mathematics for management bbmp1103
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MATHEMATICS FOR MANAGEMENT
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TOPIC 6 : APPLICATION OF DIFFERENTIATION
Objectives: 1. Grasp the rules of differentiation thoroughly 2. Apply the rules of differentiation when deriving higher degree derivatives of various functions 3. Recognize and determine the functions of total costs, total revenue and total profit in economic
and business world 4. Calculate the average function of total costs, total revenue, and total profit in economics and
business world 5. Derive the marginal or ultimate function of total costs, total revenue and total profit in economic
and business world 6. Determine how to minimize the total costs function while maximizing the total revenue and total
profit functions by using differentiation
INTRODUCTION
1. dx
dyxfy '' .is the first degree differentiation
2.
2
2''''
dx
ydxfy .is the second degree differentiation
3.
3
3''''''
dx
ydxfy .is the third degree differentiation
6.1 SECOND AND THIRD DEGREE DIFFERENTIATION
Example 1:
Given 26124 23 xxxy
Derive ''y
Solutions:
62412
621234
26124
2
2
23'
xx
xx
xxxy
24212
62412 2''
x
xxy
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Example 2:
Given xexy 3
Derive ''y
Solutions:
x
x
ex
exy
2
3'
3
x
x
ex
exy
6
3 2''
Example 3:
Given 12 21 xxy
Derive '''y
Solutions:
xx
xx
xxy
22
22
12
2
11
21'
3
12
2''
4
222
22
x
x
xxy
4
13
3'''
12
34
4
x
x
xy
Example 4:
Given xey 3
Derive '''y
Solutions:
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x
x
x
e
xdx
de
eyy
3
3
3'
3
3
x
x
x
x
e
e
xdx
de
ey
3
3
3
3''
9
33
33
3
x
x
x
x
e
e
xdx
de
ey
3
3
3
3'''
27
93
39
9
6.2 TOTAL COSTS FUNCTION C
1. Total cost function is the total cost required to produce x units of a product. In short, it is the
cost required to conduct a business. 2. There are two types of costs:
(a) Fixed Costs: unchanged or unvaried costs, flat although the number of units of a product being produced varies
(b) Variable Costs: the costs which depend on the number of units of a product produced. For example, raw material, part time worker, etc
xC = Fixed Costs + Variable Costs
= Fixed Costs + (number of units) x (cost per unit)
Example: If the production costs for one unit of a children toy are RM5 while its fixed costs are RM7000, (a) determine the cost function (b) What is the total cost for producing 100 units of the above toy? Solution:
(a) The cost function, xC = fixed costs + ((number of units) x (cost per unit))
= 7000 + ( x x 5)
= 7000 + 5 x
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(b) When 100x , xC = 7000 + 5 x
= 7000 + 5(100) = 7500 Therefore, the total cost for producing 100 units is RM7500.
6.2.1 Average Total Cost Function C
The average total cost function, C is the total cost for producing one unit of a product.
x
xCxC
Example:
Given the total cost function, 402 qqC . What is the average total cost function?
Solution:
The average total cost function, qC = q
qC
= q
q 402
= q
402
6.2.2 Marginal or Ultimate Total Cost Function 'C
1. In business, the rate of change for a function is known as marginal function.
2. Marginal or ultimate total cost function, denoted by xC' is the rate of change for total cost
function over quantity Example:
Given the average cost function, x
xxC3
10
1
(a) What is the total cost function? (b) What is the ultimate total cost function? (c) Calculate the rate of change for cost (assuming the cost is in RM) when 4 units of product are
produced.
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Solution:
(a) Total cost function, xC = xxC
= xx
x
3
10
1
= 310
1 2 x
(b) Ultimate total cost function, xC' = 310
1 2 x
= 010
12 12 x
= x5
1
(c) The rate of change for cost, 4'C = x5
1
= 45
1
= 5
4
= 8.0
Hence, the rate of change for cost when 4 units of product are produced is RM0.80 per unit.
6.2.3 Minimizing Total Cost
1. In business and economic, the cost is usually reduced (minimized) to obtain the highest
(maximized) production revenue and total profit. Example:
If 100501.0 2 qqqC is the cost function.
