3.15: Application of Differentiation to Business

Calculus offers essential techniques for businesses seeking to optimize pricing strategies and revenue. In this case, a bakery wants to determine the ideal price and daily sales volume to maximize revenue. By modeling how changes in price affect demand and revenue, the bakery can apply calculus to make data-driven decisions.

The demand function relates the price per cupcake to the number of cupcakes sold and captures how lower prices increase sales. Based on market data, the demand function can be represented as:

Where p(x) is the price per cupcake when x cupcakes are sold per day

The revenue function, which calculates total revenue, is represented as:

To maximize revenue, the bakery obtains the marginal revenue function, the derivative of the revenue function:

By solving for when marginal revenue equals zero, the bakery finds the optimal sales level of:

At this sales level, the price is approximately $2.33 per cupcake.

This example demonstrates how calculus links pricing, demand, and revenue into a cohesive optimization process, enabling businesses to make informed decisions that maximize financial outcomes using mathematical modeling.

A store wants to find the best price for a Smart TV to maximize weekly sales revenue. Revenue is the total sales income, calculated as price multiplied by units sold.

Finding the best price starts with a demand equation that shows how price and units sold relate. Two key pieces of information define this relationship. First, sales records show that 200 units are sold when the price is 350 dollars.

Second, a market study shows a linear relationship between price and units sold, with a slope of minus one-half.

Using the point-slope form with this slope and the data point gives the demand equation. Rearranging the equation expresses units sold in terms of price.

Substituting this into the revenue formula gives a single-variable quadratic function for revenue.

As price increases, revenue first rises, reaches a maximum, and then falls. The highest point on this curve shows the price that gives maximum revenue.

This point is found by differentiating the revenue function and setting the derivative to zero.

Solving this gives the best price as 225 dollars.

This example shows how calculus finds maximum values in real situations.