Group : Get A
BG1200 Mathematics for Business
Review
x-intercept means y=0
y-intercept means x=0
When the line passing through (x1,y1) and (x2,y2), you can find slope
m = y2 – y1
=
X2 – x1
change in y
change in x
There are many ways to find the Equations of Lines
>> y = mx+b When we know m or slope
>> y-y1 = m(x-x1) When we know m and (x1,y1)
Topic1 Linear Functions : Applications
 Depreciation
v = at+b
(t,v) – time,value
t – time in years
v – value of asset at time or book value
a – rate of depreciation or slope always minus
b – original value of asset or v-intercept (0,b)
Example: A new radio depreciates $30 per year, and it is worth $1200 after 5 years.
a) Find the equation that express the book value v of the radio at the end of year t
.
b) Find the original value of this radio.
c) Find the useful life.
(5, 1200)
M = -300
b = 300(5) + 1200
b = 2700
a) V = -300t + 2700
b) The original value of the radio id $2700
c) The useful life is 9 years.
 Linear Supply and Demand Functions
p = -mq+b – Demand
p = mq+b – Suply
(q,p) – quatity,price
p – selling price per unit
m – slope = change in p
change in q
q – quatity (unit)
b – demand stop in demand function , supply stop in supply function
q = 0 means No demand in demand function,
No supply in supply function
p = 0 means Maximun quantity of demand
Example:
Demand
1. A company owns an apartment building containing 100 units. If company charges
$400 per month rent, then all units can be rented out and for every increase of $20
per rent, the company will lost one units.
(q , p )
(100,400)
M=
M=
= -20
b = 20(100) + 400
b = 2400
p = -20q + 2400
Slope M =
, means for every $20 increase in price, the quantity demand
will decrease by 1 unit.
Supply
1. If the price of a certain product is $8 each 12 units can be supplied for the product
and, at $10 each 4 more units can be supplied.
(q, p )
(12, 8 )
(16, 10 )
M=
M=
=
b = - (12) + 8
=2
p= q+2
Slope M = , means for every $1 increase in price, the quantity supply will
increase by 2 units.
 Equilibrium Point
You can find the Equilibrium Point by giving
Demand function = Supply function
Revenue = Cost or Profit = 0
Examples: A group of wholesalers will buy 50 dryers per month if the price is $200 and 20
per month if the price is $300. The manufacturer is willing to supply 20 if the price is $210
and 30 of the price is $230.
a) Find the demand and the supply equation.
b) Find the equilibrium point.
c) The effect on demand if the price increases by $10.
d) The effect on supply if the price decreases by $4.
Demand (q , p )
(50, 200)
(30, 300)
M=
M=
M=-5
b = 5 (50) + 200
= 250 + 200
= 450
Demand equation: p = -5q + 450
Supply (q , p )
(20, 210)
(30, 230)
M=
M=
M=2
b = -2 (20) + 210
= -40 + 210
= 170
Supply equation: p = 2q + 170
a) Demand equation: p = -5q + 450
Supply equation: p = 2q + 170
450 – 170
=
2q + 5q
280
=
7q
40
=
q
find p
p = 2q + 170
= 2(40) + 170
= 80 + 170
p = 250
b) (40, 250 )
The effect on demand if the price increases by $10.
=
q= 2
c) If the price increases by $10, the quantity demand will decreases by 2 units.
The effect on supply if the price decreases by $4.
=
q=2
d) If the price decreases by $4, the quantity supply will decreases by 2 units.
 Linear Cost, Revenue and Profit Functions, Break-Even Analysis
>> Cost
C(X) = vx + F
C – the total cost from producting x units
X – the number of units
v- variable cost per unit such as material or fuel
F – fixed cost suh as rent, insurance and maintence
>> Revenue
R(X) = px
R(X) – the revenue from selling x units
p – selling price
>> Profit
P(X) = R(X) – C(X)
>> Break-even analysis
R(X) = C(X)
Or
Profit P(X) = 0
>> Average Cost function = number of unit produced
A(X) = C(X) X>0
X
Example:
Total cost
1.
