Assignment 2 PDF

Preview

Summary

FACULTY OF BUSINESS AND ECONOMICS Question 1 U ( q )=√ q+2 √ 10− pq, q ∈[0, 10 ] p a) The first order condition for the utility maximizing quantity of ice cream is: −1 −1 1 ' 2 U ( q )= q 2 − p ( 10− pq ) 2 b) Set the first order condition derived in question to zero, −1 −1 1 ' 2 U ( q )= q 2 − p ( ...

Description

FACULTY OF BUSINESS AND ECONOMICS Question 1

U ( q ) =√ q+2 √10− pq , q ∈[ 0,

10 ] p

a) The first order condition for the utility maximizing quantity of ice cream is: −1

1 U ( q )= q 2 − p ( 10− pq ) 2 '

−1 2

b) Set the first order condition derived in question to zero, −1

−1

1 U ' ( q )= q 2 − p ( 10− pq ) 2 =0, 2 the quantity that maximize utility function expressed as a function of p is written as ¿

−1

2

q =10(4 p + p)

.

c) When we solve the first order condition in part c, q* is the only solution, which makes it the only stationary point. Judging by the function, we know that the maximization point exists, and since q ∈[ 0,

10 ] , meaning that the maximum point must lies in either the end points or the p

stationary point. We can determine where the maximum point lies by figuring out the whether the function is concave or convex. −3

''

U (q ) =

−3

−1 2 1 2 q − p (10− pq) 2 4 2

q ∈[ 0,

breaking the equation to component. Since

10 ] , p

−3

−1 2 q 4

must be

≤ 0 . And we

know the price of the ice cream must take positive value, from the question we know that the amount spend on other items (10-pq) is a nonnegative value, which means that −1 2 p ( 10−pq ) 2

−3 2

≤0. −3

To sum up, the second order derivative

U ' ' (q ) =

−3

−1 2 1 2 2 q − p (10− pq) 2 4

is always

≤0 .

Therefore, the utility function is concave. Combining the fact that the function is concave and the stationary point exist and is within the interval is positive, we can conclude

[ 0,

10 ] p

since 4 p 2+ p

always ≥ p

when p

¿ q =10(4 p2 + p)−1 is indeed the maximum point.

d) 1

FACULTY OF BUSINESS AND ECONOMICS 1

10−pq ¿ 2 ¿

1

10−2.5 q ¿2 1 2

q −2 ¿ 1

q 2 −2 ¿

[

−1

−1 2

1 2 q −p ( 10− pq) ' 2 U (q )∗ q El q U (q )= = ¿ U (q)

]

∗q

Question 2 E ( p ,r )= A p−a r b a)

p

∂ E( p , r ) ∂ E ( p ,r ) +r =kE ( p , r ) , which means ∂p ∂r

p∗E P ( p ,r ) + r∗E r ( p , r ) =kE ( p , r ) .

−a b A∗r b∗( −a ) p−a + A∗p−a ∗b∗r b = kA p r , which means that

– a +b=k

b) When a =1.5 and b=2.08, k=-1.5+2.08=0.58 t

c)

¿ 1.08 ¿ ¿ t −1.5 1.06 ¿ ¿ r 0 ¿ p0 ¿ E ( t ) =E( p (t ) , r (t ) ) =A ¿ ¿ ¿ 1.06 ¿ −1.5 2.08 ¿ A p 0 r 0 −1.5 ln 1.06 t+ 2.08 ln 1.08 t ) ¿ ¿ ( E ( t ) )=ln ¿ ln ¿ so, if we treat p0 and r0 as constant and find the derivative with respect to t, we have ln ( E ( t ) ) '=−1.5 ln ( 1.06 )+2.08 ln ( 1.08 )=0.0727

Question 3 P ( x , y) =−0.1 x 2−0.2 x y−0.2 y 2 + 47 x +48 y−600 a)

Px ( x , y )=−0.2 x−0.2 y +47=0 P y (x , y ) =−0.2 x−0.4 y+48=0 We can express the above in matrix form

2

FACULTY OF BUSINESS AND ECONOMICS 0.2 0.2 =0.08−0.04 ≠ 0 , thus Cramer’s rule is applicable. , 0.2 0.4

][ ] = [ ] | | 47 0.2 0.2 27 | | 48 0.4| 0.2 48| x= =230, y= =5 0.2 0.2 0.2 0.2 |0.2 0.4| |0.2 0.4| [

0.2 0.2 x 0.2 0.4 y

47 48

Thus the stationary point is product

230

units

of

P ( x , y) =P(230,5) . The production level that maximize profit is to x

and

5

units

of

y.

