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\noindent
{\bf Math 105}  \hfil {\bf Answers to supplementary problems}
 
\medskip \noindent
\textbf{1.} $q = -30p + 1250$, with $p$ = price and $q$ = number of customers
(demand).

\medskip \noindent
\textbf{2.}  \ (a) $p = -(1/20)\,q + 150$

\smallskip
(b) $R = -(1/20)\,q^2 + 150q$ and $P = -(1/20)\,q^2 + 115q - 10,000$.

\smallskip
\quad Break-even points: 2209.5, 90.52.

\smallskip
(c) $R = 3000p - 20p^2$ and $C = 115,000 - 700p$, so

\smallskip \noindent
\qquad $P = -20p^2 + 3700p - 115,000$.  Break-even points: 39.52, 145.47

\medskip \noindent
\textbf{3.} \ (a) $q = -8p + 700$ \quad (b) $R = -8p^2 + 700p$

\medskip \noindent
\textbf{4.} \ (a) $-(x-3)^2 + 20$, opens downward, high point at $(3, 20)$

\smallskip
(b) $3(x+1)^2 - 4$, opens upward, low point at $(-1,-4)$

\smallskip
(c) $2\left(x + (5/4)\right)^2 - (25/8)$, opens upward, low point at
$(-5/4, - 25/8)$

\smallskip
(d) $-2\left(x - (1/4)\right)^2 + (9/8)$, opens downward, high point at
$(1/4, 9/8)$

\medskip \noindent
\textbf{5.} \ (a) 4 \quad (b) 2 \quad (c) 64

\medskip \noindent
\textbf{6.} \ (a) 75 \quad (b) 92.5

\medskip \noindent
\textbf{7.}  43.75.

\medskip \noindent
\textbf{8.} \ (a) h.a. $y=0$, v.a.  $x=-1$. \quad (b) h.a.  $y=0$, v.a.  $x=1$

\smallskip \noindent
(c) h.a.  $y=2$, v.a.  $x=-4$ \quad (d) h.a.  $y=0$, v.a.  $x=2$ and $x=-2$

\smallskip
(e) h.a. is $y = 1/3$, no v.a. \quad (f) no h.a., v.a. is $x=2$.

\medskip \noindent
\textbf{9.} 44.25 at 35 mph, 708 at 140 mph.

\medskip \noindent
\textbf{10.} With $x$ = amt. of gold purchased and $f(x)$  = fee,
$$f(x) = \cases{0.06x & if $50 \le x \le 300$ \cr 12 + 0.02x &if $300 < x
\le 600$ \cr 15 + 0.015x &if $600 < x$ \cr}$$

The function is continuous.

\medskip \noindent
\textbf{11.} \ (a) continuous \quad (b) discontinuity at $x=0$

\smallskip \noindent
(c) continuous \quad (d) discontinuity at $x=1$

\medskip \noindent
\textbf{12.}

\smallskip \noindent
(a) \
\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|}  \hline $x$ & 0 & 0.5 & 1.0 & 1.5 &
2.0 & 2.5 & 3.0 & 3.5 & 4.0 \\ \hline
$f(x)$ & 1 & 1.225 & 1.5 & 1.837 & 2.25 & 2.756 & 3.375 & 4.134 & 5.0625\\
\hline
\end{tabular}

\medskip \noindent
(b) \
\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|}  \hline $x$ & 0 & 0.5 & 1.0 & 1.5 &
2.0 & 2.5 & 3.0 & 3.5 & 4.0 \\ \hline
$f(x)$ & 1 & 0.775 & 0.6 & 0.465 & 0.36 & 0.279 & 0.216 & 0.167 & 0.130\\
\hline
\end{tabular}

\next
\textbf{13.} (a) $t = {{\ln7}\over{\ln5}} \approx 1.209$ \hfil
  (b) $t = {{\ln10}\over{\ln2}} \approx 3.322$

\bigskip
(c) $t = {{\ln100}\over3}\approx 1.535$ \hfil (d) $t = \ln(8/5) \approx 0.47$

\next
\textbf{14.} (a) $-3$ \hfil (b) $x+1$\hfil (c) $x^2$ \hfil (d) 2

\next
\textbf{15.} (a) $e^{7\ln3}$ \hfil (b) $e^{x\ln5}$ \hfil (c) $e^{5\ln x}$

\next
\textbf{16.} (a) ${{\ln12}\over{\ln5}} \approx 1.544$ \hfil
   (b) ${{\ln5}\over{\ln12}} \approx 0.6477$ \hfil (c) ${1\over{\ln2}}
\approx 1.4427$

\next
\textbf{17.} (a) ${40\over{1 + 11e^{-0.8}}} \approx 6.731$ billion

\medskip
    (b) when $t = {{\ln11}\over{0.08}} \approx 29.97$ (that is, around 2,020)

\next
\textbf{18.} ${{\ln(5/3)}\over{0.08}} \approx 6.385$ years

\next
\textbf{19.} (a) $5000e^{0.32} \approx 6885.64$ \hfil
  (b) ${{\ln(8/5)}\over8} \approx 0.059$, or 5.9 \%

\next
\textbf{20.} (a) 5 \hfil (b) $2a + 3$ \hfil (c) $-3$ \hfil (d) $1/3$

\next
\textbf{21.} (a) $2x + 1$ \hfil (b) ${{-2}\over{\left(2x+3\right)^2}}$ \hfil
(c) ${{-2}\over{x^3}}$

\next
\textbf{22.} (a) ${5\over3}\,x^{2/3}$ \hfil (b) $-3x^{-4}$, or
${{-3}\over{x^4}}$  \hfil
      (c) $-{1\over2}\,x^{-3/2}$, or $-{1\over{2x^{3/2}}}$

\next
\textbf{23.} (a) 36 ft.
\hfil
(b) 50 ft./sec.
\hfil
(c) 34 ft./sec.
\hfil
(d) 18 ft/sec.
\hfil
(e) $-14$ ft./sec.

