I have to maximize the function $$u(x_1, x_2) = x_1^{1/2} + x_2^{1/2}$$ in the set $$\{(x_1,x_2) \in \mathbb R^2 \mid x_1,x_2\geq0, (x_1p_1+x_2p_2) \leq N\},$$ where $p_1, p_2>0$ and $N>0$.
The Lagrange system is :
$1/2x_1^{-1/2}+\lambda_1-\lambda_3p_1 =0$
$1/2x_2^{-1/2}+\lambda_2-\lambda_3p_1 =0$
$\lambda_1(-x_1)=0$
$\lambda_2(-x_2)=0$
$\lambda_3(x_1p_1+x_2p_2-N)=0$
$-x_1\leq 0$
$-x_2\leq 0$
$x_1p_1+x_2p_2 \leq N$
My professor claims that the only solution of the system is: $x_1=\frac{p_2}{p_1^2+p_1p_2}N$ and $x_2=\frac{p_1}{p_1+p_2}N$. But I can not get that solution fro the system. Can you help me please?
For $x_1,x_2$ both not zero, you immediately get $\lambda_1=\lambda_2=0$. You need $\lambda_3\ne0$ for the first equation to be satisfied. So the system is reduced to \begin{align} \tfrac12\,x_1^{-1/2}&=\lambda_3p_1\\ \tfrac12\,x_2^{-1/2}&=\lambda_3p_2\ \ \ \ \textit{ (you have a typo there)}\\ x_1p_1+x_2p_2&=N \end{align} From the first and second equations you get, eliminating $\lambda_3$, $$ \frac1{2x_1^{1/2}p_1}=\frac1{2x_2^{1/2}p_2}. $$ Now the system becomes \begin{align} x_1p_1^2&=x_2p_2^2\\ x_1p_1+x_2p_2&=N \end{align} Eliminating $x_2$ we get $$ N=x_1p_1+\frac{x_1p_1^2}{p_2^2}\,p_2=x_1p_1+\frac{x_1p_1^2}{p_2}, $$ so $$ x_1=\frac{N}{p_1+\frac{p_1^2}{p_2}}=\frac{Np_2}{p_1p_2+p_1^2}, $$ $$ x_2=\frac{N-x_1p_1}{p_2}=\frac{N}{p_2}-\frac{Np_1}{p_1p_2+p_1^2}=\frac{Np_1p_2+Np_1^2-Np_1p_2}{p_1p_2^2+p_1^2p_2}=\frac{Np_1}{p_2^2+p_1p_2}. $$