(a) Obtain the average cost function (b) Determine the production level, q which minimize the average cost
(c) What is the minimum value for the average cost? Solution:
(a) Average cost function, qC = q
qC
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= q
qq 100501.0 2
= q
q100
501.0
(b) STEP 1: Find qC '
2
1111
1
10001.0
1001001.0
100501.0
100501.0'
q
qqqC
STEP 2: Find qC ''
3
3
12
2
200
200
10020
10001.0''
q
q
q
qqC
STEP 3: Find 0' qC
100
10000
01.0
100
01.0100
010001.0
2
2
2
q
q
q
q
STEP 4: Substitute the value of q into qC ''
3
3
100
200
200100''
q
C
0100
2003 , then qC will have a mininimum value when =100
(c) 100q
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7
151
100
100510001.0
100501.0
q
qqC
6.3 TOTAL REVENUE FUNCTION R
1. Total revenue function, xR is the revenue received from production and sales of x unit of the
product.
Total Revenue Function, xR = price p x quantity x
6.3.1 Average Total Revenue Function R
1. The average total revenue function, xR is the revenue received from selling one unit of a
product, x
xRxR .
6.3.2 Marginal 0r Ultimate Total Revenue Function 'R
Ultimate total revenue function is the rate of change of total revenue over quantity of a product:
Ultimate Total Revenue Function = xR'
Example: The demand function of a product is given by 500200 qp .
(a) What is the total revenue function? (b) Determine the ultimate total cost function Solution:
(a) Total Revenue Function, xR = quantity x price
= qp
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= 500200 qq
= qq 500200 2
(b) Ultimate Total Revenue Function, xR' = qq 500200 2
= 1112 5002002 qq
= 500400 q
6.3.3 Maximizing Revenue Function
1. In business and economy, the total revenue is usually maximized to achieve the maximum profit. Example 1:
The demand function of a product is given by 4
80 qqp
.
(a) Determine the quantity which maximizes the total revenue. (b) Obtain the price which maximizes the total revenue. Solution:
(a) STEP 1: Find xR
xR = pq
4
80
= 4
202q
q
STEP 2: Find xR'
xR' = 4
202q
q
= 1211
4
1220 qq
= q2
120
STEP 3 : Find xR ''
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2
1
2
10
2
120''
11
q
qxR
STEP 4: Find 0' xR
40
202
1
02
120
q
q
q
STEP 5: Substitute the value of q into xR ''
02
140'' R , then xR '' will have a maximum value when q = 40
(b) 4
80 qqp
10
4
40
4
408040
p
The price has to be fixed at RM10 in order to maximize the revenue Example 2:
The demand function of a product is given by qp 600 . Suppose the number of unit produced is
no less than 100 but not more than 400 units, determine the quantity which has to be produced to maximize the total revenue? Solution:
(a) STEP 1: Find qR
qR = pq
= qq600
STEP 2: Find xR'
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xR' = qq 600
= Let 1'
xg
qxg, Let
q
xh
qxh
6002
1'
600
Using the product rule formula
= xgxhxhxg '.'.
= 16006002
1q
=
q
q
q
6002
6002600
6002
=
q
q
q
q
6002
6002
6002
= q
6002
21200
= q
q
6002
31200
=
q
q
6002
4003
STEP 3 : Find xR ''
xR '' =
q
q
6002
4003
= Let 3'
31200
xg
qxg, Let
2
3
600'
6002
qxh
qxh
Using the division rule formula
=
2'.'.
xh
xhxgxgxh
=
22
3
6002
6003120036002
q
qqq
STEP 4: Find 0' xR
STEP 5: Substitute the value of q into xR ''
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02
140'' R , then xR '' will have a maximum value when q = 40
Example 3: A research has been conducted to determine the import tax of a unit of electronic item made in a
foreign country. The demand on that particular item is given by a function ttD 208000 , where
D denote the demand quantity (in hundreds units) and t represent the import tax (in cent unit).
(a) Determine the revenue function for tax, tR .
(b) Calculate the import tax which needs to be imposed to maximize the tax revenue. (c) What is the maximum tax revenue? (d) Obtain the quantity of the required electronic item at the tax level which maximizes its revenue. Solution:
(a) STEP 1: Find tR
tR = ttD
= tt208000
= 2208000 tt
(b) STEP 2: Find tR'
tR' = 2208000 tt
= 1211 2028000 tt
= t408000
STEP 3: Find tR ''
40
400
408000''
11
t
ttR
STEP 4: Find 0' tR
200
800040
0408000
t
t
t
When 200t , 0200'' R then the total import tax which need to be imposed is RM200 to
maximize the tax revenue.
(c) STEP 5: Substitute the value of t into tR
tR = 2208000 tt
200R = 2200202008000
= 8000001600000
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= 800000 Maximum tax revenue is RM800000.
