Problem 20 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Phillip, the proprietor of a vineyard, estimates that the first 10,000 bottles of wine produced this season will fetch a profit of $$\$ 5 /$$ bottle. But if more than 10,000 bottles were produced, then the profit/bottle for the entire lot would drop by $$\$ 0.0002$$ for each additional bottle sold. Assuming at least 10,000 bottles of wine are produced and sold, what is the maximum profit?

Step-by-Step Solution

Answer

The maximum profit that Phillip can make is $62,500 by producing and selling 12,500 bottles of wine.

Step 1: Define the profit function

First, we need to define the profit function P(x), representing the total profit as a function of the number of bottles produced and sold (x). P(x) will have two parts: one that covers the profit from the first 10,000 bottles, and the other to account for the reduced profit per bottle after producing/selling more than 10,000 bottles. \(P(x) = \begin{cases} 10,000 \times 5 & \text{if } x \leq 10,000, \\ (5 - 0.0002(x - 10,000))x & \text{if } x > 10,000. \end{cases} \)

Step 2: Simplify the profit function for x > 10,000

We will only analyze the case when x > 10,000, as producing fewer than 10,000 bottles will result in lower profits. When x > 10,000, we can simplify the profit function: \(P(x) = (5 - 0.0002(x - 10,000))x\)

Step 3: Calculate the derivative of P(x)

In order to find the maximum or minimum of the function, we need to calculate the derivative of P(x) and set it equal to zero. \(P'(x)= \frac{d}{dx} (5x - 0.0002x(x - 10,000))\)

Step 4: Set P'(x) = 0 and solve for x

Now we set P'(x) equal to zero and solve for x: \(0 = 5 - 0.0004x + 0.0002 \times 10,000 \\ x = \frac{5 - 0.0002 \times 10,000}{0.0004}\)

Step 5: Calculate the maximum profit

Calculate the value of x obtained from the equation and substitute it back into the profit function P(x) to find the maximum profit: \(x = \frac{5 - 0.0002 \times 10,000}{0.0004} = 12,500\) \(P(12,500) = (5 - 0.0002(12,500 - 10,000)) \times 12,500 = \$62,500\) So, the maximum profit that Phillip can make is $62,500 by producing and selling 12,500 bottles of wine.

Key Concepts

Profit Function

A profit function is a mathematical representation showing how profit changes with varying levels of output or sales. In the context of Phillip's vineyard, it depicts the relationship between the number of bottles produced and the total profit generated.

For the first 10,000 bottles, the function is a constant as each bottle adds a flat profit of \(5. However, beyond this point, the profit per bottle decreases by \)0.0002 for every additional bottle. This decrease accounts for potential market saturation or increased production costs which typically occur in many businesses as production increases. By understanding this function, Phillip can calculate the optimal number of bottles to produce to maximize his profits.

Derivative Application

The application of derivatives is a powerful tool in calculus used for finding the rate at which one quantity changes with respect to another. In business applications, derivatives assist in profit optimization by identifying points where the rate of change of profit is zero — these points can represent maximum or minimum profit scenarios.

In Phillip's case, the derivative of the profit function with respect to the number of bottles, denoted as P'(x), helps determine the production level that maximizes profit.

Why Set the Derivative to Zero?

Setting P'(x) to zero is critical because it signifies a point where increasing or decreasing the number of bottles produced does not change the profit – a potential maximum. To optimize, the goal is to find this 'stationary point' and confirm it's a maximum through further analysis.

Maximum Profit Calculation

Calculating the maximum profit involves finding the highest value of the profit function within the feasible range of production. After deriving the profit function's derivative and finding its zeros, these points are evaluated to determine if they correspond to a maximum or minimum.

Phillip's maximum profit occurs at the production level where the derivative equals zero. Once identified, this quantity, symbolized by x, is substituted back into the profit function to calculate the profit at this level. As seen in the exercise, by producing and selling 12,500 bottles, Phillip can achieve a maximum profit of $62,500. This calculated point is crucial for business strategy, allowing for effective production planning and resource allocation to achieve optimal financial results.