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Cost and Revenue

- Calculus
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By now, you have learned about using derivatives to analyze things such as population change and motion along a line. While that is all well and good, have you ever wondered if **math could make you money**?

Well, it turns out, it can! You can use derivatives and rates of change to describe the simple business and economic concepts of change in **cost**, **revenue**, and **profit**.

The **rate of change** of a company's profit depends on

the rate at which the company sells its products,

the cost of making each product,

the cost of labor, and

the potential competition for these resources.

You can use these parameters to create an equation that models a company's **revenue**. Then, the **derivative**, or rate of change of that equation, is the rate at which the company earns revenue.

While cost and revenue rates of change help the business world make predictions on the future prospects of investments, they can also be useful to businesses themselves to describe how well they're doing or if changes need to be made.

So, what is the meaning of **cost **and **revenue**?

** Cost **represents the amount of money a company must spend to produce a certain commodity. The derivative of a company's cost function is called its **marginal cost**.

If \(C(x)\) is the **cost **of producing \(x\) number of items, then the **marginal cost**, \(MC(x)\),** **of producing \(x\) items is \(C'(x)\).

Mathematically,

\[ \begin{align}\text{Cost} &= C(x) \\\text{Marginal Cost} &= C'(x) = MC(x).\end{align} \]

Similarly, **revenue **represents the amount of money a company obtains from selling a certain commodity. The derivative of a company's revenue function is called its **marginal revenue**.

If \(R(x)\) is the **revenue **from selling \(x\) number of items, then the **marginal revenue**, \(MR(x)\),** **of selling \(x\) items is \(R'(x)\).

Mathematically,

\[ \begin{align}\text{Revenue} &= R(x) \\\text{Marginal Revenue} &= R'(x) = MR(x).\end{align} \]

So, what is the relationship between cost and revenue?

Well, **profit **represents the amount of money pocketed by a company – once its cost and revenue are tallied up. The formula for profit is subtracting the total **cost **to the company spent on producing the commodity from the total **revenue **obtained from selling the commodity.

If \(P(x) = R(x) - C(x)\) is the **profit **from making and selling \(x\) number of items, then the** ****marginal profit**, \(MP(x)\),** **from making and selling \(x\) items is \(P'(x)\).

Mathematically,

\[ \begin{align}\text{Profit} &= P(x) \\ &= R(x) - C(x), \\\text{Marginal Profit} &= P'(x) \\ &= MP(x) \\ &= R'(x) - C'(x).\end{align} \]

Unfortunately, there are no general formulas for cost, revenue, or profit. Depending on the question, you may either be tasked with determining these formulas based on context clues within a word problem, or the formula(s) may be given to you.

That being said, using the definition of the derivative, you can approximate

\[ \begin{align}\text{Marginal Cost} &= MC(x) \\&= C'(x) \\&= \lim_{h \to 0} \frac{C(x+h) - C(x)}{h} \\\end{align} \]

and

\[ \begin{align}\text{Marginal Revenue} &= MR(x) \\&= R'(x) \\&= \lim_{h \to 0} \frac{R(x+h) - R(x)}{h} \\\end{align} \]

and

\[ \begin{align}\text{Marginal Profit} &= MP(x) \\&= P'(x) \\&= \lim_{h \to 0} \frac{P(x+h) - P(x)}{h} \\\end{align} \]

by choosing an appropriate value for \(h\). But, what is an appropriate value for \(h\)?

Considering that the independent variable \(x\) represents physical items, \(h\) must be an integer greater than zero. So, since you want \(h\) to be as small as possible for the best approximation, substituting \(h = 1\) gives you the formulas:

\[ \begin{align}\text{Marginal Cost} &= MC(x) = C'(x) \approx C(x+1) - C(x), \\\text{Marginal Revenue} &= MR(x) = R'(x) \approx R(x+1) - R(x), \text{and} \\\text{Marginal Profit} &= MP(x) = P'(x) \approx P(x+1) - P(x).\end{align} \]

Looking at these formulas, it should become clear that:

\(C'(x)\) for any value of \(x\) can be thought of as the change in cost associated with the company making one more item.

\(P'(x)\) for any value of \(x\) can be thought of as the change in revenue associated with the company selling one more item.

\(P'(x)\) for any value of \(x\) can be thought of as the change in profit associated with the company making and selling one more item.

As you might have noticed in the definitions above, the **derivatives**, also known as the **marginals**, of the cost, revenue, and profit functions measure the change in the functions over time.

Just like the other applications of derivatives, the rate of change of the cost and revenue functions are the same.

For a review of average and instantaneous rate of change, see our article Rates of Change and Amount of Change Formula.

The **average rate of change of a cost or revenue function** measures how much the cost or revenue changes.

