Question 5
F(x1x2) = (min{x1, 3×2) ^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=1.5×2^-1/2
The marginal rate of technical substitutions is given by:
MPX1MPX2=w1/w2
1/2 X1^-1/2/1.5×2^-1/2=1/1
X1=3×2
Substituting for x1 in the Cobb Douglas Production function
y = (3×2, 3×2) ^1/2
X1=3×2
X2=0.5/1.5(x1)
X2= 0.333×1
When P=8
P=3×2
x2 =2.66667
X1=1.666
X2=0.333×1
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
