Question

A retail sporting goods store estimates that weekly sales \(S\) and weekly advertising costs \(x\) are related by the equation \(S=2250+50 x+0.35 x^{2}\). The current weekly advertising costs are \(\$ 1500\), and these costs are increasing at a rate of \(\$ 125\) per week. Find the current rate of change of weekly sales.

Short answer

At the current advertising cost rate, the rate of change of weekly sales is increasing at \$137,500 per week.

Step-by-step solution

Derive the Given Function

Differentiate the given function \(S=2250+50x+0.35x^2\) with respect to \(x\). Performing the derivative results in \(S' = 50 + 0.7x\). This returns the instantaneous rate of change of sales with respect to the advertising costs.

Evaluate the Derivative Function at the Current Advertising Cost

Substitute the current advertising cost (\$1500) into the derivative function. This equates to \(S'(1500) = 50 + 0.7(1500) = 1100\). This is the rate of change of sales with respect to advertising cost at the current level of advertising cost.

Derive the Sales Rate of Change

The rate of change of sales to time, not to advertising costs, is requested. So use the chain rule: \(\frac{dS/dt} = \frac{dS/dx} * \frac{dx/dt}\). The value of \(\frac{dS/dx}\) is calculated from step 2 and the quantity \(\frac{dx/dt}\) (\$125 per week) is directly provided. Plugging the values, \(\frac{dS/dt} = 1100 * 125 = \$137,500.\)

Key concepts

Rate of Change

In Calculus, the rate of change refers to how one quantity changes with respect to another quantity. It is essentially asking "how fast" something is changing. For instance, in the given exercise, we are interested in how the sales (S) change as the advertising costs (x) change.

• The rate of change can be instantaneous or average over an interval.
• The derivative of a function provides the instantaneous rate of change at a specific point.
• It is a critical concept when interpreting results from models, allowing industries to optimize their operations.

For sales and advertising modeling, knowing the rate of change helps businesses decide how much more or less to invest in advertising to achieve desired sales targets.

Differentiation

Differentiation is the process of finding the derivative of a function. It is a key operation in calculus and is used to determine the rate of change of a function concerning its variable.
Differentiating the function necessary in our exercise involves identifying how sales output relates to inputs of advertising costs. Using the function provided, we can find:

• The derivative, represented as \(S'(x)\), describes how sales change as advertising costs increase.
• The formula \(S' = 50 + 0.7x\) from our differentiation tells us that, for every one dollar increase in advertising cost, sales increase by \(50 + 0.7x\).

This derivative is valuable as it allows calculation of the specific moment-accurate rate of change, offering insights into how sensitive sales are to changes in advertising expenditure.

Applied Mathematics

Applied Mathematics uses mathematical methods to solve real-world problems. It forms the bridge between theoretical math and practical application, making math useful in fields like business, engineering, and economics.
In the exercise, applied mathematics enabled us to use differentiation to calculate the impact of advertising changes on sales revenue:

• Using the derivative \(S'(x)\), we assessed how weekly sales are impacted by shifts in advertising cost.
• Furthermore, the formula \(\frac{dS/dt} = \frac{dS/dx} * \frac{dx/dt}\) used the chain rule. This illustrates how changes in independent variables affect dependent outcomes over time.
• This showcases a practical instance of applied mathematics where business decisions on advertising can be financially quantified.

By grounding decisions in mathematical analysis, businesses can methodically optimize their strategies for improved outcomes.