1. Problem: Compute the net present value of the three-year project that costs 12000 and yields annual operating savings 3600 and increased revenue 2000, with straight-line depreciation, tax rate 40, and discount rate (after-tax WACC) 10. 2. Formula and rules: use after-tax free cash flow per year $\text{FCF}=\text{(operating benefit)}(1-t)+\text{depreciation}\times t$ and NPV $= -\text{initial} + \sum_{k=1}^{n} \text{FCF}/(1+ r)^k$. 3. Compute annual numbers: operating benefit per year $=3600+2000=5600$. 4. Depreciation per year (straight line) $=12000/3=4000$. 5. After-tax operating income $=5600(1-0.4)=5600\times 0.6=3360$. 6. Depreciation tax shield $=4000\times 0.4=1600$. 7. Annual free cash flow $=3360+1600=4960$. 8. Present value factor for 3-year annuity at $r=0.10$ is $\displaystyle \frac{1-(1+0.10)^{-3}}{0.10}\approx 2.486852$. 9. PV of cash flows $=4960\times 2.486852\approx 12341.86$. 10. NPV $= -12000+12341.86\approx 341.86$. 11. Interpretation: NPV is positive and about 342, so the project adds value when discounted at the 10 rate. 1. Problem: Find $\dfrac{d}{dx}y$ for $y=x^3-12x+13$ using the definition of the derivative. 2. Formula and rule: $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. 3. Compute numerator: $f(x+h)-f(x)=(x+h)^3-12(x+h)+13-(x^3-12x+13)$. 4. Expand and simplify: $(x^3+3x^2h+3xh^2+h^3-12x-12h+13-x^3+12x-13)=3x^2h+3xh^2+h^3-12h$. 5. Divide by $h$ and take the limit: $\frac{3x^2h+3xh^2+h^3-12h}{h}=3x^2+3xh+h^2-12$. 6. Let $h\to0$ to get $f'(x)=3x^2-12$. 1. Problem: Find relative maxima/minima and inflection points of $y=\dfrac{x}{\sqrt{1-x^2}}$ and describe the curve. 2. Domain and rule: domain is $-10$ for all $-10$ so a local minimum at $(-1,2)$. 7. Inflection at $y''=0$ when $x=0$, giving point $(0,4)$. 8. Sketch note: cubic with negative leading coefficient, max at $x=1$, min at $x=-1$, passes through $(0,4)$. 1. Problem (MMC): $R(x)=24x-3x^2$. Part a: find maximum revenue. 2. Quadratic vertex rule: for $R=ax^2+bx+c$ with $a<0$, maximum at $x=-\tfrac{b}{2a}$. 3. Here $a=-3$, $b=24$, so $x^*= -\tfrac{24}{2(-3)}=4$. 4. Maximum revenue $R(4)=24\cdot 4-3\cdot 4^2=96-48=48$. 1. Problem (MMC) part b: average revenue function. 2. Average revenue $AR(x)=R(x)/x=\frac{24x-3x^2}{x}=24-3x$ for $x\neq 0$. 1. Problem (MMC) part c: marginal revenue function. 2. Marginal revenue $MR(x)=\dfrac{dR}{dx}=24-6x$. 1. Problem (MMC) part d: plot total, average, and marginal revenue (description and relationships). 2. Notes for graphing: $R(x)=24x-3x^2$ is a downward parabola with vertex at $(4,48)$. 3. $AR(x)=24-3x$ is a straight line with intercept 24 and slope -3. 4. $MR(x)=24-6x$ is a straight line with intercept 24 and slope -6 and crosses the $x$ axis at $x=4$. 5. On one graph, $MR$ lies below $AR$ for $x>0$ except they share value at small $x$, and $R$ is the area-under-curve type shape reaching maximum at $x=4$. 1. Problem: Demand $y_d=30-2x^2$ and supply $y_s=3+x^2$ with a per-unit tax $t$; find $t$ that maximizes tax revenue and the corresponding tax rate relative to buyer price. 2. With a unit tax $t$, equilibrium price paid by buyers satisfies $30-2x^2=3+x^2+t$. 3. Solve for $x^2$: $30-3-t=3x^2$ so $x^2=9-\tfrac{t}{3}$ and equilibrium quantity $x=\sqrt{9-\tfrac{t}{3}}$ for admissible $t\le 27$. 4. Government tax revenue $G(t)=t\cdot x = t\sqrt{9-\tfrac{t}{3}}$. 