# Applications to Economics

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Suppose that some factory produces `x` units of good. Let's denote cost of producing `x` units of good by `C(x)`.

**Cost function** `C(x)` 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(x)=(C(x))/x` represents the cost per unit when `x` units are produced.

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

`c'(x)=(C'(x)x-C(x)x')/(x^2)` or `c'(x)=(C'(x)x-C(x))/(x^2)`.

Now, `c'(x)=0` when `C'(x)x-C(x)=0` .

Therefore, `C'(x)=(C(x))/x=c(x)`.

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(x)=2800+2x+0.002x^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(x)=(C(x))/x=(2800+2x+0.002x^2)=2800/x+2+0.002x`.

Marginal cost is derivative of cost function: `C'(x)=2+0.004x` .

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'(x)=c(x)` or `2800/x+2+0.002x=2+0.004x`.

This can be simplified to `2800/x=0.002x` or `x^2=1400000` or `x=sqrt(1400000)~~1183`.

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

The minimum average cost is `c(1183)=2800/1183+2+0.002*1183~~$6.73` per item.

Now let's consider marketing.

Let `p(x)` 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(x)`, then the **total revenue** is `R(x)=xp(x)` 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(x)=R(x)-C(x)` 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'(x)=R'(x)-C'(x)`. `P'(x)=0` when `R'(x)=C'(x)`.

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''(x)=R''(x)-C''(x)<0` when `R''(x)<C''(x)` 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'(x)=C'(x)` and `R''(x)<C''(x)`.

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

The revenue function `R(x)=xp(x)=29x-0.00021x^2`.

Marginal revenue is `R'(x)=29-0.00042x`.

Marginal cost function is `C'(x)=12-0.2x+0.0015x^2`.

To maximize profit must hold `R'(x)=C'(x)` or `29-0.00042x=12-0.2x+0.0015x^2` .

Solving, we get `x~~192`.

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

`R''(x)=-0.00042` and `C''(x)=-0.2+0.003x`.

We want to find when `R''(x)<C''(x)`: `-0.00042<-0.2+0.003x` or `x>66.53` (in particular `R''(x)<C''(x)` 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 `6000/2=3000` .

Or each additional attendant will reduce price by `$1/3000`.

So, demand function is `p(x)=10-1/3000(x-27000)=19-x/3000`.

Revenue function is `R(x)=xp(x)=19x-1/3000 x^2`.

`R'(x)=19-1/1500 x` , therefore `R'(x)=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(28500)=10-1/(3000)(28500-27000)=$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 `10/20` and the demand function is `p(x)=350-10/20(x-200)=450-1/2x`.

The revenue function is `R(x)=xp(x)=450x-1/2x^2`.

Since `R'(x)=450-x`, we see that `R'(x)=0` when `x=450`.

This value of `x` gives an absolute maximum by the First Derivative Test.

The corresponding price is `p(450)=450-1/2 450=225` and the rebate is `350-225=125`. Therefore, to maximize revenue the store should offer a rebate of $125.