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Econ 407: Answers to problem set 3.

> f(t):=lambda^(2)*t*exp(-lambda*t);

> eval(f(t),{lambda=0.02,t=10});

> eval(f(t),{lambda=0.02,t=70});

> eval(f(t),{lambda=0.02,t=100});

> f(t):=0.02^(2)*t*exp(-0.02*t);

> f1(t):=diff(f(t),t);

> solve(f1(t),t);

> plot(f(t),t=0..100);

Note that if you make it to 50 years of age, your chances of dying before a certain age declines!

Problem 2: Depreciation.

> S:=500*exp(-.12*10);

> S:=500*exp(-.12*t);

S := 500*exp(-.12*t)

> plot(S,t=0..50);

Q. 3

> S:=100*exp(0.05*1);

> solve(0.05=ln(1+i),i);

Note that 5 percent compounded continously is equivalent to 5.12710.. compounded once a year.

4

> S:=p*exp(.08*1/4)+p*exp(0.08*3/4)+p*exp(0.08*2/4)+p*exp(0.08*4/4);

S := 4.206135729*p

> solve(S=10000.0,p);

2377.479151

Problem 5: US population: population 200 years ago =5, after the first two hundred years=200.

P(t):=200;

P(t) := 200

> solve(200.0=5.0*exp(r*200.0),r);

.1844439727e-1

> P(t):= 5*exp(.1844439727e-1*200);

P(t) := 200.0000000

Problem 6: How much was Manhattan worth in 1980?

The following graph shows another application of exponential functions. It is the plot of the standard normal distribution which has a mean=0 and a standard deviation equal to one.

problem 6

> Vman:=24*exp(0.05*354);

Vman := 1167410436.

Note that at the rate of 5 percent, $24 would grow to 1,167,410,436. I think the chief's children would be happy to have the money!

f(x):=exp(-0.5*x^2);

> plot(f(x),x=-infinity..infinity);

To show that this distribution has a zero mean, we can use the integration tool.

> E(x):=int(x*f(x),x=-infinity..infinity);

Another example: Find the average gas price per barrell in Radford, if the the density (probability) function is given according to:

f(x)=((x-25)^0.5)/18, given that the price of gas per barrell is 25<x<34.

> f(x):=((x-25)^0.5)/18;

> avep:=int(x*f(x),x=25..34);

>