função lagrangeana

por **jmario** » Sex Mai 21, 2010 09:23

Alguém pode me dizer se a resolução dessa função utilidade está correta

$U(x,y)={x}^{\alpha}{y}^{1-\alpha}$
derivando
$\alpha{x}^{\alpha-1}{y}^{1-\alpha}$
$(1-\alpha){x}^{\alpha}{y}^{-\alpha}$

$L={x}^{\alpha}{y}^{1-\alpha}-\lambda(xp+yq=m)$

$\alpha{x}^{\alpha-1}{y}^{1-\alpha}=\lambda\rightarrow\frac{\lambda=\alpha{x}^{\alpha-1}{y}^{1-\alpha}}{p}$
$\(1-alpha){x}^{\alpha}{y}^{-\alpha}=\lambda\rightarrow\frac{\lambda=\((1-\alpha){x}^{\alpha}{y}^{-\alpha}}{q}$

restrição orçamentária
xp+yq=m

$\frac{\alpha{x}^{\alpha-1}{y}^{1-\alpha}}{p} = \frac{(1-\alpha){x}^{\alpha}{y}^{-\alpha}}{q}$
$\frac{\alpha{x}^{\alpha-1}{y}^{1-\alpha}}{{x}^{\alpha}{y}^{-\alpha}} = \frac{(1-\alpha)p}q}\rightarrow\alpha{x}^{-1}yq=(1-\alpha)p$
$\alpha\frac{1}{x}yq=(1-\alpha)p\rightarrow\frac{\alpha}{x}yq=(1-\alpha)$
$yq=\frac{(1-\alpha)}{\alpha}px$
$xp+\frac{1-\alpha}{\alpha}px=m$
$px\left(1+\frac{1-\alpha}{\alpha} \right)=m$
$px\left(\frac{\alpha+1-\alpha}{\alpha} \right)=m$
$px\frac{1}{\alpha}=m\rightarrow$
$x=\alpha\frac{m}{p}$

como $qy=\frac{1-\alpha}{\alpha}px$
$qy=\frac{1-\alpha}{\alpha}p\alpha\frac{m}{p}=y(1-\alpha)\frac{m}{q}$
$\lambda=\frac{\alpha{x}^{\alpha-1}{y}^{1-\alpha}}p}$
$\lambda=\frac{\frac{{\alpha\alpha}^{\alpha-1}{m}^{\alpha-1}}{{p}^{\alpha-1}}.\frac{{(1-\alpha)}^{1-\alpha}{m}^{1-\alpha}}{{q}^{1-\alpha}}}{p}$
$\lambda=\left(\frac{\alpha}{p} \right)^{\alpha}\left(\frac{1-\alpha}{q} \right)^{1-\alpha}$

Será que isso?

função lagrangeana

função lagrangeana

função lagrangeana

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