Solving problem for optimal price (maximize profit) *attempt inside*

Question

Let a demandfunction be defined as $D(p)=B-bp$, where $b,B>0$. A firm has some production cost, $c$, and can set the price $p$ under the constrain given by the Demand.

• What is the optimal price?
• Is the priceelasticity higher or lower than $1$ (absolute value)
• How does price depend on $b$?

How would one do such a maximization problem?

My attempt

The elasticity would be given by

$\frac{p}{q}\cdot(-b)$ since $D'(p)=-b$

Then the problem is \begin{equation*} max_p(p-c)D'(p) \end{equation*}

which then follows that

\begin{equation*} D(p)+(p-c)D'(p)=0 \Leftrightarrow \end{equation*}

\begin{equation*} D(p)\left ( 1+(p-c)\frac{\epsilon}{p} \right )=0\Leftrightarrow \end{equation*}

and then the optimal price $p^*$ \begin{equation*} p^*=\frac{\epsilon}{1+\epsilon}c=\frac{\left | \epsilon \right |}{\left | \epsilon \right |-1}c=\frac{\left | \frac{-bp}{q} \right |}{\left | \frac{-bp}{q} \right |-1}c=\frac{c\left | b \right |\left | p \right |}{\left | b\right |\left |p \right |-\left | q \right |} \end{equation*}

and then we notice that

\begin{equation*} \left | \epsilon \right |>1 \end{equation*}

...............................................................................................

I feel like this is not correct nor the right approach. Help appreciated.

Answer 1

Start with setting up profit equation:

$$\max_p \pi = (p-c)D(q)$$

Then substitute demand in:

$$ \pi = (p-c)(B-bp)$$

Take derivative of profit and equate it to 0:

$$\frac{d \pi}{dp}=B-2pb + cb=0$$

Now solve for optimum price:

$$p^*= \frac{B+cb}{2b}$$

There is also alternative way how to get to the same solution that uses elasticity of demand.

There is well known monopoly optimal pricing rule (see Peitz and Belleflamme Industrial Organization: Markets and Strategies) says that optimal monopoly prices are given by:

$$\frac{p-mc}{p}=-\frac{1}{El}$$

In your case marginal costs (mc) are $c$, and your elasticity of demand is actually:

$$EL= -\frac{bp}{q}= -\frac{bp}{B-bp}$$

In your attempt you forgot to substitute for Q in your elasticity.

Plugging these into optimal pricing rule gives us:

$$\frac{p-c}{p}=-\frac{1}{\left( -\frac{bp}{B-bp} \right) }$$

Which when solved for $p$ gives us again:

$$p^*= \frac{B+cb}{2b}$$

Both approaches give the same solution - the first one is quicker in this case in my opinion.

Now regarding what we say about price and parameter $b$, we can just rearrange the optimal price formula as:

$$p^*= \frac{B}{2b}+\frac{c}{2}$$

So clearly we can see that when $b$ parameter increases optimum price decreases.

We can also prove it more formally by inspecting derivative of $p^*$ with respect to $b$ which gives us:

$$ \frac{d p^*}{d b} = -\frac{B}{2b^2}$$

Since the derivative is negative we confirmed that the price varies negatively with $b$.

Next, per request in comment on what is elasticity at equilibrium price we can calculate that just by substituting the equilibrium price into elasticity which we already calculated above:

$$EL(p^*)= -\frac{b p^* }{B-bp^*}= -\frac{b \left( \frac{B+cb}{2b} \right)}{B-b \left( \frac{B+cb}{2b} \right)} \\= - \frac{B+bc}{B-bc}$$

The last expression will be bigger than 1 in absolute value because the numerator will be larger than the denominator ($B+cb>B-bc$) and absolute value will make the value positive.