5. To maximize, square and differentiate: maximize $S(t)=t^2(9-\tfrac{t}{3})=9t^2-\tfrac{1}{3}t^3$. 6. $S'(t)=18t-t^2=t(18-t)=0$ gives interior maximum at $t=18$. 7. Then equilibrium quantity $x=\sqrt{9-18/3}=\sqrt{3}\approx 1.7321$ and tax revenue $G=18\sqrt{3}\approx 31.1769$. 8. Buyer price at this equilibrium: $p_b=30-2x^2=30-2\cdot 3=24$, so the tax as fraction of buyer price is $18/24=0.75$ i.e. 75. 1. Problem: Marginal cost $MC=10+24x-3x^2$, with total cost at $x=1$ equal to 25; find total cost and average cost functions. 2. Total cost $C(x)=\int MC\,dx + C_0 =\int(10+24x-3x^2)\,dx + C_0 =10x+12x^2-x^3 + C_0$. 3. Use $C(1)=25$ to get $10+12-1+C_0=21+C_0=25$, so $C_0=4$. 4. Total cost function $C(x)=10x+12x^2-x^3+4$. 5. Average cost $AC(x)=C(x)/x=10+12x-x^2+4/x$ for $x>0$. 1. Problem: Growth of a society that starts with 10 members and each member invites 2 new people at the beginning of each year; find membership after 15 years. 2. Each year the population triples because each existing member invites 2 additional people, so growth factor per year is 3. 3. After 15 years the membership is $10\times 3^{15}=10\times 14348907=143489070$. 1. Problem: To have 20000000 after 20 years at 6 annual interest compounded yearly, how much to deposit now? 2. Formula: present value $PV=\dfrac{FV}{(1+r)^n}$. 3. Compute $(1+0.06)^{20}\approx 3.207135$ so $PV\approx 20000000/3.207135\approx 6236949.9$. 1. Problem: Borrow 15000000 at 9 annual interest for 10 years to be repaid by equal monthly installments; find the monthly installment and the remaining balance after 5 years of payments. 2. Monthly rate $r=0.09/12=0.0075$, total months $n=120$. 3. Annuity payment formula $A=P\dfrac{r}{1-(1+r)^{-n}}$. 4. Compute $(1+r)^{-n}=(1.0075)^{-120}\approx 0.407644$, denominator $1-0.407644=0.592356$, so $A\approx 15000000\times\dfrac{0.0075}{0.592356}\approx 189853.5$. 5. Monthly installment approximately 189854. 6. Remaining balance after 60 payments uses $B_{60}=P(1+r)^{60}-A\dfrac{(1+r)^{60}-1}{r}$. 7. Compute $(1+r)^{60}\approx 1.5655$, then $B_{60}\approx 15000000\times 1.5655 -189853.5\times \dfrac{1.5655-1}{0.0075}\approx 9156500$. 8. Interpretation: after 5 years the outstanding principal is about 9156500. 1. Problem: Find the derivative of $y=3x^5-12x^4+13$ using the definition. 2. Using derivative rules (equivalent to the definition after algebra), $\dfrac{d}{dx}(3x^5)=15x^4$ and $\dfrac{d}{dx}(-12x^4)=-48x^3$ while constant drops out. 3. Therefore $y'=15x^4-48x^3=3x^3(5x-16)$. Final answers summary: - Project NPV $\approx 341.86$. - $\dfrac{d}{dx}(x^3-12x+13)=3x^2-12$. - For $y=\dfrac{x}{\sqrt{1-x^2}}$: monotone increasing on $(-1,1)$, no local extrema, inflection at $(0,0)$. - For $y=4+3x-x^3$: local max at $(1,6)$, local min at $(-1,2)$, inflection at $(0,4)$. - MMC: max revenue 48 at $x=4$; $AR=24-3x$; $MR=24-6x$; graph relationships described above. - Tax problem: revenue-maximizing tax $t=18$, quantity $\sqrt{3}\approx 1.7321$, revenue $\approx 31.1769$, tax as fraction of buyer price 0.75. - From $MC$: $C(x)=10x+12x^2-x^3+4$, $AC(x)=10+12x-x^2+4/x$. - Society after 15 years $=143489070$. - Present deposit to reach 20000000 in 20 years at 6 is $\approx 6236949.9$. - Loan monthly payment $\approx 189853.5$, remaining balance after 5 years $\approx 9156500$. - $\dfrac{d}{dx}(3x^5-12x^4+13)=15x^4-48x^3$.