Answer:
If the Cobb Douglas production funtion is [tex]Q(\lambda{K},\lambda{L})=A(\lambda{K})^{1.4}\times(\lambda{L})^{1.6}[/tex]
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 [tex]{\displaystyle f(\lambda L,\lambda K)}=\lambda ^{m}f(L,K)\,}[/tex]. Intuitively, this means that, when you increase your productive factors (in this case, we are talking about a production function), by a factor "[tex]\lambda[/tex]", your output increases by [tex]\lambda^m[/tex]. 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, [tex]Q(\lambda{K},\lambda{L})=A(\lambda{K})^{1.4}\times(\lambda{L})^{1.6}[/tex]. Applying distributive power's property, we get [tex]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}[/tex].
- Because of power property, we can associate terms and get [tex]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}[/tex] (remember that [tex]\lambda^{1.4}\times\lambda^{1.6}=\lambda^{(1.4+1.6)}=\lambda^3[/tex].
- Finally, [tex]Q(\lambda{K},\lambda{L})=A(\lambda{K})^{1.4}\times(\lambda{L})^{1.6}=\lambda^3AK^{1.4}L^{1.6}[/tex]. In this case the function is homogeneous of degree 3 because when multiplying K and L by [tex]\lambda[/tex], the function as a whole is multiplied by [tex]\lambda^3[/tex].
Euler's Theorem: this theorem states that, if a function is homogeneous of degree "m", the following must hold: [tex]L\frac{\partial Q}{\partial L} +K\frac{\partial Q}{\partial K}= m\times{Q(K,L)}[/tex].
- 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.
- [tex]\frac{\partial{Q}}{\partial{K}}=1.4\times{A}K^{0.4}L^{1.6}[/tex]
- [tex]\frac{\partial{Q}}{\partial{L}}=1.6\times{A}K^{1.4}L^{0.6}[/tex]
- Applying Euler's Theorem then means [tex]K(1.4\times{A}K^{0.4}L^{1.6})+L(1.6\times{A}K^{1.4}L^{0.6})[/tex] should be equal to [tex]3Q(\lambda{K},\lambda{L})=3A(\lambda{K})^{1.4}\times(\lambda{L})^{1.6}[/tex] (remember that in this case, m=3, see previous exercise).
- Solving [tex]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)}[/tex]
- Then the Euler's Theorem is verified!
