A company manufactures 2 models of MP3 players. Let $x$ represent the number (in millions) of the first model made, and let $y$ represent the number (in millions) of the second model made.
The company's revenue can be modeled by the equation
\[
R(x, y)=160 x+90 y-2 x^{2}-4 y^{2}-x y
\]
Find the marginal revenue equations
\[
R_{x}(x, y)=
\]
\[
R_{y}(x, y)=
\]
We can achieve maximum revenue when both partial derivatives are equal to zero. Set $R_{x}=0$ and $R_{y}=0$ and solve as a system of equations to the find the production levels that will maximize revenue.
Revenue will be maximized when:
\[
x=
\]
\[
y=
\]

Step 1 ：Let's denote the number of the first model of MP3 players made as \(x\) (in millions), and the number of the second model made as \(y\) (in millions).

Step 2 ：The company's revenue can be modeled by the equation \(R(x, y)=160x+90y-2x^{2}-4y^{2}-xy\).

Step 3 ：We can find the marginal revenue equations by taking the partial derivatives of the revenue function with respect to \(x\) and \(y\).

Step 4 ：The marginal revenue with respect to \(x\) is \(R_{x}(x, y)=-4x - y + 160\).

Step 5 ：The marginal revenue with respect to \(y\) is \(R_{y}(x, y)=-x - 8y + 90\).

Step 6 ：To find the production levels that will maximize revenue, we set both partial derivatives equal to zero and solve the resulting system of equations.

Step 7 ：Solving the system of equations \(R_{x}=0\) and \(R_{y}=0\), we find that revenue will be maximized when \(x=\frac{1190}{31}\) and \(y=\frac{200}{31}\).

Step 8 ：\(\boxed{R_{x}(x, y)=-4x - y + 160, R_{y}(x, y)=-x - 8y + 90, x=\frac{1190}{31}, y=\frac{200}{31}}\)