Question

The Cobb-Douglas production function is given by

Answer 1

Answer:

If the Cobb Douglas production funtion is $Q(\lambda{K},\lambda{L})=A(\lambda{K})^{1.4}\times(\lambda{L})^{1.6}$

This function is homogeneous of degree 3: To understand that, we first must know that a function f(K,L)  is homogeneous of degree "m" if ${\displaystyle f(\lambda L,\lambda K)}=\lambda ^{m}f(L,K)\,}$. Intuitively, this means that, when you increase your productive factors (in this case, we are talking about a production function), by a factor "$\lambda$", your output increases by $\lambda^m$. Depending on the value of m, the function will exhibit increasing returns to scale (m>1), decreasing returns to scale (m<1) or returns to scale equal to 1 (when m=1).

• In this case, $Q(\lambda{K},\lambda{L})=A(\lambda{K})^{1.4}\times(\lambda{L})^{1.6}$. Applying distributive power's property, we get $Q(\lambda{K},\lambda{L})=A(\lambda{K})^{1.4}\times(\lambda{L})^{1.6}=A(\lambda^{1.4})K^{1.4}\times(\lambda^{1.6})L^{1.6}$.
• Because of power property, we can associate terms and get $Q(\lambda{K},\lambda{L})=A(\lambda^{1.4})K^{1.4}\times(\lambda^{1.6})L^{1.6}=A(\lambda^3)K^{1.4}L^{1.6}$ (remember that $\lambda^{1.4}\times\lambda^{1.6}=\lambda^{(1.4+1.6)}=\lambda^3$.
• Finally, $Q(\lambda{K},\lambda{L})=A(\lambda{K})^{1.4}\times(\lambda{L})^{1.6}=\lambda^3AK^{1.4}L^{1.6}$. In this case the function is homogeneous of degree 3 because when multiplying K and L by $\lambda$, the function as a whole is multiplied by $\lambda^3$.

Euler's Theorem: this theorem states that, if a function is homogeneous of degree "m", the following must hold: $L\frac{\partial Q}{\partial L} +K\frac{\partial Q}{\partial K}= m\times{Q(K,L)}$.

• To prove it, we should then calculate the partial derivative of Q with respect to L and K respectively, and apply the previous definition to see if the statement holds.

• $\frac{\partial{Q}}{\partial{K}}=1.4\times{A}K^{0.4}L^{1.6}$

• $\frac{\partial{Q}}{\partial{L}}=1.6\times{A}K^{1.4}L^{0.6}$

• Applying Euler's Theorem then means $K(1.4\times{A}K^{0.4}L^{1.6})+L(1.6\times{A}K^{1.4}L^{0.6})$ should be equal to $3Q(\lambda{K},\lambda{L})=3A(\lambda{K})^{1.4}\times(\lambda{L})^{1.6}$ (remember that in this case, m=3, see previous exercise).

• Solving $K(1.4\times{A}K^{0.4}L^{1.6})+L(1.6\times{A}K^{1.4}L^{0.6})=1.4(AK^{1.4}L^{1.6})+1.6(AK^{1.4}L^{1.6})=3\times(AK^{1.4}L^{1.6})=3\times{Q(K,L)}$

• Then the Euler's Theorem is verified!