Nonlinear optimization in exponents $$ max \pi = 4x_1^\frac{1}4x_2^\frac{1}3 - x_1 - x_2 $$
It is not difficult to determine that this function is concave and yields a global maximum at some point for the quantities $ x_1, x_2 >= 0 $
My problem is in identifying the optimal quantities of $x_1$ and $x_2$ which yield this specific maximum.
So I need the stationary points... but I'm not getting consistent values of $x_1$ and $x_2$ with each first-order partial. I think this has to do with the exponents not being equal to 1.
Is there anything I can do to the exponents which will allow me to derive a solution? Perhaps Lagrangian multiple, or some kind of restriction? Or am I just missing something completely obvious?
If there is any other way, I'd love to hear about it. All help is greatly appreciated. Thank you!
 A: The first order equations are:$$
x^{-3/4}y^{1/3}=1\\
x^{1/4}y^{-2/3}=3/4\\
$$
Writing the $\log$:
$$
-3/4 \log x + 1/3 \log y = 0\\
1/4 \log x - 2/3 \log y = \log 3/4
$$you can then solve the system:
$$
\log x = 4/5 \log 4/3\ \ \ \ 
\log y = 9/5 \log 4/3\\
x = \left( \frac 43 \right)^{4/5}\ \  \ \ 
y = \left( \frac 43 \right)^{9/5}
$$
A: Alternatively, consider simply checking where $\nabla f = 0$ for candidates of the maximum. 
So $\frac{ \partial f}{\partial x_{1}}: x_{1}^{-\frac{3}{4}} x_{2}^{\frac{1}{3}} - 1 = 0$, which gives us $x_{2}^{\frac{1}{3}} = x_{1}^{\frac{3}{4}}$.
$\frac{ \partial f}{\partial x_{2}}: \frac{4}{3} x_{1}^{\frac{1}{4}} x_{2}^{-\frac{2}{3}} - 1 = 0$, which gives us $\frac{4}{3} x_{1}^{\frac{1}{4}} = x_{2}^{\frac{2}{3}}$. 
Now if we notice $x_{2}^{\frac{1}{3}} = x_{1}^{\frac{3}{4}}$ and square both sides, we get $x_{2}^{\frac{2}{3}} = x_{1}^{\frac{6}{4}}$. And so a very nice substitution:
$\frac{4}{3} x_{1}^{\frac{1}{4}} = x_{1}^{\frac{6}{4}}$. 
I'm sure you can take it from here.