(d) tD = t208000
= 200208000
= 40008000
= 4000
To reach the level of tax which maximizes its revenue, 4000 units of electronic item have to be imported.
6.4 TOTAL PROFIT FUNCTION
1. Total profit function, x obtain from the production of a single unit product. In general:
= Total Revenue Function – Total Cost Function
= xCxR
Example 1: The demand function for vehicle spare-parts items at ATSAS Enterprise is given by qp 2400 and
the average total cost per unit producing the item is given by a function q
qqC2000
160 .
Determine the total profit function for ATSAS Enterprise. Solution:
STEP 1: Find xR
xR = pq
= qq2400
= 22400 qq
STEP 2: Find xC
xC = qqC
q
2000160
= 20001602 qq
STEP 3 : Find
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20003240
20001602400
20001602400
2
22
22
qqqq
qqqq
xCxR
6.4.1 Average Total Profit Function
1. Total average total profit is the profit obtain from the production of a single unit product.
x = x
x
Example:
Given the demand function for a product is xxp 025.08 and the total cost function is
xxC 7500 . Find
(a) total revenue function (b) total profit function (c) function for the average total cost, average total revenue and average total profit. Solutions:
(a) xR = px
= xx025.08
= 2025.08 xx
(b) = xCxR
= xxx 7500025.08 2
= xxx 7500025.08 2
= 500025.0 2 xx
(c) xC = x
xC
= x
x7500
= 7500
x
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xR = x
xR
=x
xx 2025.08
= x025.08
x = x
x
=x
xx 500025.0 2
=x
x500
025.01
6.4.2 Ultimate Total Profit Function '
1. It is the rate of change of total profit over the quantity of a product.
x' = xCxR ''
Example:
Suppose the total cost function 500305.0 2 xxxC , and the function for total revenue is
201.03 xxxR . Obtain
(a) total profit function (b) functions for the ultimate total cost, ultimate total revenue and ultimate total profit. Solutions:
(a) = xCxR
= 500305.001.03 22 xxxx
= 500305.001.03 22 xxxx
= 50006.06 2 xx
(b) xC' = 500305.0 2 xx
= 0305.02 1112 xx
= 31.0 x
xR' = 201.03 xx
= 1211 01.023 xx
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= x02.03
x' = xCxR ''
= 31.002.03 xx
= 31.002.03 xx
= x12.06
6.4.3 Maximizing Total Profit
1. As we all know, the purpose of having a business as well as of any economy is to obtain the
maximum profit. Example 1:
The demand function of a product for a corporation is given by qqp 442 and the average total
cost function is q
qC80
2 .
(a) Obtain the total cost function qC ,
(b) Derive the total revenue function qR ,
(c) Determine the total profit function q
(d) Calculate the price which maximizes the profit (e) Find the value of the maximum profit Solutions:
(a) qC = qqC
802
= 802 q
(b) qR = pq
= qq442
= 2442 qq
(c) q = qCqR
= 802442 2 qqq
= 802442 2 qqq
= 80440 2 qq
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(d) q' = 80440 2 qq
= 04240 1211 qq
= q840
5
408
0840
q
q
q
Find q'' to determine the graph maximum or minimum value
8
840''
When 8'' q , the graph has the maximum value.
Substitue the value 5q into the demand function qqp 442 to find the price which
maximizes the profit.
22
5442
4425
qp
The product has to be priced RM22 in order to maximize the profit.
(e) 5 = 80545402
= 80100200
= 20
The maximum profit is RM20 Example 2:
The demand equation for a travel agency company qqp 240 and its function for average cost is
given by q
qC100
4 .
(a) Determine the total revenue function qR ,
(b) Determine the total cost function qC ,
(c) Determine the total profit function q
(d) Calculate the price which will maximize the profit. Show that the profit is maximized. Solutions:
(a) qR = pq
= qq240
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= 2240 qq
(b) qC = qqC
1004
= 1004 q
(c) q = qCqR
= 1004240 2 qqq
= 1004240 2 qqq
= 100236 2 qq
(d) q' = 100236 2 qq
= 02236 1211 qq
= q436
9
364
0436
q
q
q
Find q'' to determine the graph maximum or minimum value
4
436''
When 4'' q , the graph has the maximum value.
Substitute the value 9q into the demand function qqp 240 to find the price which
maximizes the profit.
22
9240
2409
qp
The product has to be priced RM22 in order to maximize the profit.