How to Calculate Maximum Revenue
Understand the relation between price and demand., Create a price function., Determine the revenue function., Find the first derivative of the revenue function., Set the derivative equal to 0., Solve for the number of items at the 0 value...
Step-by-Step Guide
-
Step 1: Understand the relation between price and demand.
Economics study shows that, for most traditional businesses, as the demand for any item increases, the price for that item should decrease.
Conversely, as the price decreases, demand should increase.
Using data from actual sales, a company can determine a supply and demand graph.
That data can be used to calculate a price function.For more information on graphing supply and demand data, see Find and Analyze Demand Function Curve. -
Step 2: Create a price function.
The price function consists of two primary pieces of information.
The first is the intercept.
This is the theoretical price if no items are sold.
The second detail is a decreasing slope.
The slope of the graph represents the drop in price for each item.
A sample price function might look like:p=500−150q{\displaystyle p=500-{\frac {1}{50}}q} p = price q = demand, in number of units This function sets the “zero price” at $500.
For each unit sold, the price decreases by 1/50th of a dollar (two cents). , Revenue is the product of price times the number of units sold.
Since the price function includes the number of units, this will result in a squared variable.
Using the price function from above, the revenue function becomes:
R(q)=p∗q{\displaystyle R(q)=p*q} R(q)=∗q{\displaystyle R(q)=*q} R(q)=500q−150q2{\displaystyle R(q)=500q-{\frac {1}{50}}q^{2}} , In calculus, the derivative of any function is used to find the rate of change of that function.
The maximum value of a given function occurs when the derivative equals zero.
So, to maximize the revenue, find the first derivative of the revenue function.Suppose the revenue function, in terms of number of units sold, is R(q)=500q−150q2{\displaystyle R(q)=500q-{\frac {1}{50}}q^{2}}.
The first derivative, therefore, is:
R′(q)=500−250q{\displaystyle R^{\prime }(q)=500-{\frac {2}{50}}q} For a review of derivatives, see the LifeGuide Hub article on how to Take Derivatives. , When the derivative is zero, the graph of the original function is at either a peak or a trough.
This will be either the maximum or minimum value.
For some higher level functions, there may be more than one solution to the zero derivative, but not a basic price-demand function.R′(q)=500−250q{\displaystyle R^{\prime }(q)=500-{\frac {2}{50}}q} 0=500−250q{\displaystyle 0=500-{\frac {2}{50}}q} , Use basic algebra to solve the derivative for the number of items to sell where the derivative is equal to zero.
This will give you the number of items that will maximize the revenue.0=500−250q{\displaystyle 0=500-{\frac {2}{50}}q} 250q=500{\displaystyle {\frac {2}{50}}q=500} 150q=250{\displaystyle {\frac {1}{50}}q=250} q=50∗250{\displaystyle q=50*250} q=12,500{\displaystyle q=12,500} , Using the optimal number of sales from the derivative calculation, you can enter that value into the original price formula to find the optimal price.p=500−150q{\displaystyle p=500-{\frac {1}{50}}q} p=500−15012,500{\displaystyle p=500-{\frac {1}{50}}12,500} p=500−250{\displaystyle p=500-250} p=250{\displaystyle p=250} , After you have found the optimal number of sales and the optimal price, multiply them to find the maximum revenue.
Recall that R=p∗q{\displaystyle R=p*q}.
The maximum revenue for this example, therefore, is:
R=p∗q{\displaystyle R=p*q} R=(250)(12,500){\displaystyle R=(250)(12,500)} R=3,125,000{\displaystyle R=3,125,000} , Based on these calculations, the optimal number of units to sell is 12,500, at the optimal price of $250 each.
This will result in a maximum revenue, for this sample problem, of $3,125,000. , Suppose that another business has collected price and sales data.
Using that data, the company has determined that the initial price is $100, and each additional unit sold will cut the price by one cent.
Using this data, the following price function is: p=100−0.01q{\displaystyle p=100-0.01q} , Recall that revenue is equal to price times quantity.
