# Applications to Economics

## Related calculator: Online Derivative Calculator with Steps

Suppose that some factory produces $$$x$$$ units of good. Let's denote cost of producing $$$x$$$ units of good by $$${C}{\left({x}\right)}$$$.

**Cost function** $$${C}{\left({x}\right)}$$$ is the cost of producing $$$x$$$ units of certain product.

**Marginal cost** is the rate of change of cost with respect to $$$x$$$. In other words marginal cost function is the derivative of cost function.

The **average cost** function $$${c}{\left({x}\right)}=\frac{{{C}{\left({x}\right)}}}{{x}}$$$ represents the cost per unit when $$${x}$$$ units are produced.

Now let's find stationary points of c(x) using quotient rule:

$$${c}'{\left({x}\right)}=\frac{{{C}'{\left({x}\right)}{x}-{C}{\left({x}\right)}{x}'}}{{{{x}}^{{2}}}}$$$ or $$${c}'{\left({x}\right)}=\frac{{{C}'{\left({x}\right)}{x}-{C}{\left({x}\right)}}}{{{{x}}^{{2}}}}$$$.

Now, $$${c}'{\left({x}\right)}={0}$$$ when $$${C}'{\left({x}\right)}{x}-{C}{\left({x}\right)}={0}$$$.

Therefore, $$${C}'{\left({x}\right)}=\frac{{{C}{\left({x}\right)}}}{{x}}={c}{\left({x}\right)}$$$.

We can formulate this result as follows:

**If the average cost is a minimum, then marginal cost =average cost.**

**Example 1.** A company estimates that the cost (in dollars) of producing x items is $$${C}{\left({x}\right)}={2800}+{2}{x}+{0.002}{{x}}^{{2}}$$$.

- Find the cost, average cost, and marginal cost of producing 1000 items, 2000 items, and 3000 items.
- At what production level will the average cost be lowest, and what is this minimum average cost?

The average cost function is $$${c}{\left({x}\right)}=\frac{{{C}{\left({x}\right)}}}{{x}}={\left({2800}+{2}{x}+{0.002}{{x}}^{{2}}\right)}=\frac{{2800}}{{x}}+{2}+{0.002}{x}$$$.

Marginal cost is derivative of cost function: $$${C}'{\left({x}\right)}={2}+{0.004}{x}$$$.

We use these expressions to fill in the following table, giving the cost, average cost, and marginal cost (in dollars, or dollars per item, rounded to the nearest cent).

x | C(x) | c(x) | C'(x) |

500 | 4300 | 8.6 | 4 |

1000 | 6800 | 6.8 | 6 |

2000 | 14800 | 7.4 | 10 |

To minimize average cost we must have $$${C}'{\left({x}\right)}={c}{\left({x}\right)}$$$ or $$$\frac{{2800}}{{x}}+{2}+{0.002}{x}={2}+{0.004}{x}$$$.

This can be simplified to $$$\frac{{2800}}{{x}}={0.002}{x}$$$ or $$${{x}}^{{2}}={1400000}$$$ or $$${x}=\sqrt{{{1400000}}}\approx{1183}$$$.

To see that this production level actually gives a minimum, we note that $$${c}'{\left({x}\right)}=-\frac{{2800}}{{{x}}^{{2}}}+{0.002}$$$ and $$${c}''{\left({x}\right)}=\frac{{5600}}{{{x}}^{{3}}}>{0}$$$ and therefore average cost function is concave upward on its entire domain.

The minimum average cost is $$${c}{\left({1183}\right)}=\frac{{2800}}{{1183}}+{2}+{0.002}\cdot{1183}\approx\${6.73}$$$ per item.

Now let's consider marketing.

Let $$${p}{\left({x}\right)}$$$ be the price per unit that the company can charge if it sells $$${x}$$$ units.

Then $$${p}$$$ is called the **demand function** (or price function). Naturally it is decreasing function, because the more units are produced the lesser price for them.

If $$${x}$$$ units are sold and the price per unit is $$${p}{\left({x}\right)}$$$, then the **total revenue** is $$${R}{\left({x}\right)}={x}{p}{\left({x}\right)}$$$ and $$${R}$$$ is called the **revenue function** (or sales function).

The derivative of the revenue function is called the **marginal revenue** function and is the rate of change of revenue with respect to the number of units sold.

If $$${x}$$$ units are sold then the **total profit** is $$${P}{\left({x}\right)}={R}{\left({x}\right)}-{C}{\left({x}\right)}$$$ and $$${P}$$$ is called **profit function.**

The **marginal profit** function is $$${P}'$$$, the derivative of the profit function.

In order to maximize profit we look for the critical numbers of $$${P}$$$, that is, the numbers where the marginal profit is 0.

$$${P}'{\left({x}\right)}={R}'{\left({x}\right)}-{C}'{\left({x}\right)}$$$. $$${P}'{\left({x}\right)}={0}$$$ when $$${R}'{\left({x}\right)}={C}'{\left({x}\right)}$$$.

