All firms have one objective in mind: to earn the maximum profit possible.

Profit can be defined as the difference between total revenue and total cost.

The profit-maximization rule is used to determine the quantity that maximizes profit.

The profit maximization rule states that a firm maximizes profit at the production level where

marginal revenue is equal to marginal cost.

**Marginal revenue** is the change in total revenue that results from a unit change in quantity sold.

It is the **first derivative of the revenue function.**

**Marginal cost** is the change in total cost that results from a unit change in quantity.

It is the **first derivative of the cost function**.

So, a firm maximizes where:

$$MR=MC$$

__Example 1__

Find the production level that maximizes profit given that cost function is $C(x)=120+2x^2$ and revenue function is $R(x)=100x$

**Solution:**

A firm maximizes profit where MR=MC

Recall that Marginal revenue is the first derivative of the revenue function.

Hence, Marginal revenue is

$$R'(x)=100$$

Recall that marginal cost is the first derivative of the cost function. Hence, marginal cost is:

$$C'(x)=4x$$

Equating MR to MC

$4x=100$

$x=25$

__Example 2__

Given that the cost function is $C(x)=420+3x+4x^2$ and the Revenue function is $R(x)=100x-x^2$ of a monopolist, Find the production level that maximizes profit.

**Solution:**

If the cost function is $420+3x+4x^2$, then marginal cost is

$C'(x)=3+8x$

Because revenue function is $100x-x^2$, marginal revenue is

$R'(x)=100-2x$

Equating MR to MC

$100-2x=3+8x$

$100-3=8x+2x$

$x=9.7$

__Example 3__

Find the production level that maximizes profit given that $R(q)=\frac{80q-q^2}{2}$ and $C(q)=120+2q^2$

Solution

The first derivative of $R(q)=\frac{80q-q^2}{2}$ is $R'(q)=40-q$

The first derivative of $C(q)=120+2q^2$ is $C'(q)=4q$

To optimize profit, the firm must produce where $MR=MC$

$40-q=4q$

$q=8$

__Example 4__

A perfect competitor sells books for N3 each and the cost associated with these books is given by $C(x)=1.25x+0.01x^2+50$. Find the

A. The production level maximizes profit.

B. The average revenue

**Solution:**

A. For a perfect competitor,

$MR=AR=P$

That is, the price of a perfectly competitive firm is always the same as its marginal revenue and its average revenue. Hence, $$MR=3$$

Marginal cost is the first derivative of the cost function. Hence,

$C'(x)=1.25+0.02x$

To maximize profit, $MR=MC$

$3=1.25+0.02x$

$3-1.25=0.02x$

$1.75=0.02x$

$x=\frac{1.75}{0.02}$

$x=87.5$

Hence, the production level that maximizes profit is approximately 88 units.

B. As noted earlier, The marginal revenue of a perfect competitor is always included in its average revenue.

Hence, the average revenue is just N3

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