Question

A firm wants to minimize cost when Q=160=2√KL ;r=4 k=2

a) write out the Lagrangian function

b) Find K* and L*

c).Is the second-order sufficient condition for minimum satisfied?

Answer 1

a) The Lagrangian can be written as

$l=4K+2L+\lambda[80-\sqrt{KL}]$

where λ is the Lagrange multiplier.

b)

The first order conditions can be written as

$\frac{\delta l}{\delta K}=4-\lambda \frac{ 1}{2}K^{\frac{-1}{2}}L^{\frac{1}{2}}=0\\or\\8=\lambda \sqrt{\frac{L}{K}}.....(1)\\ \frac{\delta l}{\delta K}=2-\lambda \frac{ 1}{2}K^{\frac{1}{2}}L^{\frac{-1}{2}}=0\\ or\\ 4=\lambda \sqrt{\frac{K}{L}}.......(2)\\ \frac{\delta l}{\delta K}=80-\sqrt{KL}.....(3)$

Now, (1)/(2) gives -

2 = L/K or L = 2K

 

Substituting this value in (3) gives -

$2 =\frac{ L}{K}\\ or\\ L = 2K$

Substituting this value in (3) gives -

$\sqrt{2K^2}=80\\or\\ K^*=\frac{80}{\sqrt{2}}=56.5\\ L^*=\frac{160}{\sqrt{2}}=113.14$

c.

Second order conditions are 

$\frac{\delta ^2 l}{\delta K^2}=-\frac{\lambda}{2}(-\frac{1}{2})K^{\frac{-3}{2}}L^{\frac{1}{2}}\\=\frac{\lambda \sqrt{L}}{4K^{\frac{-3}{2}}}>0$

$\frac{\delta ^2 l}{\delta L^2}=-\frac{\lambda}{2}(-\frac{1}{2})K^{\frac{1}{2}}L^{\frac{-3}{2}}\\ =\frac{\lambda \sqrt{K}}{4L^{\frac{3}{2}}}>0$

Since second order partial derivatives are positive, hence we can conclude that condition for cost minimisation is satisfied.