Ch 08.C Consumer Prod Surplus Final

8/11/2019 Ch 08.C Consumer Prod Surplus Final

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CHAPTER 8 : INTEGRATION & AREAS

APPLICATIONCONS & PROD SURPLUS

John Wiley and Sons 2013www.wiley.com/college/Bradley John Wiley and Sons 2013

Essential Mathematics for Economics and Business, 4thEdition
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www.wiley.com/college/Bradley John Wiley and Sons 2013

Slides on the following topics

Definite Integration and Area: Figures 8.6, 8.7, 8.8

Examples on definite integration

Consumer Surplus

Worked Example 8.12(b) Figure 8.12

Producer Surplus.

Figures 8.13, 8.14 and 8.15.

Worked Example 8.13 (b) Producer surplus


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Integration and areaFigure 8.6 Area under the curve approximately equal to the sum of areas

of rectangles.

Area y x y x y x y x n1 2 3 . . . . . .

x x x x

y 1 y 2 y 3 yn-1

x=ax=b

y = f (x )

y

x

xybi

ai

i

Area


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Integration and area

Figure 8.7 Decreasing the size of xgives a betterapproximation to area

x = a x = b

widths, x 0

y = f(x)

y

x


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Figure 8.8 Area under the curve is determined exactly by

integration.

Area =

y = f(x)

x=a x=b

y

x

Shaded area =(8.9)

bx

ax

xxd)(f

)(F)(F)(Fd)(f abxxx bx

ax

bx

ax


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x xx

x

21

3

d

x x x

x c 2 2 22

d

F( )x x

x c 2

22

F( ) ( ) ( ) .3 32

2 3 10 5

2

c c

F() ( )

( ) .1 1

22 1 2 5

2

c c

3

1

23

1

22

d2

x

x

x

x

cxxxx )1(F)3(F

)5.2()5.10( cc

cc 5.25.10 8

Worked Example 8.8 (a)

Hence

Evaluate the

integrand at the upper

and lower limits forx

The cs cancel.

No need for c

in definite

integration!
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3

1

23

1

22

d2

x

x

x

x

cxxxx )1(F)3(F

)5.2()5.10( 8

Worked Example 8.8 (a). Calculate the area geometrically

The plot of f(x) =x+ 2

2

5

31

3

1

d)2(

x

x

xx

..is the area under the line, f(x)

betweenx= 1 andx= 3

The area = area of triangle + area of rectangle

=0.5(2)(5-3)+2(3) = 8

3

Area byintegration

Area by

geometry



4

0

2d

x

x

xx

4

0

34

0

2

3d

x

x

x

x

cxxx

cx

x 3

)(F3

cc 364

3)4()4(F

2

cc3

)0()0(F

3

4

0

34

0

2

3d

x

x

x

x

cxxx )0(F)4(F

)(3

64cc

33.21

Worked Example 8.8 (b)

Hence

Evaluate the

integrand at the upper

and lower limits forx

The cs cancel.

No need for c indefinite

integration!



Consumer Surplus at P = P0

CS is represented graphically as the area of triangleAP0B

= the area under demand function (Q = 0 toQ = Q0)- area of rectangleP0 Q0 See sketch

00

0

0

dfunction)(demand QPQCS

QQ

Q

P

P0

Q

Q0

Consumer

surplus

Demand function

0

RectangleP0 Q0

Area under function Consumer surplus

CScalculated by integration:

A

B



Worked Example 8.12(a)

Consumer surplus for P= 60 - 2Q

Geometrically, CS= area of shaded triangle = 0.5(24)(48) = 576



Worked Example 8.12(a)



00

0

function)ddemand(0

QPQCS

QQ

Q

)24)(12()d260(

24

0

QQ

Q

Q

)24)(12(60 240

2

Q

QQQ

)24)(12()0()0(60)24()24(60 22

576)288(0864


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For practise -

calculate the

CSby

integration

Area under demand function = 750

Area of rectangle = 500

CS= 750500

= 250


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Figure 8.12 Consumer surplus for Worked Example

8.12(b)

Area of rectangle =P0Q0 = (20)(3) = 60

The area under the demand functionmust be calculated by integrating the

demand function, from Q = 0 to 3.2

100

QP

00

0

function)ddemand(0

QPQCS

QQ

Q

00

3

0

d2Q

100QPQCS

Q

Q
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Figure 8.12 Consumer surplus for Worked Example

8.12(b)


CS = 91.6291 - 60 = 31.6291

Q

QQ

Q

function)ddemand(functiondemandunderArea0

0

QQ

Q

Q

d2

1003

0

30

)2ln( Q

QQ

6291.91)2ln5(ln100

00

0

function)ddemand(0

QPQCS

QQ

Q


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Producer surplus (PS) P P 0

PS = Area of rectangle

- area under the supply function

whenQ = Q 0

P

QQ0

Producersurplus

Area under thesupply function

P 0

Supply function

0

A

B

= area of triangleP0AB

QQPPS

QQ

Q

function)dsupply(0

0

00


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Producer surplus

P

20

QQ 0 = 7

Producer

surplus

Revenue the producer

is willing to accept

P 0 = 55

P = 20 + 5Q

0

QQPPS

QQ

Qfunction)dsupply(

0

000

QQQPPS

Q

Q

)d520(

7

0

00

7

0

2

002

520

Q

Q

QQQP

2

)0(5)0(20

2

)7(5)7(20

22

00QP

05.262)7(55

5.1225.262385


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Producer surplus

P

A =20

QQ0= 7

Producer

surplus



P 0 = 55

P = 20 + 5 Q

0

QQPPS

QQ

Q

function)dsupply(0

0

00

ABPPS 0triangleofArea

B

heightbase5.0

)2055(75.0

5.122


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Figure 8.14 Producer surplus for Worked Example 8.13 (b)

Area under supply function= 69.3333

by integration


PS = 160 - 69.3333 = 90.6667

P

P0 = 40

0 QQ0 = 4

Producer

surplus



P = Q 2 + 6Q

The area under the supply function

must be calculated by integrating the supply function from Q= 0 to 4

Ch 08.C Consumer Prod Surplus Final

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