A company store as fixed cost of $8925 and variable cost is $850 for every computer
it sells.
a) Find cost function.
b) Find the total cost when producing 10 units.
a) C(x) = vx + F
C(x) = 850x + 8925
b) C(x) = ? , x = 10
C(x) = vx + F
C(x) = 850(10) + 8925
= $17425
2.
The total cost of producing aspirin is assumed to be linear function of the number of
bottles produced. The variable cost of producing a 100- table bottle of aspirin is
$0.60. The total cost of producing 5000 bottles of size is $4500.
a) Find the cost function.
b) What is the fixed cost of this production process?
M = 0.60
(5000, 4500)
b = -0.60 (5000) + 4500
= 1500
Fixed cost is $1500
a ) C(x) = 0.60x + 1500
b ) Fixed cost is $1500
Average cost
A(x) =
1. A company store as fixed cost of $8925 and variable cost is $850 for every computer
it sells.
C(x) = vx + F
C(x) = 850x + 8925
A(x) =
A(x) =
2. The total cost of producing aspirin is assumed to be linear function of the number of
bottles produced. The variable cost of producing a 100- table bottle of aspirin is
$0.60. The total cost of producing 5000 bottles of size is $4500.
C(x) = 0.60x + 1500
A(x) =
Total revenue
1. The store can sell the computers for $1275 each.
The revenue function is R(x) = px
R(x) = 1275x
2. A manufacturer has fixed cost of $1000 and production cost of $10 for each unit
produced. The product sells for $20 per units.
The revenue function is R(x) = px
R(x) = 20x
Profit
1. A manufacturer has fixed cost of $1000 and material costs are $3, labor costs are
$7 for each units produced. The product sells for $20 per unit.
a) How many units must the manufacturer sell to make a profit?
R(x) = 20x
C(x) = 10x + 1000
P(x) = 10x – 1000
xb =
= 100
The manufacturer must sell more than 100 units to make a profit.
b) How many units must the manufacturer sell to make a profit of $2990?
P(x) = 10x – 1000
10x -1000 = 2990
10x = 3990
x = 399
The manufacturer must sell 399 units to make a profit of $2990.
Break- even
1. Suppose the monthly cost of producing 10 stylus pens is $140. If the production
level is increased to 30 pens, the cost will be $180.
a) Find the cost function for production of stylus pens.
b) Suppose the stylus pens are sold at units price of $5, find the revenue
function.
c) Find the profit function.
d) How much stylus pens must be sold in order to break-even?
e) At least how many pens have to be sold for the company to have profit?
(10, 140)
(30, 180)
M=
M=2
b = -2(10) + 140
= -20 + 140
b = 120
a ) C(x) = 2x + 120
b ) R(x) = 5x
c ) P(x) = 3x – 120
d ) xb =
x = 40 units.
e ) At least 41 units have to be sold to make a profit.
2. A company has a linear total cost function and has determined that over the next
three months it can produce 1000 units at a total cost of $300000. This same
manufacturer can produce 2000 units at a total cost of $400000. The units sell for
$180 each.
a) Determine the revenue, cost and profit function.
b) Find the break-even quantity
c) What is the total cost at break-even?
d) At least how many units to be sold for the company to have profit?
(1000, 300000)
(2000, 400000)
M=
M = 100
b = -100(1000) + 300000
= -100000 + 300000
= 200000
a ) Revenue function is R(x) = 180x
Cost function is C(x) = 100x + 200000
Profit function is P(x) = 80x – 200000
b ) Profit function is P(x) = 80x – 200000
xb =
= 2500 units
c ) Total atbreak-even is Cb = 100(2500) + 200000
= $450000
d ) At least 2501 units should be sold to have profit.
