$. 51. Work: Differentiate $TR$: $TR'=\sqrt{9-\tfrac{t}{3}}-\dfrac{t}{6}\left(9-\tfrac{t}{3}\right)^{-1/2}$, set $TR'=0$ and simplify to $9-\tfrac{t}{3}-\tfrac{t}{6}=0$. 52. Work: Multiply by 6 to get $54-3t=0$ so $t=18$. 53. Work: Corresponding quantity $x^{2}=9-18/3=3$ so $x=\sqrt{3}$ and maximum tax revenue $TR_{\max}=18\sqrt{3}\approx31.1769$. 54. Final: Optimal tax $t=18$ with maximum revenue $18\sqrt{3}\approx31.1769$. 55. Problem 7: Given marginal cost $MC=10+24x-3x^{2}$ and total cost at $x=1$ equals 25, find total cost function and average cost. 56. Formula: Total cost $C(x)=\int MC\,dx +C_{0}$ and $AC(x)=C(x)/x$. 57. Work: Integrate $C(x)=10x+12x^{2}-x^{3}+C_{0}$. 58. Work: Use $C(1)=25$ gives $10+12-1+C_{0}=21+C_{0}=25$ so $C_{0}=4$. 59. Final: $C(x)=10x+12x^{2}-x^{3}+4$ and $AC(x)=10+12x-x^{2}+4/x$ for $x>0$. 60. Problem 8: Society growth with 10 initial members each inviting 2 people at the beginning of each year, find membership after 15 years. 61. Model and rule: Each year every member becomes 1+2=3 members at the beginning, so growth factor per year is 3 and count after n years is $10\times3^{n}$. 62. Work: After 15 years membership $=10\times3^{15}$ and $3^{15}=14348907$. 63. Final: Members after 15 years $=143489070$. 64. Problem 9 (Public assignment 2): To have 20000000 after 20 years at 6% annual interest, how much to deposit now at the beginning. 65. Formula: Present value $PV=\dfrac{FV}{(1+0.06)^{20}}$. 66. Work: $PV=20000000(1.06)^{-20}\approx20000000/3.207135472\approx6236094$. 67. Final: Deposit required $\approx6236094$. 68. Problem 10a: Borrow 15000000 at 9% annual interest for 10 years to be paid monthly in equal installments; find monthly installment. 69. Formula and rules: Monthly rate $r=0.09/12=0.0075$, $n=120$, payment $P=\dfrac{Lr}{1-(1+r)^{-n}}$. 70. Work: Compute $(1+r)^{120}\approx2.4517$ and denominator $1-(1+r)^{-n}\approx0.5922$. 71. Work: $P\approx15000000\times0.0075/0.5922\approx189926$. 72. Final: Monthly installment $\approx189926$. 73. Problem 10b: After paying 5 years (60 payments), find remaining loan balance. 74. Formula: Balance after $m$ payments $B_{m}=L(1+r)^{m}-P\dfrac{(1+r)^{m}-1}{r}$. 75. Work: Using $m=60$, $(1+r)^{60}\approx1.566$, compute $B_{60}\approx15000000\times1.566-189926\times\dfrac{1.566-1}{0.0075}\approx9156815$. 76. Final: Remaining loan portion after 5 years $\approx9156815$. 77. Problem 11: Find the first derivative of $y=3x^{5}-12x^{4}-413$ using the definition of the derivative. 78. Formula: $f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. 79. Work: Compute difference quotient and simplify to $15x^{4}-48x^{3}+\text{terms with }h$. 80. Work: Taking the limit $h\to0$ gives $f'(x)=15x^{4}-48x^{3}$. 81. Final: $y'=15x^{4}-48x^{3}$. Summary: Each problem above includes the statement, formula, intermediate algebra, and final numeric or algebraic answer.