Cumulative Revenue From Changing Sales Rates

From Sales to Cumulative Revenue

A sales rate tells us how quickly units are being sold. Revenue, however, is measured in money. Connecting the two requires knowing how much revenue is earned per unit sold at each time.

This distinction matters when sales volume changes, when the realized price changes, or when both change. A rate measured in units per day is not itself a rate measured in dollars per day. The first must be combined with revenue per unit before it describes money accumulating over time.

Define the Quantities Before Combining Them

Let \(t\) denote time. Write \(q(t)\) for the sales volume rate: the number of units sold per unit of time at time \(t\). For example, \(q(t)=40\) units per day means sales are occurring at a rate of 40 units each day at that time.

Write \(p(t)\) for the realized revenue per unit sold at time \(t\). If the price is uniform for the sales being counted, this is the selling price per unit. More generally, it can be the average revenue per unit among the units sold at that time. The model assumes \(p(t)\) applies to the units represented by \(q(t)\).

Let \(R(t)\) denote cumulative revenue up to time \(t\). It is a stock measured in currency, not currency per unit of time. If sales generate gross revenue, usually \(q(t)\geq 0\) and \(p(t)\geq 0\). The model below can also accommodate a signed monetary rate when the quantity being tracked includes signed adjustments, provided that interpretation is stated.

The integration variable \(s\) indicates that the rates are evaluated throughout the interval from \(t_0\) to \(t\). The baseline \(R_0\) matters: if the model begins after some revenue has already accumulated, setting the initial revenue to zero would discard that information. If the model begins before any relevant sales, \(R_0\) may be zero.

Why Sales Rate and Revenue Rate Must Be Kept Separate

Over a short time interval of length \(\Delta t\), a sales rate of about \(q(t)\) units per unit time corresponds to about \(q(t)\Delta t\) units sold. If each such unit generates about \(p(t)\) units of currency, the revenue over that short interval is about \(p(t)q(t)\Delta t\). Thus \(p(t)q(t)\), not \(q(t)\), is the rate at which money accumulates.

The calculus justification follows from the Fundamental Theorem of Calculus. Suppose \(p\) and \(q\) are continuous on an interval containing \(t_0\) and \(t\), and define \(R\) by the accumulation formula above. Then

The constant baseline has derivative zero. The Fundamental Theorem applies because the integrand \(p(s)q(s)\) is continuous, and it says that differentiating its accumulation recovers the integrand at the upper endpoint. Conversely, any differentiable revenue function satisfying \(R'(t)=p(t)q(t)\) and \(R(t_0)=R_0\) has this integral form, by integrating the rate from \(t_0\) to \(t\). This proves both how to construct the model and why its accumulated change matches the monetary rate.

Notice that the product is \(p(t)q(t)\), not the derivative of price times accumulated units. Units sold earlier have already generated revenue at the prices applicable when they were sold; a later price change does not automatically reprice those past sales.

Worked Example: Changing Sales Volume at a Fixed Price

Worked Example: Both Price and Sales Volume Change

Worked Example: Piecewise Operating Prices and Sales Rates

A Common Mistake: Repricing Earlier Sales

Suppose a business sells 10 units per day for 4 days, and the realized price is \(p(t)=5+t\) dollars per unit, where \(t\) is measured in days from the start. The correct revenue rate is \(10(5+t)\) dollars per day. The total is

A tempting but incorrect calculation is to multiply all 40 units sold by the price on the last day: \(40(5+4)=40(9)=\$360\). That assigns the final price of \(\$9\) to units sold earlier, even though their realized prices were lower. The correction is to use the price applicable to sales at each time and integrate \(p(t)q(t)\). As a check, the correct antiderivative \(50t+5t^2\) has derivative \(50+10t=10(5+t)\), the required revenue rate.

Another common error is to integrate \(q(t)\) and call the result revenue. Integrating a sales volume rate gives units sold, not money earned. Multiplying by revenue per unit—or, if already supplied, using the monetary revenue rate directly—is essential.

When a Monetary Revenue Rate Is Given Directly

Sometimes a model supplies a monetary rate such as \(\rho(t)=300+20t\) dollars per hour, without separately specifying sales volume and price. In that case, the accumulation model is \(R(t)=R_0+\int_{t_0}^{t}\rho(s)\,ds\). Do not multiply this rate by a price per unit again: \(\rho\) already measures money per time.

Before calculating, check what each quantity represents and its units. A sales volume rate has units such as units per hour; a per-unit price has units such as dollars per unit; their product has units of dollars per hour. Integrating dollars per hour over an interval measured in hours produces dollars. These checks help identify whether the model is accumulating the intended quantity.