Topic5 Linear Programming
 Graphing System of Linear Inequalities in Two Variables
 Solving Linear Programming Problem by Graphical Method
First, write a linear objective function
Second, find the corner of point by giving any number into the linear function to find x and
y (x,y)
Third, When you have points (x,y), write a graph
Examples:
1. Maximize Z = 25x + 30y
Subject to
2x + 3y
120
3x + 2y
360
x
80
y
120
x,y
0
(x, y)
(0,120)
(40, 120)
(80, 60)
(80, 0)
(60, 0)
(0, 40)
Z = 25x + 30y
25(0)+30(120) = 3600
25(40)+30(120) = 4600
25(80)+30(600) = 3800
25(80)+30(0) = 2000
25(60)+30(0) = 1500
25(0)+30(40) = 1200
a) The optimum solution is (40, 120)
b) The maximum value of the objective function is 4600
c) Is it possible to find the minimum value of Z yes
If yes, what is it? 1200
d) Are there any redundant constraint? No
2.Minimize Z = 4x + 2y
Subject to
8x + 2y
900
3x + 6y
600
y
2x
y
250
(x, y)
(125,250)
(100, 200)
(100, 50)
(200, 0)
a) The optimum solution is (100, 50 )
b) The minimum value of the objective function is 500
c) Is it possible to find the maximum value of Z yes
If yes, what is it? 1000
d) Are there any redundant constraint? x
0
Z = 4x + 2y
4(125)+2(250) = 1000
4(100)+2(200) = 800
4(100)+2(50) = 500
4(200)+2(0) = 800
 Special Cases in Linear Programming
1. Infeasibility ( Empty Feasible Region)
You cannot find the solution to the linear programming problem.
2. Unboundedness
No solution in a maximization problem
3. Multiple Optimal Solutions
There are same maximum value or both the same minimum value.
4. Redundancy
A constraint which does not relate the feasible region
 Formulating Linear Programming Problem
The symbol of this
>
more than , exceed , greater than
<
less than , fever than , up to
>
at least , no less than , more than or equal , minimum
<
at most , less than or equal , no more than , limit , available ,
exceed
maximum, not
Example:
1. The Pinewood Furniture Company produces chairs and tables from two resourcedlabor and wood. The company has 80 hours of labor and 36 pound of wood
available each day. A demand chair is limited to six per day. Each chair requires 8
hours of labor and 2 pounds of wood to produce, while a table requires 10 hours of
labor and 6 pounds of wood. The profit from each chair is $400 and from each table
$100. The company wants to determine number of chairs and tables to produce
each day in order to maximize profit.
x – the number of chair produced each day.
y – the number of table produced each day.
Maximize
P = 400x + 100y
Subject to
8x + 10y
80
2x + 6y
36
x
6
x,y
0
2. A man is planning to invest at most $22000 in two banks, Bank A and Bank B. He
wants to invest at least $2000 but no more than $14000 in bank A. He will invest no
more than $15000 in bank B. The interest in bank A is 6% and in bank B it is 6.5%.
How much should he invest in each bank to maximize his return?
x – the amount invest in bank A
y – the amount invest in bank B
Maximize
R = 0.06x + 0.065y
Subject to
x+y
22000
x
2000
x
14000
y
15000
x, y
0
Topic4 Exponential and Logarithmic Functions
Graphs and Properties
Solving Exponential and Logarithmic Equations
Logarithm Function
Y = f(x) = logbx  x = 𝒃𝒚
Important
1. b>0 and b ≠ 1
2. x must be positive number
3. no y-intercept
4. x-intercept is (1,0)
5. Range is real number (-∞,∞)
6. Domain is positive real
number (0, ∞)
Common logarithm log base 10
1. log105
2. log10 30
3. log10(2x-4) =2
 log 5
 log 30
 log(2x-4) =2
Natural Logarithm  log base e
1. ln7
2. ln3
3. ln(6x-3) = 5
 loge7
 loge3
 loge(6x-3) = 5
Properties of Logalithm
1.
2.
3.
4.
loga1 = 0
logaa = 1
logcax = xlogca
logcab = logca +logcb
5. logc a/b =logca-logcb
6. lne =1
Exponentail Function
Y = bx
b>0, b≠1,x ɛ R
y = ex
e = 2.71828…..
1. Y-intercept
2. X-intercept
No x-intercept
 2x+2
Has x-intercept
 2x+2
3. Domain is the set of real number (x ɛ R) or (-∞,∞)
4. Range is the
5. set of positive numbers (y>o) or (0,∞)
6. Decrease faster
Application of Exponential and Logarithmic Function
Example1 : solve
= 20
= 1+
2.5 – 1 =
=
= 0.2t
-0.980292 = 0.2t
t = -4.9041463
Topic3 Mathematics of Finance
Simple interest is calculated once for the period of loan or an investment
I = Prt
Compound interest
- Borrow = A loan , a debt
- Principle = The starting amount of money that you borrow
- Put money in = Deposit money
Future Value
S = P(1+r)n
Present Value
P = S(1+r)-n
P = present value
S = future value , compound amount
r= periodic rate = i/m
n=total number of period = t*m
m=number of period per year
annually
=
semiannually =
1
2
quarterly
monthly
weekly
daily
=
=
=
=
4
12
48
365
Compound interest or interest paid : I = S-P
Example 1 : When joey was born, his father deposited$2,000 in savings account in his name.
