PROFIT-MAXIMIZATION VIA COST FUNCTION AND DEMAND FUNCTION

In my previous post, we played scholarship to 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.

$C(Q)=5000+40Q$
$C'(Q)=40$
$MC(Q)=40$


$TR=P\times Q$
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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