Question 19 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 \( R(x, y)=180 x+160 y-4 x^{2}-3 y^{2}-x y \) Find the marginal revenue equations \( R_{x}(x, y)= \) \( R_{y}(x, y)=\square \) We can acheive 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 (Please show your answers to at least 4 decimal places): \( x= \) \( y= \)
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To find the marginal revenue equations, we first take the partial derivatives of the revenue function \( R(x, y) = 180x + 160y - 4x^2 - 3y^2 - xy \). 1. **Marginal Revenue with respect to \( x \)**: \[ R_{x}(x, y) = \frac{\partial R}{\partial x} = 180 - 8x - y \] 2. **Marginal Revenue with respect to \( y \)**: \[ R_{y}(x, y) = \frac{\partial R}{\partial y} = 160 - 6y - x \] Next, we want to maximize revenue by setting both partial derivatives equal to zero: \[ R_{x}(x, y) = 0 \implies 180 - 8x - y = 0 \quad \text{(1)} \] \[ R_{y}(x, y) = 0 \implies 160 - 6y - x = 0 \quad \text{(2)} \] Now we can solve this system of equations. From equation (1): \[ y = 180 - 8x \quad \text{(3)} \] Substituting equation (3) into equation (2): \[ 160 - 6(180 - 8x) - x = 0 \] \[ 160 - 1080 + 48x - x = 0 \] \[ 47x = 920 \implies x = \frac{920}{47} \approx 19.5745 \quad \text{(4)} \] Now substituting value of \( x \) back into equation (3): \[ y = 180 - 8\left(\frac{920}{47}\right) \approx 14.4468 \quad \text{(5)} \] Thus, the production levels that will maximize revenue are: \[ x \approx 19.5745 \quad \text{and} \quad y \approx 14.4468 \]
