Given utility maximization problem U=Q1Q2 subject to 1OQ1 +2Q2=240
a. Derive the lagrange function.
b. Derive the first order condition.
c. Using cramer's rule to find the critical value of q1q2 and lamda
a) Lagrange function:
b) first-order conditions:
ZQ1=Q2−λ10=0ZQ2=Q1−λ2=0Zλ=240−10Q1−2Q2=0.ZQ1 = Q2− λ10 = 0\\ ZQ2 = Q1− λ 2 = 0 \\ Zλ = 240 − 10Q1 −2 Q2 =0.ZQ1=Q2−λ10=0ZQ2=Q1−λ2=0Zλ=240−10Q1−2Q2=0.
Zλ=240−10Q1−2Q2=0Z\lambda=240-10Q1-2Q2=0Zλ=240−10Q1−2Q2=0
ZQ1=Q2−λ10=0ZQ1=Q2- \lambda10=0ZQ1=Q2−λ10=0
ZQ2=Q1−λ2=0ZQ2=Q1-\lambda2=0ZQ2=Q1−λ2=0
c) [0−10−2−1001−210]\begin{bmatrix} 0 & -10 & -2 \\ -10 & 0 &1 \\ -2 & 1 & 0 \end{bmatrix}⎣⎡0−10−2−1001−210⎦⎤ [λQ1Q2]\begin{bmatrix} \lambda \\ Q1 \\ Q2 \end{bmatrix}⎣⎡λQ1Q2⎦⎤ = [−24000]\begin{bmatrix} -240 \\ 0 \\ 0 \end{bmatrix}⎣⎡−24000⎦⎤
Q1M=2402[−10]=−12=\frac{240}{2[-10]}= -12=2[−10]240=−12
Q2M=2402[−2]=−60=\frac{240}{2[-2]}=-60=2[−2]240=−60
λ=2402[−10.−2]=6\lambda=\frac{240}{2[-10.-2]}=6λ=2[−10.−2]240=6
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