Problem 42 - Calculus an Applied Approach

Question

Profit A corporation manufactures candles at two locations. The cost of producing \(x_{1}\) units at location 1 is \(C_{1}=0.02 x_{1}^{2}+4 x_{1}+500\) and the cost of producing \(x_{2}\) units at location 2 is \(C_{2}=0.05 x_{2}^{2}+4 x_{2}+275\) The candles sell for 15 dollars per unit. Find the quantity that should be produced at each location to maximize the profit \(P=15\left(x_{1}+x_{2}\right)-C_{1}-C_{2}\).

Step-by-Step Solution

Answer

To maximize profit, the corporation should not produce any units at location 1 (\(x_{1} = 0\)) and produce approximately one unit at location 2 (\(x_{2} = 1\)).

Step 1: Define the Profit Function

To solve this problem, start by finding the profit function. Given the cost functions \(C_{1}=0.02 x_{1}^2+4 x_{1}+500\) and \(C_{2}=0.05 x_{2}^2+4 x_{2}+275\), and the revenue function \(R=15(x_{1}+x_{2})\), the profit function \(P=R-C_{1}-C_{2}\) stands for Profit=Revenue-Cost. We just have to replace \(R, C_{1}, C_{2}\) with their respective formulas. Doing this, we get \(P = 15(x_{1}+x_{2})-(0.02 x_{1}^2+4 x_{1}+500) - (0.05 x_{2}^2+4 x_{2}+275)\).

Step 2: Simplify the Profit Function

The next step is to simplify the profit function. Collect like terms and simplify the equation. Profit function will turn into \(P = 10.98 x_{1}^2 + 11 x_{1} - 11.95 x_{2}^2 + 11 x_{2} - 775\).

Step 3: Find the Derivatives

In order to find the maximum profit point, the first derivative of the profit function with respect to \(x_{1}\) and \(x_{2}\) need to be taken and set to zero. These are given by \(\frac{dP}{dx_{1}} = 21.96 x_{1} + 11 = 0\) and \(\frac{dP}{dx_{2}} = -23.9 x_{2} + 11 = 0\). Solving these equations gives the quantity of units that should be produced at each location to maximize the profit.

Step 4: Solve for \(x_{1}\) and \(x_{2}\)

Solve the two equations to find the production quantity at each location for maximum profit. From \(\frac{dP}{dx_{1}} = 21.96 x_{1} + 11 = 0\), we get \(x_{1} = -11 / 21.96 \approx -0.501\). We can't produce a negative amount of units, so our actual answer is \(x_{1} = 0\). From \(\frac{dP}{dx_{2}} = -23.9 x_{2} + 11 = 0\), we get \(x_{2} = 11 / 23.9 \approx 0.460\). Again, since we can't produce a fraction of a unit, our actual answer is \(x_{2} = 1\).

Step 5: Verify The Solution

Check the second derivative of the profit function to ensure that the values of \(x_{1}\) and \(x_{2}\) actually give a maximum profit. An inequality test can also be used to verify that the profit function has its maximum value at the found quantities.

Key Concepts

Profit Maximization

Profit maximization is a key goal for any business, and in this exercise, it involves determining the best production quantities at two different locations.
This process is crucial because businesses aim to achieve the highest possible profit by fine-tuning their production strategies.
The profit function is the heart of this concept, calculated as the difference between total revenue and total cost. In mathematical terms, it is represented as:

• Revenue: The total income from sales, given by the price per unit times the number of units sold.
• Cost: The expense incurred to produce the units, which includes both fixed and variable costs.

In the original problem, the business sells candles priced at $15 each. The profit equation is constructed by subtracting the total production costs (from both locations) from the revenue, specifically formulated by substituting the cost functions into the equation: \[ P = 15(x_1 + x_2) - C_1 - C_2 \]Here, maximizing profit means adjusting the production levels, denoted by \(x_1\) and \(x_2\), to achieve the highest possible \(P\) value.

Derivative Application

In optimization problems like profit maximization, calculus and derivative applications play a vital role. Derivatives are used to find the rate at which functions change, which is crucial for identifying maximum or minimum points.
In our context, we use derivatives to determine the critical points of the profit function. These points represent potential maximum profits, as they specify where the slope of the function is zero (indicating a peak or a trough).The process involves:

• Taking the first derivative of the profit function with respect to each variable \(x_1\) and \(x_2\).
• Setting these derivatives equal to zero, \(\frac{dP}{dx_1} = 0\) and \(\frac{dP}{dx_2} = 0\), to find the critical points.
• Solving the system of equations to find the production quantities \(x_1\) and \(x_2\) for maximum profit.

By solving \(\frac{dP}{dx_1} = 21.96x_1 + 11 = 0\) and \(\frac{dP}{dx_2} = -23.9x_2 + 11 = 0\), we derive the values of \(x_1\) and \(x_2\), which indicate the production plan for optimal profitability. The derivatives reveal the sensitivity of the profit concerning changes in production at each site.

Cost Function Analysis

Understanding and analyzing the cost functions are fundamental in solving optimization problems in calculus. In this exercise, we deal with two distinct cost functions, each defining the cost of producing candles at different locations.
The cost functions, \(C_1\) and \(C_2\), are quadratic. They illustrate how the cost increases with production units, showing economies or diseconomies of scale.Key components of cost function analysis include:

• Fixed Costs: Constant terms in the cost functions represent fixed costs, which don't vary with production level (e.g., $500 at location 1).
• Variable Costs: Terms that vary with production level, such as the linear and quadratic terms in \(0.02x_1^2 + 4x_1\), highlight the increasing costs with higher output.

Analyzing these functions helps businesses understand their cost structure and potential impact on profit. Simplifying the profit function by combining elements of \(C_1\) and \(C_2\) allows us to better capture how costs affect overall profitability. The goal is to adjust production so that any additional units produced yield higher revenue than their associated costs, thereby maximizing profit.