 This lesson will tell you how to write and use a special type of function called a profit function. After completing this lesson, you will also be able to evaluate profit functions and use them in real life scenarios.

## Definition of a Function

Before we show the profit function equation, we should first make sure that we understand what a general function is. It might be best to think of a function as some type of machine on a factory floor.

### Best services for writing your paper according to Trustpilot

From \$18.00 per page
4,8 / 5
4,80
Writers Experience
4,80
Delivery
4,90
Support
4,70
Price
Recommended Service
From \$13.90 per page
4,6 / 5
4,70
Writers Experience
4,70
Delivery
4,60
Support
4,60
Price
From \$20.00 per page
4,5 / 5
4,80
Writers Experience
4,50
Delivery
4,40
Support
4,10
Price
* All Partners were chosen among 50+ writing services by our Customer Satisfaction Team

We enter raw material (input) into the machine, then the machine processes the material, and finally, it creates a product (output).The input is the domain of the function and the output is the range of the function. The domain is usually represented by the variable x and it is called the independent variable. Each value used for the independent variable produces an output value that is unique to the independent variable. In other words, each input has only one output. The output, or range, of a function is often represented by the variable y.

## Function Notation

An example of a function could be y = 3x + 7.

We can replace y with the symbol f(x)f(x) = 3x + 7. The symbol f(x) can be read as ‘f of x‘ or as ‘the value of f at x‘. In this example, the name of the function is f. If we want to evaluate this function when x = 8, we would complete the following calculation:f(8) = 3(8) + 7f(8) = 24 + 7f(8) = 31We can use other letters to name functions and other letters to represent the input variable. Let’s say that Tom created a function that determines his salary for each week. Therefore, he named this function s.

His weekly salary is based on the number of hours he works and he makes \$15.00 per hour. Therefore, he used the letter h for his input or independent variable. His function is as follows:s(h) = 15hThis is a very simple function, but the main point here is that we can use alternate variables that make sense relevant to the purpose of a function.

## Profit Function Equation

A profit function is a function that focuses on business applications. The primary purpose for a business is to sell a product or service in order to make a profit, which is the revenue a company receives for selling a product or service less the cost for creating a product or service.The profit function equation is made up of two primary functions: the revenue function and the cost function.

If x represents the number of units sold, we will name these two functions as follows: R(x) = the revenue function; C(x) = the cost function. Therefore, our profit function equation will be as follows: P(x) = R(x) – C(x).

## Examples

Let’s show a simple example.

A street vendor in New York City sells hot dogs for \$3.00 each. Therefore, his revenue function is R(x) = 3x. His fixed cost for maintaining his stand each day is \$50.

00 and his variable cost, or the materials needed to make the hot dogs, is \$2.00 per hot dog sold. Therefore, his daily cost function is C(x) = 50 + 2x.We first would like to find how many hot dogs the vendor needs to sell in order to break even.

The breakeven point is when revenue equals cost or when the profit is zero. P(x) = R(x) – C(x) = 0. If the profit is zero, then R(x) = C(x). We will use this equation to find the breakeven point – R(x) = C(x).

3x = 50 + 2x. x = 50.So, the hot dog vendor needs to sell at least 50 hot dogs on a given day to break even. This is also reflected in the graph shown in Figure 1.

We will graph the revenue and cost functions instead of the profit function because this strategy will better explain the dynamics of the profit function. The vendor has a positive profit once he sells more than 50 hot dogs in a given day, and he adds \$1.00 to this profit with each hot dog sold over 50. Our profit equation and the graph show that the vendor’s profit potential is endless.

However, that is not realistic. There are only so many hours in a day and other factors can come into play, such as price discounts when street traffic is low or the spoiling of materials that are not used. Revenue and cost functions are rarely simple linear functions.Now let’s look at a more complex example.A small business has the following weekly cost and revenue functions for a new product:C(x) = 200 + 10x + 0.2x^2R(x) = 40x – 0.1x^2

Figure 2
[Image_Link]/cimages/multimages/16/Profit_Function_Graph_2.

png” alt=”graph” />

These functions are reflected in the graph shown in Figure 2. You might notice that the slope of the cost function increases as units sold increases. This could be caused by some factor such as the need for overtime to make the higher number of units. You might also notice that the slope of the revenue function decreases as units sold increases. This could be caused by some factor, such as the giving of price discounts for higher sales volumes.Let’s find the company’s profit when it sells 75 units in a week. P(x) = R(x) – C(x).

