Question

A pharmaceutical corporation has two plants that produce the same over-the- counter medicine. If \(x_{1}\) and \(x_{2}\) are the numbers of units produced at. plant 1 and plant 2 , respectively, then the total revenue for the product is given by \(R=200 x_{1}+200 x_{2}-4 x_{1}^{2}-8 x_{1} x_{2}-4 x_{2}^{2}\) When \(x_{1}=4\) and \(x_{2}=12\), find (a) the marginal revenue for plant \(1, \partial R / \partial x_{1}\). (b) the marginal revenue for plant \(2, \partial R / \partial x_{2}\).

Short answer

The marginal revenue for plant 1 is 40 and for plant 2 is 88.

Step-by-step solution

Calculate the partial derivative of R wrt \(x_{1}\)

To begin with, we take the partial derivative of the given function \(R\) with respect to \(x_{1}\). The process to do this is to treat every other variable as a constant, then take the derivative normally. The derivative of \(200x_{1}\) is \(200\), the derivative of \(-4x_{1}^2\) is \(-8x_{1}\), the derivative of \(-8x_{1}x_{2}\) is \(-8x_{2}\), and since \(x_{2}\) is being treated as a constant, the derivative of \(-4x_{2}^{2}\) is 0. Working through the mathematics gives \(\partial R / \partial x_{1} = 200 - 8x_{1} - 8x_{2}\)

Calculate the partial derivative of R wrt \(x_{2}\)

Next, we take the partial derivative of the function \(R\) with respect to \(x_{2}\). We treat every other variable as a constant. The derivative of \(200x_{2}\) is \(200\), the derivative of \(-4x_{1}x_{2}\) is \(-4x_{1}\), the derivative of \(-4x_{2}^{2}\) is \(-8x_{2}\), and \(x_{1}\) is being treated as a constant. This results in \(\partial R / \partial x_{2} = 200 - 4x_{1} - 8x_{2}\)

Substitute \(x_{1}\) and \(x_{2}\)

Now that we have the partial derivatives, we substitute \(x_{1}\) = 4 and \(x_{2}\) = 12 into the equations to find the values. This gives \(\partial R / \partial x_{1}\) = 200 - 8(4) - 8(12) = 40 and \(\partial R / \partial x_{2}\) = 200 - 4(4) - 8(12) = 88

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially useful in multivariable functions. They allow us to understand how a function changes as one variable changes, while keeping other variables constant. In the case of the total revenue function provided, where revenue depends on two variables, partial derivatives help determine how changes in production units at one plant affect overall revenue.

To find the partial derivative with respect to a variable means to take the derivative of the function as you would in single-variable calculus but treat all other variables as constants. In our problem, the partial derivative with respect to \(x_1\) tells us how revenue changes when we increase production at plant 1, while keeping the production at plant 2 constant. Similarly, the partial derivative with respect to \(x_2\) shows the revenue change with increased production at plant 2.

This method of analyzing each variable's impact is crucial for businesses that need to optimize production processes and maximize revenue through understanding their cost structures.

Total Revenue Function

The total revenue function plays a central role in determining how revenue is generated from different operational inputs across the production process. For the given function \(R = 200x_1 + 200x_2 - 4x_1^2 - 8x_1x_2 - 4x_2^2\), each term corresponds to different effects of production units from two plants.

The first two terms, \(200x_1\) and \(200x_2\), represent the basic revenue gained by selling \(x_1\) and \(x_2\) units from each plant. The coefficients (200) can depict the revenue per unit. The negative terms, \(-4x_1^2\), \(-8x_1x_2\), and \(-4x_2^2\), reflect decreasing returns or costs related to production.

Understanding the total revenue function helps companies evaluate how increasing production in one plant over another influences overall profitability, especially when resources might be limited or when efficiency must be optimized.

Economic Optimization

Economic optimization involves finding the best way to allocate resources to maximize the desired outcome, like profit or revenue. In this context, using calculus tools like partial derivatives enables businesses to make these decisions quantitatively.

Through calculating marginal revenue for each plant, the company can determine where increasing production would yield the most benefit. This approach focuses not only on maximizing total revenue but also on understanding where additional investment or production shifts deliver the highest returns. By substituting specific values for \(x_1\) and \(x_2\) into the derivatives, a firm can decide the optimal production levels, balancing cost and revenue effectively.

This sort of mathematical analysis is essential for strategic planning, ensuring resources are utilized where they offer the most economic value, and adjusting operations in response to changes in demand or production capacity.

Mathematical Modeling

Mathematical modeling allows businesses to represent real-world scenarios through mathematical expressions, such as functions and equations, aiding in decision-making and strategy development. The total revenue function given in the exercise is a simplified mathematical model representing how different production levels from two plants impact revenue.

By building such models, companies can simulate various operational scenarios and predict outcomes without resorting to costly real-world experiments. Models also enable businesses to foresee reactions to changes like increased competition, varying product demand, or changes in material costs.

A key part of successful mathematical modeling is ensuring the model accurately reflects the real system, capturing essential dynamics while remaining simple enough to offer clear insights. It fosters informed decision-making by providing quantitative metrics to illustrate potential business scenarios, allowing strategic planning with reliable data.