Problem 11 - Calculus an Applied Approach

Question

A company that manufactures sport supplements calculates that its costs and revenue can be modeled by the equations \(C=125,000+0.75 x\) and \(R=250 x-\frac{1}{10} x^{2}\) where \(x\) is the number of units of sport supplements produced in 1 week. If production in one particular week is 1000 units and is increasing at a rate of 150 units per week, find: (a) the rate at which the cost is changing. (b) the rate at which the revenue is changing. (c) the rate at which the profit is changing.

Step-by-Step Solution

Answer

The cost, revenue, and profit are changing at rates of \$112.5/week, \$7500/week, and \$7387.5/week, respectively

Step 1: Identify the given equations

The cost and revenue equations given are: \(C = 125,000 + 0.75x\) and \(R = 250x - \frac{1}{10}x^2\) respectively.

Step 2: Differentiate cost function with respect to x

The derivative of the cost function \(C = 125,000 + 0.75x\) with respect to x can be calculated as: \(C'(x) = 0 + 0.75\). Thus, the rate at which the cost is changing is \$0.75 per unit.

Step 3: Differentiate revenue function with respect to x

The derivative of the revenue function \(R = 250x - \frac{1}{10}x^2\) with respect to x can be calculated as: \(R'(x) = 250 - \frac{2}{10}x\). Thus, the rate at which the revenue is changing when 1000 units are produced is \(R'(1000) = 250 - 200 = 50\) dollars per unit.

Step 4: Identify the rate at which sport supplements units are increasing

The problem states that the number of sport supplement units, x, is increasing at a rate of 150 units per week.

Step 5: Use Chain Rule to calculate rates of change

By the chain rule, the rate of change of the cost is \(dC/dt = C'(x) \times dx/dt = 0.75 \times 150 = 112.5\) dollars per week, the rate of change of the revenue is \(dR/dt = R'(x) \times dx/dt = 50 \times 150 = 7500\) dollars per week.

Step 6: Calculate rates of change for the profit

Now the rate of change of the profit can be found by the formula \(dP/dt = dR/dt - dC/dt = 7500 - 112.5 = 7387.5\) dollars per week.

Key Concepts

Differentiation

Differentiation is the mathematical process of finding a derivative, which represents how a function changes as its input changes. In the context of cost and revenue functions, differentiation helps us understand how costs and revenues change with the production of additional units.
For the cost function given by \(C = 125,000 + 0.75x\), differentiating with respect to \(x\) yields \(C'(x) = 0.75\). This tells us that the cost increases by $0.75 for every additional unit produced.
Similarly, for the revenue function \(R = 250x - \frac{1}{10}x^2\), the derivative \(R'(x) = 250 - \frac{2}{10}x\) measures changes in revenue per unit produced. Calculating this at 1000 units, we find \(R'(1000) = 250 - 200 = 50\) dollars per unit. These calculations provide insights into costs and revenues per production level.

Cost Analysis

Cost analysis involves understanding how costs behave as production levels change. The cost function \(C = 125,000 + 0.75x\) indicates fixed and variable costs.

• **Fixed Costs:** The constant \(125,000\) represents costs that don't change with production level, such as rent and salaries.
• **Variable Costs:** The term \(0.75x\) shows costs that increase with each unit produced, like raw materials.

The differentiation gives us \(C'(x) = 0.75\), pointing out that variable costs add $0.75 for each unit increase. When units grow by 150 per week, the cost change is \(dC/dt = 112.5\) dollars per week, using the chain rule: \(dC/dt = C'(x) \times dx/dt = 0.75 \times 150\). Understanding this helps businesses predict expenses with changing production.

Revenue Analysis

Revenue analysis looks at how income from sales changes with production. The revenue function \(R = 250x - \frac{1}{10}x^2\) suggests both linear and nonlinear relationships.

• **Linear Component:** The term \(250x\) gives revenue per each unit sold, up to certain levels.
• **Nonlinear Component:** The negative part \(-\frac{1}{10}x^2\) implies decreasing returns, reducing revenue growth as production increases.

Using differentiation, \(R'(x) = 250 - 0.2x\), we calculate the revenue's sensitivity to production. At 1000 units, \(R'(1000) = 50\) dollars per unit. With growing production rates, \(dR/dt = R'(x) \times dx/dt = 7500\) dollars per week, highlighting revenue changes with production growth. Businesses can assess optimal production levels managing profit and sustainability.

Business Calculus

Business Calculus combines calculus concepts with business applications, facilitating decision-making on cost, revenue, and profit. It entails using derivatives for understanding and foreseeing changes in these economic elements.
In our scenario:

• The rate at which cost changes \((dC/dt) = 112.5\) illustrates how expenses adjust with production changes.
• The revenue change \((dR/dt) = 7500\) helps anticipate income fluctuations with varying output.
• The profit rate \((dP/dt) = 7387.5\) shows net financial growth, derived from combining both cost and revenue analysis.

This holistic view enables companies to tune production strategies, maximizing profit potential while understanding economic dynamics. Business calculus provides the tools to strategically manage resources and predict financial outcomes effectively.