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Application of derivatives
to Business and economics
Presented by:
Amal Al-Kuwari
Bashayer
Noof
Hind Nader
Presentation content:
• Introduction to Application of derivatives
and it’s importance in the Business field
• The demand function
• The cost function
• The revenue function
• The profit function
• Examples
The demand function
The demand function is the function that
relates the price p(x) of a specific unit of a
product to the number of units x produced.
p(x) is also called the demand function.
p(x)= p * x
The cost Function
The cost function is the function that
relates the total cost C(x) of producing
an x number of units of a product to
that number x
• C(x) = p(x) * x
A general Example:
A football match takes place in a stadium
that can take 60,000 people. At first the
tickets cost QR30 , it was expected that
around 30,000 people would come , then
when the tickets value were lowered to
QR20 the audience attendance was
expected to rise to 5000.
Find the demand function assuming that
this is a linear Function?
Solution
• First we find the equation of the straight
line through the following points:
(30000,30) ,(35000,20)
• Secondly we find the slope:
m = (20-30)/(35,000 - 30,000)
= -10/5000
= -1/500
So the equation of this line is:
P(x) - 20 =(-1/500).(x - 35,000)
P(x)=-1/500.(x - 35,000)+20
So the demand function is :
P(x)= -1/500x+70+20
= -1/500x+90
• Total cost example:If BMW company wants to produce 10 cars
and each car cost 200,000 QR, find total
cost gain from selling this amount of cars:-
•
•
•
•
The solution:C(x) = P(x) * x
C(x) = (10*200,000) * 200,000
C(x) = 40,000,000,000 QR
The revenue function
The revenue function which represented
in R(x) is simply the product of the
number of units produced x by the
price p(x) of the unit.
So the formula is:
R(x)=x . p(x)
Example
• The demand equation of a certain
product is p=6-1/2x dollars.
• Find the revenue:
R(x)= x.p
=x(6-1/2x)
=6x-1/2x2
The Profit Function
The profit function is nothing but the revenue
function minus the cost function
So the formula is :
P(x)= R(x)-C(x)
→x.p(x)-C(x)
Notice that :
A maximum profit is reached when:
1. the first derivative of P(x) when P‘(x) is zero or doesn’t
exist
And
2. The second derivative of P(x) is always negative P″(x)<0
*when P‘(x)=0 this is the critical point, and since P″(x)<0
the parabola will be concave downward
Another important notice is:
Since P(x)= R(x)- C(x)
Then a maximum profit can be reached when:
R'(x)-C'(x)=0 and R″ (x)-C″(x)<0
Example
• A factory that produces i-pods computed
its demand, and costs and came to realize
that the formulas are as follows:
• p=100-0.01x
• C(x)=50x+10000
Find:
The number of units that should be
produced for the factory to obtain
maximum profit.
The price of the unit.
R(x)=x.p
=x.(100-0.01)
=100x -0.01x2
Now:
P(x)=R(x)-C(x)
=100x-0.01x2 –(50x+10000)
=-0.01x2 + 50x-10000
Now the maximum point in the parabola will occur
When P'(x)=0 ; P"(x)<0
P(x) =-0.01x2 + 50x-10000
P'(x)=-0.02x+50
→ -0.02x+50=0
→ -0.02x=-50
→ x=2500
P'(2500)= -50+50=0
So the P'(X)=0 When x=2500
- Now we will find the second derivative:
P″(x)= -0.02
P″(2500)= -0.02<0
So x=2500 is at a local maximum
Graph
y
80000
70000
60000
50000
40000
30000
20000
10000
x
-500
500
1000
1500
2000
2500
3000
3500
4000
4500
Finally to find the price that is needed to be
charged per unit we return to the demand
function:
P=100-0.01(2500)
= 100-25= $75
Thank
you