I think that I can safely assume that $f$ is a function of $2$ variables, rather than $3$.
Let $g(x,y)=5x^2+6xy+5y^2$. Note that
$$
E=\{(x,y)\in\mathbb{R}^2\,:\,g(x,y)=1\}
$$
is a compact set (it is an ellipse), and therefore the restriction of $f$ to that set must have a maximum and a minimum. The points at which they are attained are such that $\nabla g=(0,0)$ or that, for some $\lambda\in\mathbb{R}$, $(x,y,\lambda)$ is a solution of the system
$$
\left\{\begin{array}{l}
2\bigl((1-5\lambda)x-3\lambda y\bigr)=0\\
2\bigl(-3\lambda x+(1-5\lambda)y\bigr)=0\\
5x^2+6xy+5y^2=1.
\end{array}\right.\label{sys}\tag{1}
$$
But $\nabla g(x,y)=(10x+6y,6x+10y)$, which is $(0,0)$ only at the origin, which does not belong to $E$. On the other hand, the first two equations of the system \eqref{sys} form a system of two linear equations in two variables depending upon a parameter $\lambda$. The matrix of the coefficients of this system is
$$
\begin{pmatrix}
1-5\lambda&-3\\
-3&1-5\lambda
\end{pmatrix},
$$
whose determinant is $0$ if and only if $\lambda=\frac12$ or $\lambda=\frac18$. If the determinant is not $0$, then the only solution of the system which consists of the first two equation is $x=y=0$, which is not a solution of the third equation.
Now, suppose that $\lambda=\frac12$. Then the system becomes
$$
\left\{\begin{array}{l}
-\frac32x-\frac32y=0(\iff x+y=0)\\
5x^2+6xy+5y^2=1,
\end{array}\right.
$$
whose only solutions are $\pm\left(\frac12,-\frac12\right)$. And if $\lambda=\frac18$, then the system becomes
$$
\left\{\begin{array}{l}
\frac38x-\frac38y=0(\iff x-y=0)\\
5x^2+6xy+5y^2=1,
\end{array}\right.
$$
whose only solutions are $\pm\left(\frac14,\frac14\right)$. But $f\left(\pm\left(\frac12,-\frac12\right)\right)=\frac12$ and $f\left(\pm\left(\frac14,\frac14\right)\right)=\frac18$. So, the maximum is $\frac12$ and the minimum is $\frac18$.
Best Answer
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.