Download Homework 4_Solutions

410 Homework 4 Solutions
June 10, 2013
This homework is to be turned in on Monday, 6/10 at the beginning of class.
You may discuss this homework with classmates, but you must write up your
own solutions. A word of warningif you rely too heavily on others and don't
learn to solve the problems yourself, you will be poorly prepared for tests.
Problem I
For each of the following production functions, determine what returns to scale
they exhibit.
1/3
1. f (x1 , x2 ) = x1 x2 2/3
Solution:
tf (x1 , x2 ).
1/3
1/3
f (tx1 , tx2 ) = (tx1 )1/3 (tx2 )2/3 = t1/3 t2/3 x1 x2 2/3 = tx1 x2 2/3 =
So we have constant returns to scale.
2. f (x1 , x2 ) = x1 x2 2
Solution:
tf (x1 , x2 ).
f (tx1 , tx2 ) = (tx1 )(tx2 )2 = t ∗ t2 x1 x2 2 = t3 x1 x2 2 > tx1 x2 2 =
So we have increasing returns to scale.
3. f (x1 , x2 ) = 3x1 + x2
Solution:
f (tx1 , tx2 ) = 3tx1 + tx2 = t(3x1 + x2 ) = tf (x1 , x2 ).
So we have constant returns to scale.
4. f (x1 , x2 ) = min(3x1 , 2x2 )
Solution:
f (tx1 , tx2 ) = min(3tx1 , 2tx2 ) = t∗min(3tx1 , 2tx2 ) = tf (x1 , x2 ).
So we have constant returns to scale.
1/3
5. f (x1 , x2 ) = x1 x2 1/3
1
1/3
t
2/3
1/3
f (tx1 , tx2 ) = (tx1 )1/3 (tx2 )1/3 = t1/3 t1/3 x1 x2 1/3 = t2/3 x1 x2 1/3 =
f (x1 , x2 ) < tf (x1 , x2 ).
Solution:
So we have decreasing returns to scale.
6. f (x1 , x2 ) = (axp1 + bxp2 )1/p
f (tx1 , tx2 ) = (a(tx1 )p + b(tx2 )p )1/p = (atp (x1 p + bx2 p ))1/p =
+ bx2 p ))1/p = tf (x1 , x2 ).
Solution:
t(a(x1
p
So we have constant returns to scale.
7. f (x1 , x2 ) = x1 + x2 + 10
Solution:
f (tx1 , tx2 ) = tx1 + tx2 + 10 = tx1 + tx2 + t10 + 10 − t10 =
tf (x1 , x2 ) + 10 − t10 = tf (x1 , x2 ) + 10(1 − t) < tf (x1 , x2 ).
So we have decreasing returns to scale.
Problem II
For each of the following production functions, solve the cost minimization problem and report the minimal cost.
1. x1/3 y 2/3 , px = 2, py = 1, q=10
First, we solve for x and y, using fx /fy = px /py and the
isoquant equation 10 = x1/3 y 2/3 :
Solution:
(y 2/3 /3x2/3 )/(2x1/3 /3y 1/3 ) = y/(2x) = px /py
y = 2px /py ∗ x
We can substitute into the isoquant equation:
q = x1/3 (2px /py ∗ x)2/3
q = (2px /py )2/3 x
10 = 42/3 x
10 ∗ 4−2/3 = x
Now we can solve for y
2
y = 4 ∗ 10 ∗ 4−2/3 = 41/3 10
Cost is then
C = px x + py y = 2 ∗ 10 ∗ 4−2/3 + 41/3 10 = 5 ∗ 41/3 + 41/3 10 = 15 ∗ 41/3
2. xy 2 , px = 1, py = 16, q=8
Solution:
First, we solve for x and y, using fx /fy = px /py and the
isoquant equation 8 = xy 2 :
y 2 /(2xy) = y/(2x) = px /py
y = 2px /py ∗ x
We can substitute into the isoquant equation:
q = x(2px /py ∗ x)2
q = 4x3 (1/16)2
8 = x3 /64
83 = x3
8=x
Now we can solve for y
y = 2/16 ∗ x = 1
Cost is then
C = px x + py y = 1 ∗ 8 + 16 ∗ 1 = 24
3. 3x + y , px = 2, py = 5, q=9
This is a linear production function, so we know we have a
corner solution. There are two possibilities:
Solution:
1) only use x: 3x=9, so x=3 and cost is 3*2=6.
2)only use y: y=9 and cost is 9*5=45.
Clearly, only using x is cheaper so the cost is 6.
4. min(3x, 2y), px = 1, py = 10, q=6
3
Solution:
The kink of min(3x,2y) occurs where 3x=2y or y=3/2x.
We can substitute into the isoquant equation:
q = min(3x, 3x)
q = 3x
x = 6/3 = 2
Now we can solve for y
y = 3/2 ∗ 2 = 3
Cost is then
C = 1 ∗ 2 + 10 ∗ 3 = 32
Problem III
For each of the following problems, assume that there is an interior solution.
Solve for the demand for x for arbitrary px , py , and q , nd the cost function for
quantity q, and graph a Hicksian demand curve for x using px =1/4, 1/2, 1, 2,
and 4 using py = 1 and q=16.
