BUSSINESS MATHEMATICS. LINEAR FUNCTION y = a + b x Cost Revenue Profit BEP . Is there any relationship between linear function with buss.problem. Recall : COST / TOTAL COST Two kind of cost : Fixed cost not depend on the item produced
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y = a + bx
Cost Revenue Profit BEP
COST / TOTAL COST
Two kind of cost :
Fixed cost not depend on the item produced
not a function of item produced
Variable cost a function of item produced
the more item produced, the more is the cost
Cost function :
C(x) = Fixed cost + (Variable cost).x
the cost of producing x items
The “parameters” (numbers that are specific to a particular business situation), are a and b.
a = constant Fixed cost
b = slope
= a rate of change
change of the y-variable with respect to the x-variable
(If the x-variable is time in hours and the y-variable is distance in kilometers , then the slope would be expressed as e.g. 50 kilometers per hour. Indeed the familiar rate of speed!
Can you draw the function ?
If x = 0 C = ?
If x = 20 C = ?
draw a line through the points
what does the constant stand for ?
what does the coefficients stand for
What about this function :
C(t) = 25 – 5 t
positive trend or negative trend
what does the negative sign mean
give an example
The total cost equation is a function of 350000 and 5000 x
So that the equation becomes : C = 350000 + 5000 x
What is the (total) cost for 100 items C = 350000+5000(100)
= 850 000
It is not allowed to find C(150) why ?
Most real-world functions will have restricted domains
A = 6t + 10, 0 ≤ t ≤ 100
The “0 ≤ t ≤ 100” is a domain restriction, meaning that the function is valid only for values of t between 0 and 100, inclusive.
To graph it, calculate the points at the extreme left and right:
if t = 0, A = 10 point (0, 10)
if t = 100, A = 610 point (100, 610)
Graph the points and draw the line:
f(x) = 3x2 +1 , if x>1
find : f(2)
What is the cost for the company that produced q unit material in a certain comodity, using a function
C(q) = q3 – 30 q2 + 500 q + 200
a. cost for 10 unit
b. Cost for the 10 th
we can say that R(x) = 500 x
then we can find what is the Revenue for 200 being sold 100 000
Can you draw a line on that ?
The revenue is 200000 for 75 units sold
What is the function of Revenue ?
y = a + bx
( y-y1 )/(y2-y1) = (x-x1)/x2-x1) ?
What is the interpretation of the function you’ve got
Profit = Revenue – Cost
P = R – C
What does Revenue mean ?
Revenue (=R) is a function of item produced R = f(x)
(It could be linear or non linear as shown below)
For our example:
C(x) = 100 + .50x
R(x) = x(1 - .0001x)
P(x) = R(x) – C(x)
= x(1 - .0001x) – (100 + .50x)
= - 0.0001x2 + 0.50x - 100
At break Even there is no profit
the costs equal the revenue : R(x) = C(x)
If the total costs are C(x) = 500 + 90x
and total revenues are R(x) = 150x – x2
Find the breaking points
R(x) = C(x) 150 x – x2 = 500 +90 x
x2 – 60x + 500 = 0 (x-10)(x-50) = 0 x=10 ; x=50
Bring this x value to either C(x) or R(x)
BEPs are : (10,5000) and (10,1400)
Or you can find the BEP in units
BEP = (Fixed cost)/ (Price-variable cost)
R < C ?
R > C ?
Find another example
1. If total cost of a certain company C is equal to 250000 + 250 x and total Revenue is 500 x.
When is the BEP occur ?
BEP R = C
500 x = 250000 + 250 x
250 x = 250000 x = 1000
It means that BEP occurs when 1000 has been sold
and at that time the Revenue is 1000 * 500 = 500000
When x = 1500
what would you find ? Loss or profit ?
how much is that ?
show by a graph
(the Revenue is in linear function)
a. What is the Revenue function draw a line
b. What is the cost function draw
c. Find out when it would be loss
d. When will you get profit and how much is that,
Show it in a graph as well
usually in percentage
Find out an example about Av Fixed cost & Av Var cost
Using the example : R(x) = (150x –x 2) and C(x) = (500 + 90x)
a. write the profit function
b. what level production maximizes the profit
c. what is the maximum profit
= (150x –x 2) – (500 + 90x) = -x2 +60x – 500
b. At vertex , function reaches maximum point
V(x) = -b/2a = -60/(2 * -1) = 30
means producing 30 units maximizes the profit
So, the maximum profit is 2200
Assume the demand is Qd = 150 - 5P and that supply is given by Qs = 90 + 10P. Mathematically, what we are looking to find is the point where the quantity demanded (Qd) is equal to the quantity supplied (Qs)
as price goes down, demand goes up
prices , etc
P is price, a as a constant, x is quantity demanded
P and x are always positive
0 ≤ P ≤ a
0 ≤ x ≤ b
Look at this simple example :
a. Find the interval of P and x
b. When does the free goods occur (free goods is the conditon where no value for the unit, the price P = 0)
P must be in the interval of 0 – 8
x must be in the interval of 0 – 4, to get P=0
as x = 0 P = 8
x = 1 P = 6 ...... etc, up to x = 4 P=0
so when free goods occur, x = 4
When P is $6.0 x should be 1.5 and when P is $4, x should be 3.0
When quantity demanded is 3 units, the price is $4
When quantity demanded is 4 units, the price is $2.65
when the price goes down, the demand goes down
For example :
1. Given the function x=2p – 10, the lowest price for the consumer is at x=0, it means that no supply at all. At that time the price is $5
When the price is $10, the units sold should be 10
Can you draw the function ?
For 250 units, the supply is $1500
Remember : (y-y1)/(y2-y1) = (x-x1)/(x2-x1)
When the ordinate is P (p-p1)/(p2-p1) = (x-x1)/(x2-x1)
p = 10x – 1000
Draw the supply function
At market equilibrium
Demand = Supply
E.g. The demand function for a product is given by p2 + 2q = 1600
The supply function is given by 200 – p2 + 2q = 0
(p = price ; q = quantity)
Find : a. The equilibrium quantity
b. The equilibrium price
Demand function q = (-1/2) p2 + 800
Supply function q = (1/2) p2 – 100
ME D = S (-1/2) p2 + 800 = (1/2) p2 – 100 p = 30
The equilibrium price is 30
Bring this price to either S or D function
q = (1/2)(30)2 – 100 = 350
The equilibrium quantity is 350 units