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Here is an explanation as to why the change is in the opposite direction for price and revenue if it is elastic at
$p_0$. \hb
\noindent $\bullet$ $p$=price, $q$=quantity, Demand function is $f(p)$ {\bf expressed in terms of } $p$.\hfil\break
\noindent $\bullet$ Elasticity of demand $E(p)=\frac{-pf'(p)}{f(p}.$
\smallskip
{\bf Demand is elastic at some price} $p_0$ if $E(p)>1$, or what is the same, \\ if
$(1-E(p_0))<0.$\hb
{\bf Demand is inelastic at some price} $p_0$ if $E(p_0)<1,$ or what is the same, \\ if $(1-E(p_0)>0).$
\smallskip
$R(p)=$ Revenue function {\bf expressed in terms of price}, i.e. {\bf as a function of} $p$.
$$R(p)=f(p).p.$$
By the product rule,
$$R'(p)= f(p)+p.f'(p) =f(p)\left(1 + \frac{pf'(p)}{f(p)}\right)= f(p)(1-E(p)).$$
If demand is {\bf elastic} at some price $p_0$, then, as $f(p)$ is always positive, we have
$$
E(p_0)>1\Longleftrightarrow(1-E(p_0))~{\rm is ~negative~}
\Longrightarrow
R'(p_0)<0 \Longrightarrow R(p) ~{\rm is ~decreasing~at~} p_0.
$$
\bigskip
Note also that for some $p_0$, if $E(p_0)=1$, then $R'(p_0)=f(p_0)(1-E(p_0)) =0$. So
when you look at the {\bf graph of the Revenue function as a function of the price $p$}, it
means that the slope is zero at $p_0$ if $E(p_0)=1$; hence the graph (which is an inverted parabola) has {\bf maximum value at $p_0$}, which means that the revenue is maximum at $p_0.$
We will use this later while we do {\bf PROFIT MAXIMISATION} .
\bigskip
\n{\bf MORE PROBLEMS ON EXPONENTIAL GROWTH}:
\medskip
\n This is in continuation with the first problem on oil consumption. {\bf You should always integrate over the required period when you are asked to find something over a period of time.}
If you just substituted a value for $t$ in the formula, this will only give you the amount consumed at
that point of time $t$. In the oil consumption problem, note that {\bf we were asked to find the amount of oil consumed over the period of a year} and NOT how much oil was being consumed at the end of one year. Recall that we computed the function as
$$
y=1.2 e^{\rm ln}(1.015)t.
$$
\medskip
\n b) Find the function that gives the amount of oil consumed between $t=0$ and a future time $t$.
\medskip
\n {Ans:} So we have to integrate between time $s=0$ and a future time $s=t.$ {\bf I have changed the variable here to $s$ because the future time is denoted by $t$.} Thus the answer is
$$
\int^t_0 \, 1.2 e^{{\rm{ln}}(1.015)s} \, ds = \frac{1.2}{{\rm ln}(1.015)} \times (e^{{\rm ln}(1.015)s}\,\LARGE{\mid}^t_0 = \frac{1.2}{{\rm ln}(1.015)} \times (e^{{\rm ln}(1.015)t}-1) ~{\rm million ~barrels}.
$$
\medskip
\n c) After how many years will the amount of oil consumed reach 10 million barrels?\hb
\n{Ans:} We want $\frac{1.2}{{\rm ln}(1.015)}\times (e^{\rm ln}(1.015)t-1)=10.$ Thus
$$
\begin{array}{clc}
e^{{\rm ln}(1.015)t}&=& 10 \times \frac{{\rm ln} (1.015)}{1.2} +1\\
& &\\
(1.015)^t& =& 10 \times \frac{{\rm ln} (1.015)}{1.2} +1\\
& &\\
t \times {\rm ln}(1.015) &=& 10 \times \frac{{\rm ln} (1.015)}{1.2} +1\\
& &\\
t & = &\frac{{\rm ln}\left((10{\rm ln}(1.015)/2) +1\right)}{{\rm ln}(1.015)}\\
\end{array}
$$
which gives $t \approx 7.85551 $ years.
\medskip
PROBLEM: One thousand dollars is invested at $5\%$ interest compounded continuously.\hb
\n (a) Give the formula for the amount $A$ as a function of time $t$, after $t$ years.\hb
\n{ANS:} See the notes sent yesterday; we have $P=1000$, $r=0.05$. Hence
$$
A(t)\,=\, 1000\times e^{0.05 t}.
$$
\bigskip
\n (b) How much money will be in the account after 6 years?\hb
\n{ANS:} Here we just substitute as we are asked to find the value at the end of 6 years, so
we substitute $t=6$ to get
$$
A(6) =1000\times e^{0.05\times 6} =1000\times e^{0,3}\approx \$1349.86.
$$
\bigskip
\n (c) After six years, at what rate will $A(t)$ be growing?\hb
\n{ANS:} Need to measure {\bf RATE OF GROWTH}, which is the derivative
$A'(t)$.
$$
A'(t)=1000\times (0.05)\times e^{0.05 t}=50e^{0.05}t.$$
$$
A'(6)\approx \$ 67.49.
$$
Therefore after six years, the amount will be growing at the rate of \$67.49
per year.
\medskip
\n (d) How long is it required for the initial investment to double?\hb
\n{ ANS:} We must find $t$ so that $A(t)=\$2000.$ So we look at $1000e^{.05 t} =2000$ and solve
for $t$.
$$
\begin{array}{lcl}
1000\,e^{.05t} &=& 2000\\
&&\\
e^{0.05t} &=&2\\
&&\\
{\rm ln}\,e^{.05t}& =&{\rm ln} \,2\\
&&\\
0.05 t& =& {\rm ln} 2\\
& &\\
t&=&\frac{\rm ln\, 2}{.05}\approx 13.86~{\rm years}.
\end{array}
$$
Therefore the amount will double after approximately 13.86 years.
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