1. xy
Solution:
First, we solve for x and y, using fx /fy = px /py and the
isoquant equation q = xy :
y/x = px /py
y = px /py ∗ x
We can substitute into the isoquant equation:
q = x2 px /py
qpy /px = x2
q
qpy /px = x
Now we can solve for y
y = px /py
q
qpy /px =
4
q
qpx /py
Cost is then
C = px ∗
q
q
√
qpy /px + py ∗ qpx /py = 2 qpx py
10
8
6
4
2
0
0
1
2
3
4
5
p is on the vertical axis, x is on the horizontal axis.
2. x1/2 + y 1/2
Solution:
First, we solve for x and y, using fx /fy = px /py and the
isoquant equation q = x1/2 + y 1/2 :
1/(2x1/2 )/(1/(2y 1/2 )) = y 1/2 /x1/2 = px /py
y = (px /py )2 ∗ x
We can substitute into the isoquant equation:
q = x1/2 +
q
(px /py )2 ∗ x = x1/2 + (px /py )x1/2 = (1 + (px /py ))x1/2
q/(1 + (px /py )) = x1/2
q 2 /(1 + (px /py ))2 = x
Now we can solve for y
y = (px /py )2 ∗ x = q 2 (px /py )2 /(1 + (px /py ))2
Cost is then
C = px ∗ q 2 /(1 + (px /py ))2 + py q 2 (px /py )2 /(1 + (px /py ))2
5
C = q 2 (px + p2x /py )/(1 + (px /py ))2
C = q 2 px (1 + px /py )/(1 + (px /py ))2
C = q 2 px /(1 + (px /py ))
C = q 2 /(1/px + 1/py )
10
8
6
4
2
0
0
1
2
3
4
5
p is on the vertical axis, x is on the horizontal axis.
3. 17 − 1/x − 1/y
Solution:
First, we solve for x and y, using fx /fy = px /py and the
isoquant equation q = 17 − 1/x − 1/y :
1/x2 /(1/y 2 )) = y 2 /x2 = px /py
y = (px /py )1/2 ∗ x
We can substitute into the isoquant equation:
q = 17−1/x−1/((px /py )1/2 ∗x) = 17−1/x−(px /py )−1/2 /x = 17−1/x(1+(px /py )−1/2 )
(q − 17)/(1 + (px /py )−1/2 ) = −1/x
(1 + (py /px )1/2 )/(17 − q) = x
Now we can solve for y
y = (px /py )1/2 ∗ x = (px /py )1/2 (1 + (py /px )1/2 )/(17 − q)
y = (1 + (px /py )1/2 )/(17 − q)
6
Cost is then
C = px (1 + (py /px )1/2 )/(17 − q) + py (1 + (px /py )1/2 )/(17 − q)
C = (px + (py px )1/2 )/(17 − q) + (py + (px py )1/2 )/(17 − q)
C = (px + py + 2(py px )1/2 ))/(17 − q)
10
8
6
4
2
0
0
1
2
3
4
5
p is on the vertical axis, x is on the horizontal axis.
4. 2x1/2 + y
Solution:
First, we solve for x and y, using fx /fy = px /py and the
isoquant equation q = xy :
1/x1/2 = px /py
x = (py /px )2
We can substitute into the isoquant equation:
q = 2(py /px ) + y
Now we can solve for y
y = q − 2(py /px )
Cost is then
7
C = px (py /px )2 + py ∗ (q − 2(py /px ))
C = p2y /px + py q − 2(p2y /px )
C = py q − p2y /px
C = py (q − py /px )
10
8
6
4
2
0
0
1
2
3
4
5
p is on the vertical axis, x is on the horizontal axis.
Problem IV
Using the following cost functions and demand functions, solve for the optimal
output good price and quantity
1. C(q) = q , D(p) = 9 − p
Solution:
∂Π
= D0 (p)p + D(p) − C 0 (D(p))D0 (p) = 0
∂p
∂Π
= −p + 9 − p − 1 ∗ −1 = 0
∂p
10 − 2p = 0
5=p
q = D(5) = 4
2. C(q) = q 2 , D(p) = 8 − p
8
Solution:
∂Π
= D0 (p)p + D(p) − C 0 (D(p))D0 (p) = 0
∂p
∂Π
= −p + 8 − p − 2(8 − p) ∗ −1 = 0
∂p
24 − 4p = 0
6=p
q = D(6) = 2
3. C(q) = 2ln(q), D(p) = 4 − p
Solution:
∂Π
= D0 (p)p + D(p) − C 0 (D(p))D0 (p) = 0
∂p
∂Π
= −p + 4 − p − 2/(4 − p) ∗ −1 = 0
∂p
4 − 2p + 2/(4 − p) = 0
2(p − 2) = 2/(4 − p)
−2(p − 2)(p − 4) = 2
−2p2 + 12p − 16 = 2
p2 − 6p + 9 = 0
p=
6±
√
√
36 − 4 ∗ 9
6± 0
=
=3
2
2
q = D(3) = 1
4. C(q) = 0, D(p) = 2 − ln(p)
Solution:
∂Π
= D0 (p)p + D(p) − C 0 (D(p))D0 (p) = 0
∂p
∂Π
= D0 (p)p + D(p) = 0
∂p
∂Π
= (−1/p)p + 2 − ln(p) = 0
∂p
−1 + 2 = ln(p)
1 = ln(p)
e=p
9