Solving Economics Derivatives: Complementary Inputs & APL Minimization

[annalynne]

Hey everyone! This is an Economics question that's stumping me; but it requires a lot of calculus. Any help would be appreciated!

Homework Statement

Y = A[K^α][L^(1-α)/2][H^(1-α)/2] where 0 < α < 1.

A) At what level of L is APL (average productivity of labour) minimized?
B) Show K and L are complementary inputs in that more capital increases MPL and APL.

Homework Equations

MPL = marginal productivity of labor = ∂Y/∂L
APL = average productivity of labor = Y/L

The Attempt at a Solution

A) APL = Y/L = A[K^α][L^(α/2)][H^(1-α)/2]
minimize: therefore, take derivative and set to 0
∂APL/∂L = α/2[A][K^α][L^(α-1)/2][H^(1-α)/2] = 0
(I have no idea how to set solve this; would it also be possible to use the quotient rule to solve ∂APL/∂L from APL=A[K^α][L^(1-α)/2][H^(1-α)/2]/L ?)

B) prove that an increase in K causes and increase in MPL and APL
Y = A[K^α][L^(1-α)/2][H^(1-α)/2]
condition 1: assume A = 1, α = 0.6, K = 2, L = 3, H = 2
MPL = ∂Y/∂L = (1-α)/2 [A][K^α][L^(α/2)][H^(1-α)/2]
MPL = (0.25)(1)(1.41)(1.32)(1.19)
MPL = 0.55
APL = Y/L = A[K^α][L^(α/2)][H^(1-α)/2]
APL = (1)(1.41)(1.32)(1.19)
APL = 2.21

condition 2: assume A = 1, α = 0.6, K = 4, L = 3, H = 2
MPL = ∂Y/∂L = (1-α)/2 [A][K^α][L^(α/2)][H^(1-α)/2]
MPL = (0.25)(1)(2)(1.32)(1.19)
MPL = 0.79
APL = Y/L = A[K^α][L^(α/2)][H^(1-α)/2]
APL = (1)(2)(1.32)(1.19)
APL = 3.14

therefore, K and L are complements because APL and MPL increase as K increases

Any help would be appreciated! Thanks guys!

 

[CompuChip]

For the first one, I think you have made a small mistake. If you take L^(1-α)/2 and divide it by L, you get L^(-α)/2 instead of L^(α)/2. Then taking the derivative you will get something like
1/L^(-α-1)
which can only become zero if α < -1.

For the second one, you could take the derivative w.r.t K and show that it is positive everywhere (I don't know if you are allowed to plug in numbers, in mathematics that usually means checking for a specific case instead of proving it generally).