P(x) = 40x – 0.1x^2 – (200 + 10x + 0.2x^2).

It might be easier to evaluate the revenue and cost functions separately:R(75) = 40(75) – 0.1(75)^2 = 3000 – 562.5 = 2437.5 C(75) = 200 + 10(75) + 0.2(75)^2 = 200 + 750 + 1125 = 2075 P(75) = 2437.

5 – 2075 = 362.50At a weekly volume of 75 units, the company makes a profit of \$362.50. Similar to our first example, the graph in Figure 2 shows that the cost is higher than revenue before any units are sold. At some point, revenue begins to exceed cost. At a higher volume, however, cost begins to exceed revenue again. Therefore, there is a certain volume at which the company will maximize its revenue.

In order to be successful, the company would like to find this volume.

## Marginal Revenue and Marginal Cost

To help the company find the volume at which it will maximize its revenue, we will need to find the marginal revenue function and the marginal cost function. The marginal revenue is the rate of change in revenue at a certain volume, and the marginal cost is the rate of change in cost at a certain volume.

In other words, we need to find the derivatives (slopes) of the original revenue and cost functions. We can do this by using the power rule. The marginal revenue and marginal cost functions are as follows:R'(x) = 40 – 0.2x C'(x) = 10 + 0.4xThe prime marks denote these special functions. If you are unfamiliar with the power rule and with the derivative of polynomials, you can just accept the functions as shown above.

However, knowledge of these concepts will be helpful as you move forward through your learning of mathematics.We want to know at which volume are the marginal revenue and marginal cost functions equal? This volume is the point at which the company will maximize its profit. Here is why: if the marginal revenue is higher than the marginal cost, then the company will want to increase the units sold because their profit will increase with each unit sold. If the marginal revenue is lower than the marginal cost, then the company will want to decrease the units sold because they are losing money with each unit sold.Therefore, the company’s maximum profit will occur when its marginal profit (marginal revenue less marginal cost) is zero, P'(x) = R'(x) – C'(x) = 0. We will make the marginal revenue and marginal cost functions equal to each other and solve for x.R'(x) = C'(x) 40 – 0.

2x = 10 + 0.4x 0.6x = 30 x = 50The company will maximize its profit for this product by selling 50 units. This solution agrees with the graph in Figure 2. You’ll notice that the positive difference between the graph of the revenue function and the graph of the cost function is highest at 50 units.

## Lesson Summary

Let’s review.The profit function is similar to any other function in mathematics in that each input from a designated domain results in a unique output. However, the profit function also has unique features. Primarily, the profit function is often represented, as in this lesson, as a composition of the revenue function and the cost function.

All the complexities of these sub-functions can be incorporated into their mathematical expressions.The main feature of this lesson is the equation P(x) = R(x) – C(x). Other important equations to remember are as follows:

• The breakeven point is when revenue equals cost, or when the profit is zero. R(x) = C(x) P(x) = R(x) – C(x) = 0
• Maximum profit will occur when marginal revenue equals marginal cost, or when the marginal profit is zero. R'(x) = C'(x) P'(x) = R'(x) – C'(x) = 0

## Learning Outcomes

After you have finished, you should be able to:

• Identify the parts of a function
• Write an equation in function notation
• State the profit function equation
• Calculate maximum profit
A Project Report On “CURRENT MODE BIQUAD FILTER” Submitted to The Department of Electronics Engineering In partial fulfilment of requirements for the award of the Degree of Bachelor of Technology In Electronics and Communication Engineering By
• The Ohio State University (OSU) Academic Calendar
• EFFECT of this project has been submitted to
• 181102039624000 WOLAITA SODO UNIVERSITY SCHOOL OF GRADUATE STUDIES DETERMINANTS OF FINANCEIAL PERFORMANCE OF MICROFINANCE INSTITUTION IN ETHIOPIA
• Learn Variables can either be dependent or independent
• This canine. It’s a little like saying,
• -485775-342900COMSATS experience I have picked up at
• CHAPTER from information technology investment programs sporadically
• EMBEDDED SOLUTION FOR VIDEO CAPTURE
• Declaration We Hagazi Abrha 