At that time the bank was paying 6% interest compounded semiannually.
A. If the rate did not change, what was the value of the account after 15 years ?
B. If the money had been invested at 6% compounded quarterly, what would the
value of the account have been after 15 years?
A.
p = $2,000 , r =
S=
= 0.03
1
S=
S = $ 4854.52
B. P = $2,000 , r =
S=
1
S = 2,000
S = $4886.44
= 0.015,
Example 2 : Over the past 10 years you have made the following investment :
1. Deposited $10,000 at 7% compounded semiannually, in a 3-year certificate of
deposit.
2. After the 3 years, you took the maturity value (principal and interest) of the CD and
added another $5,000 to buy a 4- year, 6% certificate compounded monthly.
3. When that certificate matured, you added another $8,000 and bough a 3-year, 6%
certificate compounded quarterly.
a. What was the total worth of your investment when the last certificate
matured ?
b. What was the total amount of compound interest earned over the 10year period ?
① P = 10,000 , r =
S=
= 0.035 , n = 2×3 = 6
1
S₁ = 10,000
= 122,292.55
② P = S₁ + 5,000 = 17,292.55 , r =
S=
1
S₂ = 17292.55
= 21,969.997
③ P = S₂+8,000 = 29,969.997 , r =
S=
= 0.005 , n = 12×4 = 48
= 0.015 , n = 4×3 = 12
1
S₃ = 29,969.997
= 35,832.67
A . Total amount is $35832.67
B. Total interest = S₃ - (10,000 + 5,000 + 8,000)
= $12832.67
Effective Rate
reflects the real rate of return on an investment
Re = [
- 1] x 100
Net Present Value (NPV) การที่ประเมินว่าเราควรลงทุนดีไหม ใช้สูตร
P = S(1+r)-n
โดยมี Keywords : initial , cash flow , guarantee , return
NPV > 0 means the investment is profitable
NPV < 0 means the investment is not profitable
NPV = 0 means the investment is desirable cash flow initial investment
Example 1 : The car worth $4,000 depreciates in value by 40% each year. Write an
exponential function that gives the value of the car after x years of depreciation
Decay : y =
V(t) = 4,000
Example2 : The following demand equation for a product is given by D(P) =
where p is a retail price in cents per unit and D(P) is the demand in thousands of units .
a. What demand can be expected at 35 cents per unit ?
b. What is the total revenue expected to be at that price ?
a)
When p = 35 cents
P(35) =
=
= 547(0.32628)
= 178,475 thousand units
b)
Total revenue = selling price (P) × demand (units)
=P
When p = 35 cents = $0.35
R = 0.35(178,475)
= $62,466.25
Example3 : If the supply function for a product is given by p = 100ln(q+1) ,then
a.
At what price will 9 items be supplied ?
b.
How many units, to the nearest unit, will be supplied if the price is
$346.57 ?
a.
b.
p = 100ln(q+1)
= 100ln10
= 230.2585098
346.57 = 100ln(q+1)
3.4657 = ln(q+1)
3.4657 =
= q+1
31.9989-1 = q
Q ≈ 31
1. Ordinary Annuity **** (At the End)
• Present Value (A)
A=Ran r

A=R[
] Interest earn (I) = n(R) – A
• Future value (S)
S=Rsn r

S = R[
]
Interest paid (I) = S – n(R)
2. Annuity Due **** (At the beginning)
• Present value

A = R (a n-1 r )+ R
• Future value

S = R (s n+1 r) – R
Sinking Fund
a fund that is accumulated for the purpose of paying off a financial obligation.
