Problem:
$f_1(x,y)=100,000-800x+10y \\ f_2(x, y)=150,000+10x-800y \\ \text{Find some combination $x$ and $y$ that provides the maximum of the system}$
\begin{align*} \text{I used } f_1 x+f_2 y&=R, \text{where $R$ is the maximum.} \\ R &= x(100,000-800x+10y)+y(150,000+10x-800y) \\ R &= -800x^2-800y^2+20xy+150,000y \end{align*} Then I took the partial derivative with respect to $x$, then again with respect to $y$
\begin{align*} \cfrac{\partial}{\partial y}=-1600y+150,000, \hspace{0.5cm} \cfrac{\partial}{\partial x}=-1600x \end{align*}
At this point I'm not sure what else to do. I tried to solve for $x$ and $y$ from the partial derivatives and plug them back into the original system, but doing this has not worked. Do I find the second derivative and go on from there? Or is my approach totally wrong?
If you replace $p_1$ by $x$ and $p_2$ by $y$ then your your function of the total revenue is right.
$$R = x(100000-800x+10y)+y(150000+10x-800y)$$
You can multiply out the brackets to make it easier calculating the partial derivatives.
$$R = 100000x-800x^2+10xy+150000y+10xy-800y^2$$ $$=-800x^2-800y^2+20xy+100000x+150000y$$
The partial derivatives are
$$\frac{\partial R}{\partial x}=100000-1600x+20y=0$$
$$\frac{\partial R}{\partial y}=150000-1600y+20x=0$$
Now solve this little equation system to obtain the maximum of the total revenue.