How to Find The Maximum Profit

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  1. General Maximization
  2. Maximizing Profits

General Maximization

how to maximize in calculus
To maximize a function means to find its maximum value in a given range. A global maximum means to find the maximum to find the maximum over the entire range of the function. Local maximums occur at inflection points (where the graph changes direction). At the maximum of a function, the gradient or slope of the function is zero. You can differentiate the function to find points in the function where the gradient is zero, but these could be either maxima or minima.

  • If the slope is increasing at the turning point, it is a minimum.
  • If the slope is decreasing at the turning point, then you have found a maximum of the function.

How to Maximize in Calculus: Steps

Sample problem: Find the local maximum value of y = 4x3 + 2x2 + 1.

First Step: Differentiate the function, using the power rule. Constant terms disappear under differentiation.
d/dx (4x3 + 2x2 + 1) = 12x2 + 4x
The result, 12x2 + 4x, is the gradient of the function. This has two zeros, which can be found through factoring.
12x2 + 4x = 4x(3x+1), which equals zero when x = 0 or x = -1/3

Second Step: Check each turning point (at x = 0 and x = -1/3)to find out whether it is a maximum or a minimum. To do this, differentiate a second time and substitute in the x value of each turning point.
d/dx (12x2 + 4x) = 24x + 4
At x = 0, 24x + 4 = 4, which is greater than zero. This is a minimum.
At x = -1/3, 24x + 4 = -4, which is less than zero. This is a maximum.

Third Step: Find the corresponding y-coordinates for the x-value (maximum) you found in Step 2 by substituting back into the original function.
At x = -1/3, y = 4x3 + 2x2 + 1 = -4/27 + 2/9 + 1 = 29/27
Therefore the function has a maximum value at (-1/3, 29/27).
That’s how to maximize in calculus!
Tip: You can check your answer by sketching the graph and looking for the highest and lowest points.

Profit Maximization

Profit maximization is one of the topics that are likely to be tested in the short-answer section of the AP Calculus exam. It is equal to a business’s revenue minus the costs incurred in producing that revenue. Profit maximization is an important concern because businesses are run in order to earn the highest profits possible. Calculus can be used to calculate the profit-maximizing number of units produced.

Find the maximum profit in calculus: Steps

find maximum profit in calculus

Sample question: Find the profit equation of a business with a revenue equation of 2000x – 10x2 and a cost equation of 2000 + 500x.

First Step: Set profit to equal revenue minus cost. For example, the revenue equation 2000x – 10x2 and the cost equation 2000 + 500x can be combined as profit = 2000x – 10x2 – (2000 + 500x) or profit = -10x2 + 1500x – 2000.

Second Step: Find the derivative of the profit equation (here’s a list of common derivatives). For example, the profit equation -10x2 + 1500x – 2000 becomes -20x + 1500.

Third Step: Set the equation equal to zero:
-20x + 1500 = 0

Fourth Step: Use algebra to find how many units are produced from the equation you wrote in Step 3.
20x = 1500
x = 75.

Fifth Step: Calculate the maximum profit using the number of units produced calculated in the previous step. In this example, inserting x = 75 into the profit equation -10x2 + 1500x – 2000 produces -10(75)2 + 1500(75) – 2000 or 54,250 in profit.
That’s how to find the maximum profit in calculus!

For answering this type of question on the AP calculus exam, be sure to record this figure using the unit of measurement presented in the short-answer problem. For profit maximization short-answer problems on the AP Calculus exam, this unit of measurement is almost certainly US dollars or $. Some equations might present more than one possible answer. Some of these answers can be picked out and discarded using common sense but most often cannot be treated the same. In these cases, insert all possible answers into the profit equation to calculate their profits and then select the answer that produces the highest profit as the profit maximizing number of units produced.

Sources: Retrieved July 12, 2015.