  Assignment 73

 "Algebra is generous; she often gives more than is asked of her" - (Jean Le Rond d'Alembert, 1717-1783) George David Birkhoff (1884-1944): One of the first great American mathematicians to be recognized abroad, Birkhoff's work on dynamical systems has been used extensively by modern mathematicians and scientists. His many contributions include developing theories relating to differential equations and coloring maps. He received many worldwide honors, including the Querini-Stampalia Prize of the Royal Institute of Science,Arts and Letters of Venice in 1923. He held positions at Princeton, Harvard, and the University of Wisconsin. Why does Herkimer think that coffee gets its name from a famous childrens' story? Answer: He thought it was Snow White's eighth dwarf, the first cousin of Sneezy. Herky wants to know: Should you volunteer for submarine duty if you don't want to get on a ship that sinks on purpose? Is it true that those who exercise regularly die healthier? ASSIGNMENT #73 Reading: Read the Items for reflection portion below. Written: Problems provided on Algebra II Financial Worksheet. Mathematical word analysis:AREA : Taken directly from the Latin word area , which referred to a flat piece of level ground.
Financial example #1:

I want to have \$50,000 available in ten years. I will save for this through beginning-of-year equal annual deposits for 10 years. What deposit should I make if I assume a 9% interest rate?

Response: Here is a display of the situation, with x representing the unknown deposit accumulating at 9%.

 \$50,000 Deposit x x x x x x x x x x Year 0 1 2 3 4 5 6 7 8 9 10

The linear equation to be solved is

x(1.09)[(1.0910-1)/.09] = \$50,000

The solution is x = \$3019.27.

Financial example #2:

I make four beginning-of-year deposits, and then just let the money accumulate at 7% interest. What amount, A, will I have at the end of ten years?

 A Deposit \$1,000 \$1,000 \$1,000 \$1,000 Year 0 1 2 3 4 5 6 7 8 9 10

We have A = \$1000(1.07)7 + \$1000(1.07)8 + \$1000(1.07)9 + \$1000(I.07)10 . This is a geometric series with a = \$1000(1.07)7 and r = 1.07. Using the formula for the sum of a geometric series, we have \$1000(1.07)7[(1.074 - 1)/(1.07 - 1)] = \$7,129.58.

We could also use our TI-83 and calculate sum(seq(1000*(1.07)^x,x,7,10,1)) = 7,129.58 (dollars).

 Problem: I deposit \$5000 now, \$3000 one year from now, and \$7000 two years from now. If I make no more deposits, what will the accumulation of the three deposits after 20 years at an annual interest rate of 7.5%? Solution (with communication): The accumulation is \$5000(1.075)20 + \$3000(1.075)19 + \$7000(1.075)18 = \$58,824.35. Problem: I invest \$10,000 now and \$20,000 one year from now. At the end of 5 years, the accumulated value of these two investments is \$48,000. What is the annual interest rate earned? Solution (with communication): If x is the annual interest rate, then 10000(1+x)5 + 20000(1+x)4 = 48000 ==> (1+x)5 + 2(1+x)4 = 4.8. Using the TI-83 and entering Y1 = (1+x)^5+2(1+x)^4 and Y2 = 4.8, and then using INTERSECT, we find that x = 0.1142248, or about 11.42%.