So, **if the profit is a maximum, then marginal revenue=marginal cost.**

To ensure that this condition gives a maximum we could use the Second Derivative Test. Note that $$${P}''{\left({x}\right)}={R}''{\left({x}\right)}-{C}''{\left({x}\right)}<{0}$$$ when $$${R}''{\left({x}\right)}<{C}''{\left({x}\right)}$$$ and this condition says that the rate of increase of marginal revenue is less than the rate of increase of marginal cost. Thus, the profit will be a maximum when $$${R}'{\left({x}\right)}={C}'{\left({x}\right)}$$$ and $$${R}''{\left({x}\right)}<{C}''{\left({x}\right)}$$$.

**Example 2.** The cost, in dollars, of producing yards of a certain fabric is $$${C}{\left({x}\right)}={1200}+{12}{x}-{0.1}{{x}}^{{2}}+{0.0005}{{x}}^{{3}}$$$ and the company finds that if it sells yards, it can charge $$${p}{\left({x}\right)}={29}-{0.00021}{x}$$$ dollars per yard for the fabric. Find the production level for maximum profit.

The revenue function $$${R}{\left({x}\right)}={x}{p}{\left({x}\right)}={29}{x}-{0.00021}{{x}}^{{2}}$$$.

Marginal revenue is $$${R}'{\left({x}\right)}={29}-{0.00042}{x}$$$.

Marginal cost function is $$${C}'{\left({x}\right)}={12}-{0.2}{x}+{0.0015}{{x}}^{{2}}$$$.

To maximize profit must hold $$${R}'{\left({x}\right)}={C}'{\left({x}\right)}$$$ or $$${29}-{0.00042}{x}={12}-{0.2}{x}+{0.0015}{{x}}^{{2}}$$$.

Solving, we get $$${x}\approx{192}$$$.

To check that this gives a maximum we compute the second derivatives:

$$${R}''{\left({x}\right)}=-{0.00042}$$$ and $$${C}''{\left({x}\right)}=-{0.2}+{0.003}{x}$$$.

We want to find when $$${R}''{\left({x}\right)}<{C}''{\left({x}\right)}$$$: $$$-{0.00042}<-{0.2}+{0.003}{x}$$$ or $$${x}>{66.53}$$$ (in particular $$${R}''{\left({x}\right)}<{C}''{\left({x}\right)}$$$ for $$${x}={192}$$$).

Therefore, a production level of 192 yards of fabric will maximize the profit.

**Example 3.** A baseball team plays in a stadium that holds 55000 spectators. With ticket prices at $10, the average attendance had been 27000. When ticket prices were lowered to $8, the average attendance rose to 33000. Find the demand function, assuming that it is linear. How should ticket prices be set to maximize revenue?

Let $$${x}$$$ be number of attendants, then growing of price from $8 to $10 (by $10-$8=$2) will reduce number of attendants by 33000-27000=6000. Therefore, assuming demand function is linear, each increase in price by $1 will reduce number of attendants by $$$\frac{{6000}}{{2}}={3000}$$$.

Or each additional attendant will reduce price by $$$\$\frac{{1}}{{3000}}$$$.

So, demand function is $$${p}{\left({x}\right)}={10}-\frac{{1}}{{3000}}{\left({x}-{27000}\right)}={19}-\frac{{x}}{{3000}}$$$.

Revenue function is $$${R}{\left({x}\right)}={x}{p}{\left({x}\right)}={19}{x}-\frac{{1}}{{3000}}{{x}}^{{2}}$$$.

$$${R}'{\left({x}\right)}={19}-\frac{{1}}{{1500}}{x}$$$, therefore $$${R}'{\left({x}\right)}={0}$$$ when $$${x}={28500}$$$. This value of $$${x}$$$ gives an absolute maximum by the First Derivative Test (or simply by observing that the graph of is a parabola that opens downward).

Therefore, ticket price should be $$${p}{\left({28500}\right)}={10}-\frac{{1}}{{{3000}}}{\left({28500}-{27000}\right)}=\${9.5}$$$.

**Example 4.** A store has been selling 200 compact disc players a week at $350 each. A market survey indicates that for each $10 rebate offered to buyers, the number of units sold will increase by 20 a week. Find the demand function and the revenue function. How large a rebate should the store offer to maximize its revenue?

If $$${x}$$$ is the number of CD players sold per week, then the weekly increase in sales is $$${x}-{200}$$$. For each increase of 20 players sold, the price is decreased by $10. So for each additional player sold the decrease in price will be $$$\frac{{10}}{{20}}$$$ and the demand function is $$${p}{\left({x}\right)}={350}-\frac{{10}}{{20}}{\left({x}-{200}\right)}={450}-\frac{{1}}{{2}}{x}$$$.

The revenue function is $$${R}{\left({x}\right)}={x}{p}{\left({x}\right)}={450}{x}-\frac{{1}}{{2}}{{x}}^{{2}}$$$.

Since $$${R}'{\left({x}\right)}={450}-{x}$$$, we see that $$${R}'{\left({x}\right)}={0}$$$ when $$${x}={450}$$$.

This value of $$${x}$$$ gives an absolute maximum by the First Derivative Test.

The corresponding price is $$${p}{\left({450}\right)}={450}-\frac{{1}}{{2}}{450}={225}$$$ and the rebate is $$${350}-{225}={125}$$$. Therefore, to maximize revenue the store should offer a rebate of $125.