Micro-Theory

Published: 2017-11-28 15:17:30
529 words
2 pages
5 min to read
letter-mark
B
letter
University/College: 
Harvey Mudd College
Type of paper: 
Essay
This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.

Question 5

F(x1x2) = (min{x1, 3x2) ^1/2

W1≥ 1

W2≥ 1

The cost minimizing combination requires that:

MRTS=MPx1 /MPK=w1/w2

Let x1 be units of labor while x2 be units of capital

The MPx1 = Change in quantity/Change in labor

MPL=1/2 x1^-1/2

The MPx2 = Change in quantity/Change in capital

MPx2=1.5x2^-1/2

The marginal rate of technical substitutions is given by:

MPX1MPX2=w1/w2

1/2 X1^-1/2/1.5x2^-1/2=1/1

X1=3x2

Substituting for x1 in the Cobb Douglas Production function

y = (3x2, 3x2) ^1/2

X1=3x2

X2=0.5/1.5(x1)

X2= 0.333x1

When P=8

P=3x2

x2 =2.66667

 

X1=1.666

X2=0.333x1

X2= (0.333*1.666)

X2=0.5444

Q = (1.666, 3{0.5444}) ^1/2

Q= (2.7216)1/2

Q*=1.65 Units

 

Question 6

Question 6a

The long run formula will be :

Q(x1, x2)=min(x1,x2)^1/2 

In the short run, the labor which is greater than 16 has no effect on the level of out put.

The short run production function is given by:

Y=

Y= 4 Therefore, L>4 or L≤ 4

The formula for the marginal product of labor

MPL= (4L)^-1/2 :L>4 or L≤ 4

Question 6b

When w=1 and price =$ 4

The labor demand which is optimal=PMPL=w

It thus means 4*(1/2(L)^1/2=1

2/(L)^1/2=1

(L^1/2)^2=(2)^2

L=4

 

Question 6c

When w=1 and price =$ 10

The labor demand which is optimal=PMPL=w

It thus means 4*(1/2(L)^1/2=1

5/(L)^1/2=1

(L^1/2)^2= (5)^2

L=25 Units

Question 6d

The equilibrium demand for labor PMPL=w

And in this situation w=p/2(L) ^1/2

Therefore, the equation can be written as:

L=min {(p/2w) ^2, 4}

Question 7

Question 7a

Total cost function=c=y^2+10 for y>0 and c=0

MR=MC

MC=First derivative of cost function

MC=MR=2y

The average cost= TC/Q

= (y^2+10)/y

= y + (10/y)

Question 7b

The minimum average cost 

Av=(y^2+10)/y

First order derivatives

AV’=y+(10/y)

1+10/y

The minimum of MC

MC’=2y

But at Minimums MC=AV

1+10/y=2y

Y+10=2y2

2y2-y-10=0

y(2y-1)=10

y=10

0r

2y-1=10 Units 

Y= 4.5 units

The quantity at which the quantity is minimized in the long run

The first derivative equated to 0

MC=MR=2y=0

Y=0/2= ∞

Hence the minimum is absolute minimum quantity.

Question 7c

MC=2y=MR

The variable cost= y2

Average variable cost= y2/y

=y which is minimized at y=0

Marginal function is given as

First derivative

oVC/dy=2y

P=2y which is the supply curve

At the quantity of 4.5

P= (2*4.5)

P = 9 Units

 

At the quantity of 10

P=2*(10)

P=20 Units

Therefore, the minimum price in which they can produce a positive quantity is 9 Units

Question 8

Question 8a

Given that s1(p) = p, s2(p) = 2p, s3(p) = 3p.

Then the market supply function is given as

S(P)= (S1+S2+S3)P

=S(p)=( p+2p+3p)

S(p)=6p

Question 8b

Given that s1(p) = 2p, s2(p) = p – 1

Then the market supply function is given as:

S(P)= (S1+S2)P

S(P)= (p – 1)+2p

S(P)=3P-1

Question 8c

200 firms each having a supply function of s1 (p) = 2p – 8 and other 100 firms having a supply function of s2 (p) = p – 3

The first 200 firms supply function

S1(p)=200(2p-8)

S(p)=400p-1600 …..1

The 2nd 200 firms supply function

The supply function = s2(p) = p – 3

(s(p) =100( p – 3)

(s(p)=100p-300………2

Total supply for the market is the summation of equation 1 and 2

Market’s s(p)= {400p-1600} +{100p-300}

s(p)= 500p-1900

s(p)=5p-19

sheldon

Request Removal

If you are the original author of this essay and no longer wish to have it published on the SpeedyPaper website, please click below to request its removal: