ECONOMICS DUE IN 6 HRS
XWALKER
functionfandango1_1.docFUNCTION FANDANGO
Name ___________________________ Due __________________ (worth 50points)
This assignment exercises your knowledge of fundamental algebra. Rusty algebra gives more problems than economics. You are hereby warned that some problems have nice, clean answers and some do not (even if calculations are carried to several decimal places). Please carry out calculations to two decimals before rounding off, as a general rule. Also, please show your work where possible.
1. Here is a linear demand function: Q = 20 -5P. Find its price function by inverting the demand function. Then find its total revenue function by multiplying through by Q. EXAMPLE: The linear demand function Q = 400 -250P inverts into the price function P = 1.6 -0.004Q. Multiplying this by Q gives its total revenue function TR = 1.6Q -0.004Q2. This skill will be useful in assignment 4. Show the algebra involved.
P =4 – 0.2Q
Q=20 -5P 4 -0.2Q=P P=4 -0.2Q
Q -20=-5P TR= 4Q -0.2Q2 TR = 4Q -0.2Q2
2. Evaluate the following expression. Y = 3X + 5(X + 2)2 EXAMPLE: When X = 1, Y = 48.
a. Y=3(0) +5(0 +2)2 b. Y= 3(2) +5(2 +2)2 c. Y=3(3) +5(3 +2)2 d. Y= 3(5) +5(5 +2)2
Y=0 +5(4) Y=6 +5(4)2 Y=9 +5(5)2 Y=15 +5(7)2
Y=20 Y= 6 +80 Y=9 +5(25) Y= 15 +5(49)
Y=9 +125 Y= 15 +245
a. When X = 0, Y = 20 b. When X = 2, Y = 86
c. When X = 3, Y = 134 d. When X = 5, Y = 260
3. Evaluate the following exponentials. You may need to use a calculator with a Yx key. EXAMPLE: X -1/4. If X = 16, this gives _0.5__. Compute all to two decimal places or more.
a. 3-1/3 = 0.6934 b. 20 =1 c. 31/3 =1 d. 43/2 =32
a. X-1/3, When X = 3, this gives 0.6934.
b. X0 , When X = 2, this gives 1.
c. X1/3, When X = 3, this gives 1.
d. X3/2, When X = 4, this gives 32.
4. Find the two roots of each of the following quadratic functions (that is find the two X values making Y = (0). This skill may be useful in assignment 11. Example: Y= 3X2 -11X +6 is the product of (3X -2)(X – 3). If you let Y = 3X-2 then X = 2/3 will make Y = 0. If you let Y = X -3 then X = 3 will make Y = 0. Thus both X = 2/3 and X = 3 are roots. Show the algebra involved.
a. Y = 2X2 -2X -12. The two roots are X = 3 and X = -2.
Y= 2X2 +4X -6X -12 Y=2X(X+2) -6(X+2) Y= (2X -6)(X+2)
Y=0 0=2X-6 2X=6 0=X+2
X=3 X= -2
b. Y = 2X2 + X -15. The two roots are X= 2.5 and X = -3.
Y=2X2 +6X-5X-15 Y=2X(X +3) -5(X+3) Y= (2X -5)(X +3)
Y=0 2X-5=0 2X=5 X+3=0
X=2.5 X= -3
5. Exponential functions are useful in business and economics. Lesson 7 discusses them. Show how the values are entered into your functions and also calculate the amounts for each of the following:
a1. You learn on the business channel that inflation was about 0.2% last month. Assume this rate is maintained each month for a year. What will the annualized rate be? EXAMPLE: A rate of 0.1% per month represents (1 + 0.001)12 -1 = 0.0121 or 1.21% annually.
(1 +0.002)12 -1= 0.02427 or 2.427%
a2. Suppose you set a goal of 5% sales growth for the year. What monthly growth rate would this require, assuming a level (the same) average rate for each month? EXAMPLE: A 2% growth rate for the year would require 1.02 = (1 + r)12. Solve this for r:
1
02
.
1
12
-
=
r
; r = .00165 or .165% per month on average.1.05= (1 +r)12 r =12√1.05 -1 ; r = 4.074 or 407.4%
b1. F = Pert , which assumes continuous compounding, says that the Future value (F) of an amount (P) invested today at an annual rate (r), expressed as a decimal for the time (t), in years is given by the function (assuming continuous compounding). EXAMPLE: invest $100 at the annual rate of 5 1/2% for 6 years and 3 months and you should get back (at the end of the time), F = $100e (0.055)(6.25) = $100e(0.3438) = $100(1.4102) = $141.02. If you deposit $10000 today into your saving account, what amount would you have in 10 years if it can earn 6% per annum?
$10000e (0.06)(10) = $10000e(0.6) =$10000(1.822) =$18220.1
b2. Alternatively, P = F/ert. P tells you the most you should loan. EXAMPLE: A borrower tells you that he will give you back $141.02 at the end of a loan that for 6 years and 3 months with an interest rate of 51/2% per annum. The most you should have loaned him was P = $100. Here is a loan proposition more in line with current interest rates. A borrower agrees to pay you 4 3/4% annually for 7 years and 6 months. At the end of the term, he will make a balloon payment of $20000 to repay the loan and the interest. What amount (P) should loan this prospect?
$20000= (P)e(4.75)(7.5) (P)e(0.35625) e=2.718 2.718(0.35625) 1.4279P=$20000
P=$14006.58
FunctionFandango1.doc August 29, 2016