The

maximum

profit

is

−0.1 ¿ 2302−0.2∗230∗5−0.2 ¿ 52+ 47∗230 + 48∗5−600= 4925 b) Now that we want to maximize constraints that

P ( x , y ) =−0.1 x 2−0.2 xy−0.2 y 2+47 x +48 y−600

under the

x+ y=200

We can find the production level that maximize the profit using the Lagrange 1 method, F ( x , y , ) =−0.1 x 2 −0.2 xy−0.2 y 2 +47 x + 48 y −600 + ( 200−x− y ) , x , y ,∈ D , where D {( x , y , ) : x , y >0, ∈ ( −∞ ,+∞ ) } F x ( x , y, ) =−0.2 x−0.2 y+47−¿ 0

①

F y ( x , y ,) =−0.2 x−0.4 y +48−¿ 0 F❑ (x , y , ) =200−x− y

②

②- ①

③

we get y=5, then by plug it in ③, we get x=195. Thus the stationary point is

P ( x , y ) =P(195,5) . The production level that maximize profit is to product 195 units of x and 5 units of y. c)

at the maximum point is 7 and the maximum profit is 4802.5

d) Now that the constraint is

x + y =250 , applying the envelop theory, we know that if the

constraint change from its original level by 1 unit, the maximum value will change by approximately

( 250−200)=350

units, which means that the new maximum profit is

4802.5+350=5152.5 e) Because in part a, x+y=2000 .

Question 6 a) Question: For all a , b ∈ (0,100 ) , prove f ( a )≥ f ( b ) if a < b f (a) ≥ f (b)

f ( a )−f ( b ) ≥ 0 ,

means that

Knowing that

f ' ( x) ≤ 0 ,

which

means

b

f ( x ) is a decreasing function∧∫ f '( x ) dx ≤ 0 , which means that for any given

a , b ∈( 0,100 ) ,

a

and satisfy

a< b ,

f ( x ) ¿ba=f ( b) −f (a)≤ 0 , f ( a ) ≥ f ( b ) ¿

x

b)

F ( x ) =∫ f ( t ) dt ''

0

Question: For all a , b ∈ (0,100 ) , prove F

1 ≥ (F (a ) +F( b) ) if a < b ( a+b ) 2 2

5

FACULTY OF BUSINESS AND ECONOMICS a+b ≥ ( F ( a) +F ( b) ) , which is equivalent Multiply 2 to both side of the inequation, we have 2 F 2

( )

F

to

( a+2 b ) −F ( a ) ≥ F ( b ) −F( a+2 b )

,

expressing

this

with

integral

we

have

a+ b 2 a +b a+b x ¿a 2 f 2 ¿ a+b a+b −a) ∗( 2 2 a+b f( )dx=¿ 2 ), a+ b

( )

( ) a+b 2

b

∫ f ( x ) dx≥ ∫ a

f ( x ) dx , so we can prove the above inequality by prove

a+b 2

2

f (¿¿)dx, ∫ ¿ a a +b 2

f ( x )dx ≥ ∫ ¿ a

a +b 2

∫¿ a

a+b 2

)dx ∫ f ( a+b 2

,

a

6

FACULTY OF BUSINESS AND ECONOMICS a+ b 2 a+b x ¿ba +b f 2 2 ¿ a+b a+b ∗(b− ) 2 2 a+b )dx=¿ f( 2 ) b

( )

( ) and

f (¿¿)dx , ∫ ¿ a+ b 2 b

f ( x )dx ≤ ∫ ¿ a +b 2

b

∫¿ a +b 2

a+b 2

we know that

a+b 2

meaning

),

b

∫ f ( x ) dx≥ ∫ f ( x ) dx a

F

is the middle point of a and b, thus

) ( a+b2 )∗(b− a+b 2 a+ b a+b −a ) =f ¿ ∗( f( ) ) 2 2

,

which

is

equivalent

to

say

a+b 2

( a+2 b ) −F ( a ) ≥ F (b ) −F( a+b2 )

, F

( a+2 b ) ≥ 12 (F ( a )+F (b ) )

7...