Part a) The Cob Douglas production function is given as:
[tex]Q(K,L)=AK^{1.4} L^ {1.6 } .[/tex]
To show that this function is homogeneous with degree 3, we introduce be a parameter, t.
[tex]Q(tK,tL)=A(tK)^{1.4} (tL)^ {1.6 } .[/tex]
Using properties of exponents, we on tinder:
[tex]Q(tK,tL)=At^{1.4}K^{1.4} t^ {1.6 }L^ {1.6 } .[/tex]
This implies that:
[tex]Q(tK,tL)=t^{1.4} \times t^ {1.6 }(AK^{1.4} L^ {1.6 } )[/tex]
[tex]Q(tK,tL)=t^{1.4 + 1.6}(AK^{1.4} L^ {1.6 } )[/tex]
Simplify the exponent of t to get;
[tex]Q(tK,tL)=t^{3}(AK^{1.4} L^ {1.6 } )[/tex]
Hence the function is homogeneous with degree, 3
Part b) To verify Euler's Theorem, we must show that:
[tex]K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L}=3AK^{1.4}L^{1.6}[/tex]
Verifying from the left:
[tex]K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L} =K(1.4AK^{0.4} L^{1.6}) + L(1.6AK^{1.4} L^{0.6})[/tex]
[tex]K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L} =1.4(AK^{1.4} L^{1.6}) + 1.6(AK^{1.4} L^{1.6})[/tex]
[tex]K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L} =(1.4 + 1.6)(AK^{1.4} L^{1.6})[/tex]
[tex]K\frac{\partial Q}{\partial \: K}+L\frac{\partial Q}{\partial \: L} =3(AK^{1.4} L^{1.6})[/tex]
Q•E•D
