# 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 ${C}{\left({x}\right)}={2800}+{2}{x}+{0.002}{{x}}^{{2}}$.

1. Find the cost, average cost, and marginal cost of producing 1000 items, 2000 items, and 3000 items.
2. 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.