**components of economic cost.**We are going to use that idea to solve these three complex problems.

__Example 1__Rachel company manufactures and sells books. Rachel knows that the cost of each book comes from the cost function $C(Q)=5000+40Q$. In addition to that, Her price is derived from this function $P(Q)=300-2Q$. Find

1. The quantity that maximizes profit

2. The profit-maximizing price

3. Find the maximum profit.

**Solution:**

$MC$, as explained in this post, is the partial derivative of the total cost. Hence, we take the derivative of the cost function.

$C(Q)=5000+40Q$

$C'(Q)=40$

$MC(Q)=40$

Marginal revenue is the partial derivatives of revenue. You know that Total revenue is:

$TR=P\times Q$

By substitution

$TR=(300-2Q)Q$

$TR=300Q-2Q^2$.....👈revenue function

Now, we can easily get marginal revenue by taking the derivative of the revenue function

$R(Q)=300Q-2Q^2$

Taking the derivative

$R'(Q)=300-4Q$

$MR=300-4Q$

Equating MR to MC

$300-4Q=40$

$300-40=4Q$

$260=4Q$

$\frac{260}{4}=\frac{4Q}{4}$

$65=Q$

$Q=65$

**Therefore, the 65th quantity maximizes profit.**

To determine the price that maximizes profit, we substitute the value of $Q$ in the price function($P(Q)=300-2Q$)

$P(65)=300-2(65)$

$=300-130$

$=170$

**Hence, the price that maximizes profit is €170.**

To determine the maximum profit, we subtract the cost from revenue. Remember that

Profit=Revenue–cost

$P=300Q-2Q^2-(5000+40Q)$

By substitution

$P=300(65)-2(65)^2-(5000+40(65))$

$P=19500-8450-(5000+2600)$

$P=3450$

**Therefore, the maximum profit that Rachel company can earn is €3450**

To a more complex problem now.

__Example 2__A firm has the following demand and the average cost-functions:

$x=480-20p$ and $AC=10+\frac{x}{15}$

Determine the profit-maximizing output and price of the monopolist.

**Solution:**

To get the price function, we make p the subject of the formula in the demand function.

$x=480-20p$

$20p=480-x$

$p=\frac{480-x}{20}$

Total revenue is

**price times quantity**. Hence,$TR=\left(\frac{480-x}{20}\right)x$

$TR=\frac{480x-x^2}{20}$

Marginal revenue is the

**first derivative of the revenue function**, hence$TR=\frac{480x-x^2}{20}$

$MR=\frac{480-2x}{20}$

$MR=24-\frac{x}{10}$

Now, to the cost function

$AC=10+\frac{x}{15}$

The total cost can be derived by

**multiplying average cost and quantity,**hence$TC=\left(10+\frac{x}{15}\right)x$

$TC=10x+\frac{x^2}{15}$

Marginal cost is the

**derivative of the cost function**, hence,$MC=10+\frac{2x}{15}$

A firm maximizing profit where MR=MC

$24-\frac{x}{10}=10+\frac{2x}{15}$

$\frac{240-x}{10}=\frac{150+2x}{15}$

$3600-15x=1500+20x$

$3600-1500=20x+15x$

$2100=35x$

$x=60$

To get the price that maximizes profit, we simply insert the value of x in the price function

$p=\frac{480-x}{20}$

$p=\frac{480-60}{20}$

$p=21$

Hence the quantity and price that maximizes profit are 60 and 21 respectively

Example 3

Given the cost function of a perfectly competitive firm is $C=0.01Q^3-0.3Q^2+5Q+1$. And given also that the marginal revenue is given as €2. Find

1. The quantity that maximizes profit

2. What is the value of the firm average variable cost at the profit-maximization level

3. What is the firm AC at that quantity

4. Should the firm continue production

5. What is the value of the firm's fixed cost?

**Solution:**

1. To maximize profit, the golden rule is to produce where MR=MC.

We know that $MR=2$, But we do not know the value of $MC$.

But, we know that $MC$ is the partial derivative of the total cost. Therefore, we will take the derivative of the cost function.

$C(Q)=0.01Q^3-0.3Q^2+5Q+1$.

Taking the derivative

$C'(Q)=0.03Q^2-0.6Q+5$

$MC(Q)=0.03Q^2-0.6Q+5$

Now let's equate it to marginal revenue

$0.03Q^2-0.6Q+5=2$

$0.03Q^2-0.6Q+5-2=0$

$0.03Q^2-0.6Q+3=0$

$0.03Q^2-0.6Q+3=0$

For simplicity, let's multiply through by $100$

$0.03Q^2\times100-0.6Q\times100+3\times100=0\times100$

$3Q^2-60Q+300=0$

By factorization

$3Q^2-30Q-30Q+300=0$

$3Q(Q-10)-30(Q-10)=0$

$(3Q-30)(Q-10)=0$

$3Q=30$ or $Q=10$

$\frac{3Q}{3}=\frac{30}{3}$ or $Q=10$

$Q=10$ or $Q=10$

**Therefore, the tenth quantity maximizes profits.**

2. To derive the average variable cost, we divide the variable cost by quantity.

**But, what is the variable cost?**

Variable cost is the cost of input that varies directly with the quantity produced.

Looking at the cost function given above

Ths $0.01Q^3-0.3Q^2+5Q$ can all be varied. Hence, the variable cost function is $0.01Q^3-0.3Q^2+5Q$

Therefore, to obtain average variable cost. We would divide this variable cost by quantity like this

$AVC=\frac{0.01Q^3-0.3Q^2+5Q}{Q}$

Since the quantity is $10$, let's substitute it accordingly.

$AVC=\frac{0.01(10)^3-0.3(10)^2+5(10)}{10}$

$AVC=\frac{0.01(1000)-0.3(100)+50}{10}$

$AVC=\frac{10-30+50}{10}$

$AVC=3$

**Thus, the average variable cost of producing the quantity that maximizes profit is €3**

3. The average cost can easily be computed by dividing total cost by quantity

Since the total cost is given as $C=0.01Q^3-0.3Q^2+5Q+1$. We will divide this function by $Q$. Accordingly,

$AC=\frac{0.01Q^3-0.3Q^2+5Q+1}{Q}$

The quantity is $10$, let's substitute it into the equation

$AC=\frac{0.01(10)^3-0.3(10)^2+5(10)+1}{10}$

$AC=\frac{10-30+50+1}{10}$

$AC=3.1$

**Therefore, the average cost of producing the 10 units is €3.1.**

4. Now, to the question of whether the firm should shut down or not?.

All firms will shut down at the point where Price is lesser than variable cost. Since the firm is a perfect competitor, its marginal revenue(€2) will be equal to the price.

The variable cost can easily be computed by multiplying quantity and average variable cost. Accordingly,

$VC=3.0\times 10$

$VC=30$

**Since variable cost(€30) is greater than price(€2). The firm should, as a matter of fact, and as a matter of urgency, shut down immediately. This is because the price it is receiving is not even covering its variable cost let alone its fixed cost**.

5. Fixed cost is easily obtainable from the cost function: $C=0.01Q^3-0.3Q^2+5Q+1$

As can be seen, only $1$ is constant in the cost function. Hence, $1$ is the fixed cost.

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