# Applications to Economics

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

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

The **marginal cost** is the rate of change of cost with respect to $$$x$$$. In other words, the marginal cost function is the derivative of the 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 the stationary points of $$$c(x)$$$ using the 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 the marginal cost = the 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 the lowest, and what is this minimum average cost?

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

The marginal cost is the derivative of the 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 the 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, the 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 a company can charge if it sells $$${x}$$$ units.

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

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

In order to maximize the 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 the marginal revenue = the marginal cost.**

To ensure that this condition gives a maximum, we could use the second derivative test: $$${P}''{\left({x}\right)}={R}''{\left({x}\right)}-{C}''{\left({x}\right)}<{0}$$$ 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.** Suppose that a company is selling fabrics and its cost function is $$${C}{\left({x}\right)}={1200}+{12}{x}-{0.1}{{x}}^{{2}}+{0.0005}{{x}}^{{3}}$$$. The company finds out that if it sells $$${x}$$$ yards of the fabrics, it can charge $$${p}{\left({x}\right)}={29}-{0.00021}{x}$$$ dollars per yard. Find the production level for maximum profit.

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

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

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

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

Solving it, 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 fabrics will maximize the profit.

Let's work another useful example.

**Example 4.** A baseball team is playing at a stadium that can host 55000 spectators. With the ticket prices at $10, the average attendance had been 27000. When the ticket prices were lowered to $8, the average attendance rose to 33000. Find the demand function assuming that it is linear. What should be the ticket prices to maximize the revenue?

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

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

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

The 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 the absolute maximum by $$${p}{\left({28500}\right)}={10}-\frac{{1}}{{{3000}}}{\left({28500}-{27000}\right)}=\${9.5}$$$.

And now, our final example.

**Example 5.** A store has been selling 200 smartphones a week at $350 each. A market survey indicates that for each $10 rebate offered to the buyers, the number of units sold will increase by 20 a week. Find the demand function and the revenue function. What rebate should the store offer to maximize its revenue?

If $$${x}$$$ is the number of smartphones sold per week, then the weekly increase in sales is $$${x}-{200}$$$. For each increase of 20 telephones sold, the price decreases by $10. So, for each additional phone 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 the 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 its revenue, the store should offer a rebate of $125.