Let \(d = C(x)\) represent the total dollar amount spent on producing \(x\) items. The average rate of change in cost from the first item produced \(x_1\) and the last item produced \(x_2\) is

\[ \begin{align} \mbox{Average Rate of Change of Cost } &= \frac{\Delta d}{\Delta x} \\ &= \frac{C(x_2) - C(x_1)}{x_2-x_1}. \end{align}\]

What about the average rate of change of revenue?

Similarly, let \(p = R(x)\) represent the total dollar amount made from selling \(x\) items. The average rate of change in revenue from the first item sold \(x_1\) and the last item sold \(x_2\) is

\[ \begin{align} \mbox{Average Rate of Change in Revenue } &= \frac{\Delta p}{\Delta x} \\ &= \frac{R(x_2) - R(x_1)}{x_2-x_1}.\end{align} \]

For a walk through on how the average rate of change formula is derived, please refer to the Population Change article.

To find the exact rate of change of a cost or revenue at the point of producing or making one certain item, you find the instantaneous rate of cost or revenue change. By now, you should recognize that the instantaneous rate of cost or revenue change is synonymous with the derivative (or marginal) of the cost or revenue equation.

Again, let \(d = C(x)\) represent the total dollar amount spent on producing \(x\) items. The instantaneous rate of change of cost when producing item \(x\) is

\[ \begin{align} \mbox{Instantaneous Rate of Change of Cost } &= \lim\limits_{\Delta x \to 0} \frac{\Delta d}{\Delta x} \\ &= \frac{dd}{dx}. \end{align}\]

What about the instantaneous rate of change?

Similarly, let \(p = R(x)\) represent the total dollar amount made from selling \(x\) items. The instantaneous rate of change in revenue when selling item \(x\) is

\[ \begin{align} \mbox{Rate of Change in Revenue } &= \lim\limits_{\Delta x \to 0} \frac{\Delta p}{\Delta x} \\ &= \frac{dp}{dx}. \end{align}\]

These concepts are easier to understand with some examples.

Let's take a look at some examples involving cost, revenue, and things like the break even point.

A newspaper company has a fixed cost of production of \(\$80\) per edition, as well as a marginal distribution and printing materials cost of \(40¢\) per copy. Newspapers are sold at \(50¢\) per copy.

- Formulate functions for the cost, revenue, and profit of the newspaper.
- Then, find out whether the newspaper makes a profit or not from selling \(600\) copies.
- How many copies of the newspaper have to be sold to break even?

**Solution****s**:

- Formulate functions for the cost, revenue, and profit of the newspaper.
- As mentioned in the Cost and Revenue Formula section above, you need to use the context of the given problem to determine the formulas for cost, revenue, and profit.
- Based on the costs that the problem provides you, the cost function can be written as the fixed cost plus the cost per copy.\[C(x) = 80 + 0.4x\]
- Similarly, the revenue equation can be written as the amount the newspaper company earns per copy.\[R(x) = 0.5x\]
- Finally, since profit is the company's revenue minus its cost:\[ \begin{align}P(x) &= R(x) - C(x) \\&= 0.5x - (0.4x + 80) \\P(x) &= 0.1x-80\end{align}\]

- As mentioned in the Cost and Revenue Formula section above, you need to use the context of the given problem to determine the formulas for cost, revenue, and profit.
- Find out whether the newspaper makes a profit or not from selling \(600\) copies.
- Plug \(x = 600\) into your profit equation.\[P(600) = 0.1(600)-80 = -20\]
**Since the profit is negative, the newspaper company does not make a profit from selling \(600\) copies, but rather incurs a loss.**

- Find out how many copies of the newspaper have to be sold to break even.
- Breaking even means that the company does not earn any profit. So, the cost of production and the earned revenue are equal.
- To find out how many newspapers the company must sell to break even, you need to set the profit equation equal to \(0\), or \(P(x) = 0\) and then solve for \(x\).
- Set \(P(x) = 0\).\[ P(x) = 0.1x - 80 = 0 \]
- Solve for \(x\).\[ \begin{align}0.1x - 80 &= 0 \\0.1x &= 80 \\x &= \frac{80}{0.1} \\x &= 800\end{align} \]

**Therefore, the newspaper must sell \(800\) copies to break even**.

Let's take a look at another example.

Suppose a toy company produces \(x\) toys with a cost of \(C(x) = 10000 + 3x + 0.01x^2\) that encompasses the total overhead costs (like rent for the factory), materials, labor, and so on.

- Find the marginal cost of producing \(500\) toys.
- What does the company have to charge for the \(500^{th}\) toy to make a profit of \(\$10\)?
- Then, find the average marginal cost from \(200\) toys to \(300\) toys.

**Solutions**:

- Find the marginal cost of producing \(500\) toys.
- Find the marginal cost equation.
- Remember, the marginal cost equation is the derivative of the cost equation. So, use the power rule to take the derivative of \(C(x)\).\[ C'(x) = 3 +0.02x\]

- Plug in the number of toys for \(x\) to find the cost of producing a toy.
- Plugging in \(x = 500\) to \(C(x)\) gives you\[ C(500) = 3 + 0.02(500) = 13.\]
- What does this value actually mean?\(C(500) = 13\) means that when producing the \(500^{th}\) toy, the company is building toys at a cost of \(\$13\) per toy.
**Therefore, the marginal cost of producing the \(500\) toys is \(\$13\)**.

- Find the marginal cost equation.
- What does the company have to charge for the \(500^{th}\) toy to make a profit of \($10\)?
- From solution A, you know that the \(500^{th}\) toy cost \(\$13\) for the company to build. And, since you know the equation for profit is \(P(x) = R(x) - C(x)\), you can solve this equation for revenue to determine how much the company should charge for the toy.
- Use the equation for profit:\[ P(x) = R(x) - C(x) \]
- Solve for revenue.\[ \begin{align}P(x) &= R(x) - C(x) \\P(x) + C(x) &= R(x) \\R(x) &= P(x) + C(x)\end{align} \]
- Plug in your known values for profit \(($10)\) and cost \(($13)\), and solve for revenue.\[ R(x) = \$10 + \$13 = \$23 \]

**In order for the company to make a profit of \(\$10\) for the \(500^{th}\) toy built, they must charge \(\$23\) for it**.

- From solution A, you know that the \(500^{th}\) toy cost \(\$13\) for the company to build. And, since you know the equation for profit is \(P(x) = R(x) - C(x)\), you can solve this equation for revenue to determine how much the company should charge for the toy.
- Find the average marginal cost from \(200\) toys to \(300\) toys.
- To find the average marginal cost from \(200\) toys to \(300\) toys, you need to use the average rate of change formula:\[ \mbox{Average Rate of Change } = \frac{\Delta d}{\Delta x} = \frac{C(x_2) - C(x_1)}{x_2-x_1}\]where,\[ x_1 = 200 \text{ and } x_2 = 300. \]
- Next, you need to use the cost function given to find \(C(x_1)\) and \(C(x_2)\).
- For \(x_1 = 200\):\[ \begin{align}C(x_1) &= 10000 + 3(x_1) + 0.01(x_1)^2 \\C(200) &= 10000 + 3(200) + 0.01(200)^2 \\C(200) &= 10000 + 600 + 400 \\C(200) &= 11000\end{align} \]
- For \(x_2 = 300\):\[ \begin{align}C(x_2) &= 10000 + 3(x_2) + 0.01(x_2)^2 \\C(300) &= 10000 + 3(300) + 0.01(300)^2 \\C(300) &= 10000 + 900 + 900 \\C(300) &= 11800\end{align} \]

- Now, plug all your values into the average rate of change formula.\[ \mbox{Average Rate of Change } = \frac{\Delta d}{\Delta x} = \frac{C(x_2) - C(x_1)}{x_2-x_1}\]where,\[ x_1 = 200, x_2 = 300, C(x_1) = 11000, \text{ and } C(x_2) = 11800. \]\[ \begin{align}\mbox{Average Rate of Change } &= \frac{\Delta d}{\Delta x} \\&= \frac{C(x_2) - C(x_1)}{x_2-x_1} \\&= \frac{11800 - 11000}{300 - 200} \\&= \frac{800}{100} \\&= 8\end{align} \]
**So, on average, the cost of producing a toy between the \(200^{th}\) toy built and the \(300^{th}\) toy built is \(\$8\) per toy**.

- You can use derivatives to describe simple economics like change in cost, revenue, and profit.
**Cost**is the amount of money a company must spend to produce a certain commodity.**Revenue**is the amount of money a company obtains from selling a certain commodity.**Profit**is the amount of money pocketed by a company.- It is calculated by subtracting the total cost to the company spent on producing the commodity from the total revenue obtained from selling the commodity.
- \(P(x) = R(x) - C(x)\)

- Marginal cost can be used to predict the cost of producing one more item.
**Marginal cost**is the derivative of a company's cost function.

- Marginal revenue can be used to predict the revenue earned by selling one more item.
**Marginal revenue**is the derivative of a company's revenue function.

- Marginal profit can be used to predict the profit gained by producing and then selling one more item.
**Marginal profit**is the derivative of a company's profit function.

As cost increases, more revenue is needed in order to produce a profit.

More about Cost and Revenue

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