Lecture 1. Applying Calculus in Business and Economics I.pdf

Applying Calculus in Business and Economics

1. Applying Calculus in
Business and Economics I
Lecture 1
1

2. 1.Understand rates of changes and the process of differentiation.
2.Be able to use differentiation to calculate turning points and find
maximum or minimum of a function
3.Apply differentiation to solve business problems:
•Maximizing profits
•Minimizing costs
2
Learning Objectives

3. 3
The slope/gradient of a straight line
A straight line expresses a linear relationship.
Linear function: y = a + bx21
21
yyy
slope
xxx
−
==
−
A
B
x
y
∆y
∆x

4. Examples:
Find the slope of the straight line passing through
(a) A(1,2) and B(3,4)
(b) A(1,2) and C(4,1)
(c) A(1,2) and D(5,2)
4
The slope of a straight line

5. Examples: Find the slope of the straight line passing through
(a) A(1,2) and B(3,4) (b) A(1,2) and C(4,1) (c) A(1,2) and D(5,2)
5
The slope of a straight line

6. The slope is:
• positive if the line is uphill/upward
• negative if the line is downhill/downward
• zero if the line is horizontal/flat
6
The slope of a straight line
How about the slope of
a vertical line?

7. •Because the slope is the increase in y divided by the increase in x, it
gives us the change in y per unit of change in x. It is called the rate
of change of y.
•‘Slope’, ‘gradient’, and ‘rate of change’ are identical in meaning.
7
The slope of a straight line21
21
yyy
slope
xxx
−
==
−

8. •A tangent is a straight line that
passes through a point on a
curve and just touches the curve
at this point.
8
The slope/gradient of a curve

9. The slope of a curve at �=�
0 is defined to be the slope of the tangent at �=�
0
9
The slope/gradient of a curve

10. 10
Constant slope vs variable slope
Fig. 1 Fig. 3Fig. 2

11. •The derivative of � at the point �
0 is the slope of the tangent
line to � at the point �
0.
•For each value of �
0, there is a uniquely defined derivative
�’(�
0).
•The slope of a curve is also determined by a function, called the
derived function.
11
Slope and the derivative

12. •The derivative of a function
•Slope of a curve at point �=�
0: �′(�
0)
•The process of finding the derived function is known as
differentiation.
12
The derivative ()
0
lim
x
dy y
fx
dx x
→

==


13. If �=��=�
??????
then
??????�
??????�
=�′�=??????�
??????−1

The constant rule:
•If ℎ�=??????.�(�) then ℎ′(�)=??????.�
′
�; c: constant
•If ℎ�=?????? then ℎ′(�)=0; c: constant
The sum rule:
If ℎ�=��+�� then ℎ′�=�′�+�′�
The difference rule:
If ℎ�=��−�� then ℎ
′
(�)=�
′
(�)−�′�
13
Rules of differentiation

14. Differentiate these functions:
a.y = 5x
2
b.y = 3/x
c.y = 2x + 3
d.y = x
2
+ x + 1
e.y = x
2
– 3x + 2
f.y = 3x – 7/x
g.y = 2x
3
– 6x
2
+ 49x – 54
h.y = 4� – 3/x + 7/x
2
EXAMPLES
14

15. For each of the curves or lines given below, find the value of the
gradient at x = 2
a.y = 5x
2
→
b.y = 3/x →
c.y = 2x + 3 →
d.y = x
2
+ x + 1 →
e.y = x
2
– 3x + 2 →
f.y = 3x – 7/x →
g.y = 2x
3
– 6x
2
+ 49x – 54 →
h.y = 4� – 3/x + 7/x
2
→
EXAMPLES
15

16. 16
Strictly increasing/decreasing function

17. 17
Optimization: Finding maximum and minimum values

18. 18
The maximum and minimum value of the function

19. The second-order derivative is a
function derived from the slope
function of the function f(x).
19
Maximum or minimum?() ()( )
2
2
=
dy
fx fx
dx

