<p><span class="math-container">$\textbf{In short}$</span></p>

<p>No, there are a number of issues. see section 1 below for the solution, and section 2 for some intuition behind why this process works. </p>

<p><span class="math-container">$\textbf{section 1 - Potential Problems}$</span></p>

<p><span class="math-container">$\underline{Objective \ function \ setup}$</span></p>

<p>Your objective function is the function that you are interested in, in your question it is profit.</p>

<p>It appears that there is a wrong sign on your cost function.  Costs take away from profits, they don't add to profits.  i.e. <span class="math-container">$$ \pi = revenue \color{red}{-} costs, \ where\ \pi \ is  \ profit$$</span></p>

<p><span class="math-container">$\underline {Inverse \ Demand, y(p)=a-bp}$</span></p>

<p>Another issue is that in a typical monopoly question, output(i.e. y) is a function of price (i.e. <span class="math-container">$y(p) $</span>) and inversely price is a function of output (i.e. <span class="math-container">$p(y) $</span>).</p>

<p>This occurs because there is one firm in the market, and thus the firm faces the entire market demand. This means that the firm <span class="math-container">$\textbf{is not a price taker}$</span>, but every price it sets has an effect on the quantity demanded, and results in the firm's price <span class="math-container">$\textbf{being the demand curve}$</span> (called the inverse demand curve). </p>

<p>For example. </p>

<p>Suppose we had a market with consumers having a demand function of 
<span class="math-container">$$y_d(p)=a-bp $$</span></p>

<p>Thus, this would result in the  objective function</p>

<p><span class="math-container">$$ max_{p} \ \pi = p\times y_d(p) - C(y) $$</span></p>

<p>One issue with this is that your costs function is now a function of price, because it is a function of output which is a function of price. </p>

<p>i.e. <span class="math-container">$ C(y(p)) $</span> ,  to solve this you would need to a.) implicitly derive or b.) substitute y(p) for it's function (e.g <span class="math-container">$C(a-bp)$</span>, if available). I think young's theorem might have something to say on this. </p>

<p><span class="math-container">$\textbf{Now we have a valid profit function for a monopoly!!}$</span></p>

<p><span class="math-container">$$ max_{p} \ \pi = \color{red}{ p \times y_d(p)} - \color{blue}{C(a-bp)} $$</span></p>

<p>now we can optimise; </p>

<p><span class="math-container">$\underline{First \ order \ conditions}$</span></p>

<p><span class="math-container">$$ \frac{\partial \pi}{\partial p} = \color{red}{(1\times y_d(p) + p \times y_d'(p) )} - \color{blue}{\frac{d}{dp}(C(a-bp))}$$</span>
NB, you have to apply chain rule to derive <span class="math-container">$ p \times y_d(p) $</span> hence the term in red on the left. </p>

<p>Set <span class="math-container">$ \frac{\partial \pi}{\partial p} = 0$</span> and solve for p.</p>

<p>By setting this equal to zero, <span class="math-container">$\color{red}{marginal \ revenue}$</span> equals <span class="math-container">$\color{blue}{marginal \ cost}$</span> with respect(in terms of) to p. Thus, this is where profit is maximised. </p>

<p><span class="math-container">$ \textbf{voila} $</span></p>

<p>Now you have the optimal price</p>

<p><span class="math-container">$\textbf{section 2 - Intuition}$</span></p>

<p>This works because you are using the process of optimisation on the profit function, to find the best price. </p>

<p>The process of taking the max or min of a function is referred to as optimisation in mathematics. It helps find the largest (max) or smallest (min) value in the objective function (profit in your case). This works because taking the derivative of a function essentially treats your objective function as a curve with respect to the variable you are interested in, and in taking the derivative allows you to find the peaks(max values) and the troughs(min values) of your objective function. Hence in your example, you are treating profit as a curve and finding the point on that curve with the largest p value. </p>

CC BY-SA 4.0
2018-10-20T00:56:17.753Z
2018-10-20T00:56:17.753Z
J.N
<p>How do I find the optimal price for a monopolist given the monopolist's cost function and market demand?</p>

<p>I have <span class="math-container">$Profit(y) = p*y + C(y)$</span> where <span class="math-container">$p$</span> is price, <span class="math-container">$y$</span> is output, and <span class="math-container">$C(y)$</span> is total cost.</p>

<p>Then I set the derivative of <span class="math-container">$Profit$</span> (with respect to y) equal to 0 and solve for <span class="math-container">$p$</span>. Is that correct?</p>

CC BY-SA 4.0
2018-10-20T00:54:56.193Z
If $C(y)$ is the cost function, why are you adding it to profits?  Shouldn't you be subtracting it?  As stands, this suggests that as costs increase, you make more profit.  I take it that $y$ is the quantity of output.  $profit = p*y - C(y)$ would make more sense as an equation.  Also, what is your demand function?  You say that you have one, but you don't include it.  It should look something like $y = D(p)$ only with the right side an actual function on $p$.  Consider the possibility that marginal profit might be zero at multiple points.
Brythan
<microeconomics><monopoly>
How do I find optimal price or maximise profit in a monopolistic market?
Guest
# How do I find optimal price or maximise profit in a monopolistic market?

