Question

Given utility maximization problem U= Q1Q2 subject to 10Q1 +2Q2=240 a. Derive the Lagrange function b. Derive the first order conditions c. Use Cramer’s rule to find the critical values of Q1, Q2 and �

Answer 1

i.) Deriving the Lagrange function:

$Z = Q1Q2+ λ(240-10Q1-2Q2)$

ii.) first-order conditions:

$ZQ1 = Q2− λ10 = 0\\ ZQ2 = Q1− λ 2 = 0 \\ Zλ = 240 − 10Q1 −2 Q2 =0.$

$Z\lambda=240-10Q1-2Q2=0$

$ZQ1=Q2- \lambda10=0$

$ZQ2=Q1-\lambda2=0$

iii.) $\begin{bmatrix} 0 & -10 & -2 \\ -10 & 0 &1 \\ -2 & 1 & 0 \end{bmatrix}$ $\begin{bmatrix} \lambda \\ Q1 \\ Q2 \end{bmatrix}$ = $\begin{bmatrix} -240 \\ 0 \\ 0 \end{bmatrix}$

Q1^M$=\frac{240}{2[-10]}= -12$

Q2^M$=\frac{240}{2[-2]}=-60$

$\lambda=\frac{240}{2[-10.-2]}=6$