Application of Differentiation to Business

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.