(Asked by: username_0 on 2018-10-19)

How do I find the optimal price for a monopolist given the monopolist's cost function and market demand?

I have $Profit(y) = p\*y + C(y)$ where $p$ is price, $y$ is output, and $C(y)$ is total cost.

Then I set the derivative of $Profit$ (with respect to y) equal to 0 and solve for $p$. Is that correct?

**Tags:** microeconomics, monopoly

> **Comments:**
> **username_1**: If $C(y)$ is the cost function, why are you adding it to profits?  Shouldn't you be subtracting it?  As stands, this suggests that as costs increase, you make more profit.  I take it that $y$ is the quantity of output.  $profit = p*y - C(y)$ would make more sense as an equation.  Also, what is your demand function?  You say that you have one, but you don't include it.  It should look something like $y = D(p)$ only with the right side an actual function on $p$.  Consider the possibility that marginal profit might be zero at multiple points.  

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## 1 answers

### Answer by username_2 on 2018-10-20. Score: 1

$\textbf{In short}$

No, there are a number of issues. see section 1 below for the solution, and section 2 for some intuition behind why this process works.

$\textbf{section 1 - Potential Problems}$

$\underline{Objective \ function \ setup}$

Your objective function is the function that you are interested in, in your question it is profit.

It appears that there is a wrong sign on your cost function. Costs take away from profits, they don't add to profits. i.e. $$ \pi = revenue \color{red}{-} costs, \ where\ \pi \ is \ profit$$

$\underline {Inverse \ Demand, y(p)=a-bp}$

Another issue is that in a typical monopoly question, output(i.e. y) is a function of price (i.e. $y(p) $) and inversely price is a function of output (i.e. $p(y) $).

This occurs because there is one firm in the market, and thus the firm faces the entire market demand. This means that the firm $\textbf{is not a price taker}$, but every price it sets has an effect on the quantity demanded, and results in the firm's price $\textbf{being the demand curve}$ (called the inverse demand curve).

For example.

Suppose we had a market with consumers having a demand function of
$$y\_d(p)=a-bp $$

Thus, this would result in the objective function

$$ max\_{p} \ \pi = p\times y\_d(p) - C(y) $$

One issue with this is that your costs function is now a function of price, because it is a function of output which is a function of price.

i.e. $ C(y(p)) $ , to solve this you would need to a.) implicitly derive or b.) substitute y(p) for it's function (e.g $C(a-bp)$, if available). I think young's theorem might have something to say on this.

$\textbf{Now we have a valid profit function for a monopoly!!}$

$$ max\_{p} \ \pi = \color{red}{ p \times y\_d(p)} - \color{blue}{C(a-bp)} $$

now we can optimise;

$\underline{First \ order \ conditions}$

$$ \frac{\partial \pi}{\partial p} = \color{red}{(1\times y\_d(p) + p \times y\_d'(p) )} - \color{blue}{\frac{d}{dp}(C(a-bp))}$$
NB, you have to apply chain rule to derive $ p \times y\_d(p) $ hence the term in red on the left.

Set $ \frac{\partial \pi}{\partial p} = 0$ and solve for p.

By setting this equal to zero, $\color{red}{marginal \ revenue}$ equals $\color{blue}{marginal \ cost}$ with respect(in terms of) to p. Thus, this is where profit is maximised.

$ \textbf{voila} $

Now you have the optimal price

$\textbf{section 2 - Intuition}$

This works because you are using the process of optimisation on the profit function, to find the best price.

The process of taking the max or min of a function is referred to as optimisation in mathematics. It helps find the largest (max) or smallest (min) value in the objective function (profit in your case). This works because taking the derivative of a function essentially treats your objective function as a curve with respect to the variable you are interested in, and in taking the derivative allows you to find the peaks(max values) and the troughs(min values) of your objective function. Hence in your example, you are treating profit as a curve and finding the point on that curve with the largest p value.


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