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.