1
$\begingroup$

$$\mathcal{L}=\int_{i=0}^Ap(i)L(i)di-\lambda([\int_{i=0}^AL(i)^\phi di]^{\frac{1}{\phi}}-1)\\ s.t.\int_{i=0}^{A}L(i)^{\phi}di=1 $$ $$ \begin{aligned} \frac{d\mathcal{L}}{dL}&=\int_{i=0}^{A}p(i)di-\lambda\frac{1}{\phi}\bigg[\int_{i=0}^{A}L(i)^{\phi}di\bigg]^{\frac{1}{\phi}-1}\bigg[\phi\int_{i=0}^{A}L(i)^{\phi-1}di\bigg] \\ &=\int_{i=0}^{A}p(i)di-\lambda\int_{i=0}^{A}L(i)^{\phi-1}di=0 \end{aligned} $$ so we get $$p(i)=\lambda L(i)^{\phi-1} $$

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.