Question

Minimize Q=1159+2L^0.5+5K^0.5 using Lagrangian method and constraint 1200=20L+30K

Answer 1

$$Solution$$

maximizing

$$Q=f(L,K)=1159+2L^{0.5}+5K^{0.5}$$

Subject to the constraint

$$20L+30K=1200$$

Step 1: Create a new Lagrangian equation from the original information;

The Lagrangian is

$$\theta=f(L,K)=1159+2L^{0.5}+5K^{0.5}+\lambda (1200-20L-30K)$$

Step 2: Then follow the same steps as used in regular minimization problem.

The first order conditions are;

$$\frac{\delta \theta}{\delta L}=\theta_L=l^{-0.5}-20\lambda=0\\ \frac{\delta \theta}{\delta K}=\theta_K=2.5K^{-0.5}-30\lambda=0\\ \frac{\delta \theta}{\delta \lambda}=1200-20L-30K=0\\$$

From the first two equations we get

$$3L^{-0.5}=5K^{-0.5}\\ 5L^{0.5}=3K^{0.5}\\\\ L=\frac{\sqrt{15}}{5}K$$

Substitute this result into the third equation

$$1200-20(\frac{\sqrt{15}}{5})-30K=0\\ K=3.484$$

Therefore

$$L=2.699,\ \lambda=0.03$$

Therefore, the combination $2.699\ units$ of labor and $3.383\ units$ of capital minimize the total cost of producing $1,200\ units$ . In addition, $λ$ equals $0.03$ .If the firm wants to produce one more unit of the good, the total cost increases by $0.03.