• S = Rsn r

R = s/sn r
• S = R[s n+1 r - 1]

R = s / s n+1 r -1
Example:
1. A company’s new corporate headquarters will be completed in 3.5 years.
At that time, $750,000 will be needed for office equipment. How should be invested
monthly to fund that expense? Assume 9% interest.
S = 750,000, r = 0.0075, n =42
S = Rs_( n r)
R = s / s_( n r)
R = 750000 / 49.15329148
R = 15258.39
R = $15258.39
2. Parents have set up a sinking fund in order to have $35000 in 12 years for
their children’s college education. How much should be paid quarterly into and account
paying 10% compounded quarterly.
S = 35,000, r = 0.025, n = 48
S = Rs_( n r)
R = s / s_( n r)
R = 35000 / 90.85958243
R = 385.21
R = $385.21
Example:
1.Mr. Tim deposit $1500 at the beginning of each six months in the account for 12
years. If interest rate is 5% compounded semiannually. He then puts the total amount on
deposit in another account paying 6% compounded semiannually for another 8 years. Find
the final amount on deposit after the entire 20 year period.
R = 1500, r = 0.025, n = 24
S = R ( s_( n+1 r) – 1)
S = 1500 (34.15776393) – 1500
S = 49736.65
S = $49736.65
S = 49736.65, r = 0.03, n = 16
S = S (1+r)^n
S = 49736.65 (1+0.03)^16
S = 79812.72
S = 479812.72
Amortization
- Relevant for larger purchases made over longer periods of time
- Loans to pay for homes and automobiles are usually amortized
- There may or not may dowm payment
• A = Ra n r

R=A/an r
• A = R a n-1 r+1

R = A / a n-1 r +1
Amortization Schedule
1. Principle outstanding at the beginning of k period
Ak = R a (n-k+1)
2. Total payment = R(n)c
3. Total finance charge = R(n)-A
4. Interest save = R(n pay off) – A pay off
5. Interest paid = total interest – interest save
r
1. ***A ( Present value )
- Principle outstanding
- Unpaid balance
- Amount finance
- Loan amount
A = cash – down payment
2.
3.
4.
-
r( periodic rate )
n(total number of payment)
R
Annual payment
- Semiannual payment
- Quarterly payment
- Monthly payment
- Per month/year/quarter
- Each month/year/quarter
- Every month/year/quarter
- A month/year/quarter
5. I ( Interest )
- Total interest charged
6.
-
Total finance charged
Total interest paid
A ( Present value )
Principle outstanding
- Unpaid balance
- Amount finance
- Loan amount
A = cash – down payment
7. r( periodic rate )
8. n(total number of payment)
9. R
- Annual payment
- Semiannual payment
- Quarterly payment
- Monthly payment
- Per month/year/quarter
- Each month/year/quarter
- Every month/year/quarter
- A month/year/quarter
10. I ( Interest )
- Total interest charged
- Total finance charged
- Total interest paid
Example:
1. Paul takes out a loan for $36000 to purchase a new car. If the interest rate on this
loan is 8%, and he wants to amortize the loan in four annual payments would be? How
much interest will he have paid for the car loan at the end of four years? Also construct the
amortization schedule for Paul.
A = 36000, r = 0.08, n = 4, R =
Period
1
2
3
4
Principle
outstanding at
the beginning of
period.
36000
28010.85
19382.57
10064.03
= 10869.15
Interest for
period.
Payment at the
end of period.
Principle repaid
at the end of
period.
2880
2240.87
1550.61
805.12
10869.15
10869.15
10869.15
10869.15
7989.15
8628.29
9318.54
10064.63
Total
7476.60
43476.60
36000
2. Josh takes out a five-year auto loan for $11500 that calls for annual payments at a rate of
8% compounded annually. Construct the amortization schedule for Josh
A = 11500, r = 0.08, n = 5, R = 2880.25
Period
1
2
3
4
5
Total
Principle
outstanding at
the beginning of
period.
11500
9539.75
7422.68
5136.24
2666.89
Interest for
period.
Payment at the
end of period.
Principle repaid
at the end of
period.
920
763.18
593.81
410.90
213.35
7476.60
2880.25
2880.25
2880.25
2880.25
2880.25
14401.24
1960.25
2117.07
2286.44
2469.35
2666.89
36000