\smallskip
(f) At $t=1$ second the ball is moving upward. At $t=2$ seconds the ball is
moving downward.

\smallskip
(g) At $t= 25/16$ seconds.  The ball is at its highest point.

\next
\textbf{24.} (a) $6t^2 + (3/t)$ \hfil (b) $2 + 4 + (1/x)$ \hfil (c) $-1/x$

\next
\textbf{25.} (a) $1/x$ \hfil (b) $-1/x$ \hfil (c) $1/(5x)$

\medskip
(d) $1/(2x)$ \hfil (e) $3/t$ \hfil (f) $-3/t$

\next
\textbf{26.} (a) $\left(1/2, \ln(1/2)\right)$ \hfil (b) $(3, \ln3)$

\next
\textbf{27.} (a) $y = x-1$ \hfil (b) lin. approx. gives 0.1, 0.02, and $-0.01$;
calculator gives 0.9531, 0.1980, and $-0.01005$.

\next
\textbf{28.} (a) $3e^t - 2t + 5$ \hfil (b) $\ds 2e^x + {1\over{2x^{3/2}}}$
\hfil (c) $\ds e^x - {1\over x}$

\next 
\textbf{29.} a) $(\ln 3, 3)$
\hfil 
b) $(\ln (\frac{1}{2}), \frac{1}{2})$
\hfil
c) $(-1,-0)$

\next 
\textbf{30.} a) $f'(x) = \frac{1}{2} e^{x/2}$
\hfil
b) $f'(x) = 4e^{4x} = 4(e^x)^4$
\hfil
c) $f'(t) = -e^{-t} = -\frac{1}{e^{t}}$

d) $\frac{dy}{dt} = \frac{1}{2} e^{t/2} = \frac{1}{2} \sqrt{e^{t}}$
\hfil
e) $f'(x) = 3e^{3x} - 10e^{2x} + 3e^x$
\hfil
f) $f'(x) = 2e^{2x} - 6e^x$

\next 
\textbf{31.} a) $f'(t) = (\ln 3) 3^{t} + 3t^{2}$
\hfil
b) $\frac{dy}{dx} = \frac{\ln 2}{3}2^x - \frac{2}{3x^2}$

c) $f'(x) = (\ln 5)5^x + 5(\ln 4)4^x$

\next 
\textbf{32.} a) $\displaystyle (-\frac{\ln(\ln 2)}{\ln
2},2^{-\frac{\ln(\ln 2)}{\ln 2}}) \approx (.539,1.443)$
\hfill
b) $(-\frac{1}{\ln 2},0)$

\next 
\textbf{33.} At $t=1$, $1.05\ln(1.05)P_0 \approx .0512P_0$

At $t=2$, $(1.05)^2\ln(1.05)P_0 \approx .0538P_0$

\next 
\textbf{34.}
Population is approximately 10,586,000, 
and decreasing at approximately 21,000 people per year.

\next 
\textbf{35.} a) $k = 2\ln(\frac13) = -2\ln 3 \approx -2.20$

b) $\frac{dH}{dt} = 30ke^{kt}  = -66e^{-2.2t}$.
Negative --- Why?

c) When $t=0$.  Why?

\next 
\textbf{36.}
$A = -80e^{\frac34} \approx -169.36$
\hfil 
$B = -1/40 = -.025$

\next 
\textbf{37.}
a) demand is 15,000 \hfil b) 1000 \hfil c) positive; increased

\next 
\textbf{38.} a) $R = 1000pe^{-0.02p}$
\hfil
b) $\frac{dR}{dq} = -50(\ln (\frac{q}{1000}) + 1)$,
or equivalently, $\frac{dR}{dq} = \frac{1-0.02p}{-0.02}$.

c) revenue is $10,000e^{-0.2} \approx \$8187$, 
marginal revenue is $-40$ dollars per unit

\next 
\textbf{39.} a) crit points: $x=-3, 2$; increasing on 
$(-\infty, -3)$ and $(2,\infty)$; 
decreasing on $(-3,2)$; relative max at $x=-3$; relative min at
$x=2$

b) crit points: $x=0,1$; increasing on 
$(1,\infty)$; decreasing on $(-\infty,1)$; 
relative min at $x=1$.

c) crit points: $x = -2, 0, 2$;
increasing on $(0,\infty)$; decreasing on $(-\infty,0)$;
relative min at $x=0$.

d) crit points: $x=1$; increasing on 
$(1,\infty)$; decreasing on $(0,1)$; relative min at $x=1$

e) crit points: $x=-1$; increasing on 
$(-1, \infty)$; decreasing on $(-\infty, -1)$;
relative min at $x=-1$.

f) crit points: $x = -2/5, 0$; increasing on 
$(-\infty, -2/5)$, $(0, \infty)$;
decreasing on $(-2/5,0)$; relative max at $x=-2/5$; relative min at
$x=0$.

\next 
\textbf{40.} one solution (no critical points)

\next 
\textbf{41.} You should find that the min of $f(x)$ occurs
at $x=1$.

\next 
\textbf{42.}
a) You should find that the min of $f(x)$ occurs at $x=2$

b) 

c) no, why?




\end{document}