  =
•Graph bends downwards when f ″(x) < 0
•Graph bends upwards when f ″(x) > 0.
concave
convex

20. For any function &#3627408486;=&#3627408467;(&#3627408485;), at any point &#3627408485;
0 on that function:
•If &#3627408467;′(&#3627408485;
0)=0 and &#3627408467;′′&#3627408485;
0<0 then &#3627408486; reaches its maximum at &#3627408485;
0
•If &#3627408467;′(&#3627408485;
0)=0 and &#3627408467;
′′
&#3627408485;
0>0 then &#3627408486; reaches its minimum at &#3627408485;
0
20
Maximum or minimum?

21. •The first order condition:
•The second order condition:
•If then reaches its minimum at x
0
•If then reaches its maximum at x
0
21
Summary()
0
0fx xx== ()
0
?fx= ()
0
0fx ()
0
0fx f f

22. Find the maximum of the below function:
&#3627408486;=&#3627408467;&#3627408485;=&#3627408485;
2
−3&#3627408485;+2
•The first order condition:
&#3627408486;′=&#3627408467;′&#3627408485;=2&#3627408485;−3=0
→&#3627408485;=
3
2
→
3
2
,−
1
4
is a turning point.
•The second order condition:
&#3627408486;′′=&#3627408467;′′&#3627408485;=2>0
Because &#3627408467;′′&#3627408485;>0, &#3627408467;&#3627408485; reaches its minimum at x=
3
2
22
Example

23. The total revenue refers to the total amount of money that a firm
received through the selling of its goods and services. The total
revenue (TR) received from the sale of Q goods at price P is given by
(The total revenue function):
23
Applications of differentiation - RevenueTRPQ=
The average revenue is defined as the measurement of the revenue
that is generated per unit.TRPQ
AR P
QQ
===

24. The total cost (TC) is the amount of money that the firm has to spend to
produce these goods.
•Fixed costs (FC) include the cost of land, equipment, rent, and possibly
skilled labor.
•Variable costs change according to the amount of output produced. Include
the cost of raw materials, components, energy, and unskilled labor.
•VC denotes the variable cost per unit of output
•TVC denotes the total variable cost, TVC = VC×Q
Average cost function
24
Applications of differentiation - CostTCFCTVCFCVCQ=+=+ TCFCVCQFC
AC VC
QQQ
+
== =+

25. Applications of differentiation - Profit
The profit function is defined as the difference between total revenue and total cost.
25TRTC=−

26. •Demand curve:
•Average cost curve:
Where:
x = output in units
AR = average revenue ($)
AC = average cost ($)
•Find the output (number of units) so that:
(a) maximizes total revenue
(b) minimizes average cost
(c) maximizes total profit
Applications of differentiation - Example
26

27. (a) Find the output (number of units) that maximizes total revenue.
•Total revenue (TR) = AR (price) x quantity (output)
•Finding the turning point:
•At this turning point, we have f”(x)=-2, then this is a maximal turning
point.
•Conclusion: to maximize total revenue, the firm needs to sell 10.5 units of
product.
Applications of differentiation - Example
27

28. (b) Find the output that minimizes average cost
•Finding the turning point:
•Taking the second-order derivative:
•At x = 4.5, f”(4.5) = 2/3>0 → this is a minimal point
•Conclusion: to minimize average cost, the firm needs to sell 4.5 units of
product.
Applications of differentiation - Example
28

29. (c) Find the output that maximizes total profit.
•Total profit (TP) = TR – TC
•Finding the turning point:
•Taking the second-order derivative: f”(x) = -2.x+4
•At x = 6, f”(6) = -2.6+4=-8 → This is a maximal point
•At x = -2, f”(-2)=-2.-2+4=8 → This is a minimal point
•Conclusion: to maximize total profit, the firm needs to sell 6 units of
product.
Applications of differentiation - Example
29