Econ 407: Problems dealing with optimal timing using exponential and log functions. See the folllowing example. Suppose you have a property that is worth $1000 and increases in value according to 1000*exp(t^1/3). Further, assume you could invest your $1000 at a rate of 9 percent

or borrow the same amount at this rate. How many years should you hold on to this property?

> v:=1000*exp(t^(1/3));

> plot(v,t=0..10);

Form the following equation, take the first deriavtive; set it equal to zero and solve for time--t.

as shown below.

Nv:=1000*exp(t^(1/3)-0.09*t);

> nv1:=diff(Nv,t);

> solve(nv1,t);

The answer clearly is the second one--the positive root which is about 7.12 years.

plot(Nv,t=0..10);

Another example when the interest rate is 8 percent.

V:=100*exp(t^(1/2));

> nv:=100*exp(t^(1/2)-0.08*t);

> nv1:=diff(nv,t);

> solve(nv1,t);

> nv:=100*exp(t^(1/2)-0.12*t);

> nv1:=diff(nv,t);

> solve(nv1,t);

Assignments: 9.23, 9.25,9.33

The price of agricultural goods is up by 4 poercent each year;the quantity by 2 percent. What is the annual rate of growth

derived from the agricultural sector?

TR:= P*Q

> TR:=P*Q;

Percennathe Change in TR= Percentage Change in Price + Percentage in Quantity. Using this definition, the Percentage

Change in TR= 4% +2% =6%.

> ln(TR):=ln(P) +ln(Q);

Equation of Exchange: M*V=P*T: The dominant model in classical macroeconomics. If you believe in this model the following

relation holds:

%M+%V=%P +%T, where M is money, V is velocity, P is the price and T is the number of transactions.

> Y:=ln(x);

> y1:=diff(Y,x);

If you take the time derivative of a natural log, you will get percentage change. Also, you can find the first difference in

natural log to approximate percentage change.

> ln(6.15);

> ln(5.15);

> pchminw:=1.816452082-1.638996715;

Preoblem 9.25, page 209

> PCE:=E/P;

> %changPCE:=0.04-.025;

> ln(PCE):=ln(E/P);

> %PCE=%E-%P;

> %PCE=.04-.025;

>