Problem 50 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The monthly profit (in dollars) of Bond and Barker Department Store depends on the level of inventory \(x\) (in thousands of dollars) and the floor space \(y\) (in thousands of square feet) available for display of the merchandise, as given by the equation $$\begin{aligned}P(x, y)=&-0.02 x^{2}-15 y^{2}+x y \\ &+39 x+25 y-20,000\end{aligned}$$ Compute \(\partial P / \partial x\) and \(\partial P / \partial y\) when \(x=4000\) and \(y=150\). Interpret your results. Repeat with \(x=5000\) and \(y=150\).

Step-by-Step Solution

Answer

For (x=4000, y=150), the profit decreases by 120 dollars for every additional 1000 dollars of inventory and increases by 3650 dollars for every additional 1000 square feet of floor space. For (x=5000, y=150), the profit decreases by 220 dollars for every additional 1000 dollars of inventory and increases by 4650 dollars for every additional 1000 square feet of floor space.

Step 1: Compute the partial derivatives

To find the partial derivatives, we need to differentiate the profit function \(P(x, y)\) with respect to \(x\) and \(y\). Using the following notation, compute the partial derivatives: 1) \(\frac{\partial P}{\partial x}\): Differentiate \(P(x, y)\) with respect to \(x\). 2) \(\frac{\partial P}{\partial y}\): Differentiate \(P(x, y)\) with respect to \(y\).

Step 2: Calculate the partial derivatives

Differentiate the profit function with respect to \(x\) and \(y\): 1) (\(\frac{\partial P}{\partial x}\)): \(-2\times 0.01x - 15y + y + 39\) 2) (\(\frac{\partial P}{\partial y}\)): \(-2\times 15y + x + 25\)

Step 3: Evaluate the partial derivatives at the given points

Now, we will evaluate the partial derivatives at the points: (x=4000, y=150) and (x=5000, y=150). For (x=4000, y=150): 1) \(\frac{\partial P}{\partial x}(4000, 150) = -2 \times 0.01 \times 4000 - 15 \times 150 + 150 + 39 = -120\) 2) \(\frac{\partial P}{\partial y}(4000, 150) = -2 \times 15 \times 150 + 4000 + 25 = 3650\) For (x=5000, y=150): 1) \(\frac{\partial P}{\partial x}(5000, 150) = -2 \times 0.01 \times 5000 - 15 \times 150 + 150 + 39 = -220\) 2) \(\frac{\partial P}{\partial y}(5000, 150) = -2 \times 15 \times 150 + 5000 + 25 = 4650\)

Step 4: Interpret the results

We have evaluated the partial derivatives at the given points: For (x=4000, y=150): 1) \(\frac{\partial P}{\partial x}(4000, 150) = -120\): This means that at this point, the profit decreases by 120 dollars for every additional 1000 dollars of inventory. 2) \(\frac{\partial P}{\partial y}(4000, 150) = 3650\): This means that the profit increases by 3650 dollars for every additional 1000 square feet of floor space. For (x=5000, y=150): 1) \(\frac{\partial P}{\partial x}(5000, 150) = -220\): This means that at this point, the profit decreases by 220 dollars for every additional 1000 dollars of inventory. 2) \(\frac{\partial P}{\partial y}(5000, 150) = 4650\): This means that the profit increases by 4650 dollars for every additional 1000 square feet of floor space.

Key Concepts

Applied Mathematics

Applied mathematics is an essential field that utilizes mathematical theories and techniques to solve complex real-world problems. In the context of the original exercise, applied mathematics takes center stage as we apply derivatives to maximize a store's profit. The profit function is a mathematical representation of profit in relation to inventory levels and floor space.

When we computed the partial derivatives of the store's profit function, we did so by taking into account how changes in inventory (\( x \) and floor space (\( y \) affect profit (\( P(x, y) \) individually. This approach is quintessential in applied mathematics, which often involves breaking down a problem into simpler, solvable parts. Through applied mathematics, we're able to quantify the impact of running out of room for merchandise or over-investing in inventory, ensuring that businesses can make data-driven decisions to optimize their operations.

Profit Function Optimization

Profit function optimization within the realm of business management is a critical activity. It helps in finding the combination of variables that leads to the maximum or desired profit. This optimization often employs partial derivatives, as seen in the original exercise, to determine how small changes in each variable affect the overall profit.

Understanding the significance of \( \frac{\( \text{partial} \)}{{\text{partial} x}} \) and \( \frac{\( \text{partial} \)}{{\text{partial} y}} \) in our context involves recognizing that these values indicate the slope of the profit function in relation to inventory and floor space, respectively. If the partial derivative with respect to inventory is negative, increasing inventory will actually decrease profit beyond a certain point due to factors like storage costs or depreciating stock value. Conversely, a positive partial derivative with respect to floor space suggests that expanding this area could lead to greater profits, perhaps due to enhanced customer experiences or display capabilities.

Inventory and Floor Space Relationship

The relationship between inventory levels and floor space is pivotal in retail management, as too much inventory can lead to increased holding costs, while too little can result in stockouts and lost sales. In our exercise, the partial derivatives helped clarify this relationship. When \( x=4000 \) dollars and \( y=150 \) square feet, an increase in inventory decreases the profit, a sign that there might already be an excess. However, an increase in floor space significantly increases profit, indicating that additional space could be utilized more effectively.

Moreover, as the inventory level rises to \( x=5000 \) dollars, the negative impact on profit is even more pronounced, while the positive effect of increasing floor space continues to grow. This suggests that there's a delicate balance between how much stock to hold and how much floor space to allocate. To optimize profit, a company must consider not just individual effects but also how these factors interact, emphasizing the intricate balance between inventory management and floor space utilization.