Using the price function above, the revenue function is:
R(q)=∗q{\displaystyle R(q)=*q} R(q)=100q−0.01q2{\displaystyle R(q)=100q-0.01q^{2}} , Using basic calculus, find the derivative of the revenue function:
R(q)=100q−0.01q2{\displaystyle R(q)=100q-0.01q^{2}} R′(q)=100−(2)0.01q{\displaystyle R^{\prime }(q)=100-(2)0.01q} R′(q)=100−0.02q{\displaystyle R^{\prime }(q)=100-0.02q} , Set the derivative equal to zero and solve for q{\displaystyle q} to find the optimal number of sales.
This calculation is as follows:
R′(q)=100−0.02q{\displaystyle R^{\prime }(q)=100-0.02q} 0=100−0.02q{\displaystyle 0=100-0.02q}
0.02q=100{\displaystyle
0.02q=100} q=100/0.02{\displaystyle q=100/0.02} q=5,000{\displaystyle q=5,000} , Use the optimal sales value in the original price formula to find the optimal sales price.
For this example, this works as follows: p=100−0.01q{\displaystyle p=100-0.01q} p=100−0.01(5,000){\displaystyle p=100-0.01(5,000)} p=100−50{\displaystyle p=100-50} p=50{\displaystyle p=50} , Using the relationship that revenue equals price times quantity, you can find the maximum revenue as follows:
R(q)=p∗q{\displaystyle R(q)=p*q} R(q)=50∗5,000{\displaystyle R(q)=50*5,000} R(q)=250,000{\displaystyle R(q)=250,000} , Using this data and based on the price function p=100−0.01q{\displaystyle p=100-0.01q}, the company’s maximum revenue is $250,000.
This assumes a unit price of $50 and a sale of 5,000 units. -
Step 3: Determine the revenue function.
-
Step 4: Find the first derivative of the revenue function.
-
Step 5: Set the derivative equal to 0.
-
Step 6: Solve for the number of items at the 0 value.
-
Step 7: Calculate the maximum price.
-
Step 8: Combine the results to calculate maximum revenue.
-
Step 9: Summarize the results.
-
Step 10: Begin with the price function.
-
Step 11: Determine the revenue function.
-
Step 12: Find the derivative of the revenue function.
-
Step 13: Find the maximum value.
-
Step 14: Calculate the optimal price.
-
Step 15: Combine the maximum sales and optimal price to find maximum revenue.
-
Step 16: Interpret the results.
Detailed Guide
Economics study shows that, for most traditional businesses, as the demand for any item increases, the price for that item should decrease.
Conversely, as the price decreases, demand should increase.
Using data from actual sales, a company can determine a supply and demand graph.
That data can be used to calculate a price function.For more information on graphing supply and demand data, see Find and Analyze Demand Function Curve.
The price function consists of two primary pieces of information.
The first is the intercept.
This is the theoretical price if no items are sold.
The second detail is a decreasing slope.
The slope of the graph represents the drop in price for each item.
A sample price function might look like:p=500−150q{\displaystyle p=500-{\frac {1}{50}}q} p = price q = demand, in number of units This function sets the “zero price” at $500.
For each unit sold, the price decreases by 1/50th of a dollar (two cents). , Revenue is the product of price times the number of units sold.
Since the price function includes the number of units, this will result in a squared variable.
Using the price function from above, the revenue function becomes:
R(q)=p∗q{\displaystyle R(q)=p*q} R(q)=∗q{\displaystyle R(q)=*q} R(q)=500q−150q2{\displaystyle R(q)=500q-{\frac {1}{50}}q^{2}} , In calculus, the derivative of any function is used to find the rate of change of that function.
The maximum value of a given function occurs when the derivative equals zero.
So, to maximize the revenue, find the first derivative of the revenue function.Suppose the revenue function, in terms of number of units sold, is R(q)=500q−150q2{\displaystyle R(q)=500q-{\frac {1}{50}}q^{2}}.
