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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 after the age of 50 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?

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!

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.

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);

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