The first derivative, therefore, is:
R′(q)=500−250q{\displaystyle R^{\prime }(q)=500-{\frac {2}{50}}q} For a review of derivatives, see the LifeGuide Hub article on how to Take Derivatives. , When the derivative is zero, the graph of the original function is at either a peak or a trough.
This will be either the maximum or minimum value.
For some higher level functions, there may be more than one solution to the zero derivative, but not a basic price-demand function.R′(q)=500−250q{\displaystyle R^{\prime }(q)=500-{\frac {2}{50}}q} 0=500−250q{\displaystyle 0=500-{\frac {2}{50}}q} , Use basic algebra to solve the derivative for the number of items to sell where the derivative is equal to zero.
This will give you the number of items that will maximize the revenue.0=500−250q{\displaystyle 0=500-{\frac {2}{50}}q} 250q=500{\displaystyle {\frac {2}{50}}q=500} 150q=250{\displaystyle {\frac {1}{50}}q=250} q=50∗250{\displaystyle q=50*250} q=12,500{\displaystyle q=12,500} , Using the optimal number of sales from the derivative calculation, you can enter that value into the original price formula to find the optimal price.p=500−150q{\displaystyle p=500-{\frac {1}{50}}q} p=500−15012,500{\displaystyle p=500-{\frac {1}{50}}12,500} p=500−250{\displaystyle p=500-250} p=250{\displaystyle p=250} , After you have found the optimal number of sales and the optimal price, multiply them to find the maximum revenue.
Recall that R=p∗q{\displaystyle R=p*q}.
The maximum revenue for this example, therefore, is:
R=p∗q{\displaystyle R=p*q} R=(250)(12,500){\displaystyle R=(250)(12,500)} R=3,125,000{\displaystyle R=3,125,000} , Based on these calculations, the optimal number of units to sell is 12,500, at the optimal price of $250 each.
This will result in a maximum revenue, for this sample problem, of $3,125,000. , Suppose that another business has collected price and sales data.
Using that data, the company has determined that the initial price is $100, and each additional unit sold will cut the price by one cent.
Using this data, the following price function is: p=100−0.01q{\displaystyle p=100-0.01q} , Recall that revenue is equal to price times quantity.
Using the price function above, the revenue function is:
R(q)=∗q{\displaystyle R(q)=*q} R(q)=100q−0.01q2{\displaystyle R(q)=100q-0.01q^{2}} , Using basic calculus, find the derivative of the revenue function:
R(q)=100q−0.01q2{\displaystyle R(q)=100q-0.01q^{2}} R′(q)=100−(2)0.01q{\displaystyle R^{\prime }(q)=100-(2)0.01q} R′(q)=100−0.02q{\displaystyle R^{\prime }(q)=100-0.02q} , Set the derivative equal to zero and solve for q{\displaystyle q} to find the optimal number of sales.
This calculation is as follows:
R′(q)=100−0.02q{\displaystyle R^{\prime }(q)=100-0.02q} 0=100−0.02q{\displaystyle 0=100-0.02q}
0.02q=100{\displaystyle
0.02q=100} q=100/0.02{\displaystyle q=100/0.02} q=5,000{\displaystyle q=5,000} , Use the optimal sales value in the original price formula to find the optimal sales price.
For this example, this works as follows: p=100−0.01q{\displaystyle p=100-0.01q} p=100−0.01(5,000){\displaystyle p=100-0.01(5,000)} p=100−50{\displaystyle p=100-50} p=50{\displaystyle p=50} , Using the relationship that revenue equals price times quantity, you can find the maximum revenue as follows:
R(q)=p∗q{\displaystyle R(q)=p*q} R(q)=50∗5,000{\displaystyle R(q)=50*5,000} R(q)=250,000{\displaystyle R(q)=250,000} , Using this data and based on the price function p=100−0.01q{\displaystyle p=100-0.01q}, the company’s maximum revenue is $250,000.
This assumes a unit price of $50 and a sale of 5,000 units.
About the Author
Margaret Hamilton
Brings years of experience writing about organization and related subjects.
Rate This Guide
How helpful was this guide? Click to rate: