<p>Here are some hints. Either solve the profit function in terms of $p$, substituting out all references to $q$ (recall that $D(p) = q$), or solve the profit function in terms of $q$, substituting out all references to $p$. Then you'll find that the profit function is concave and the differentiated with respect to $p$ in the first case or $q$ in the second case will give you the profit maximizing solution.</p>

<p>EDIT: Yeah, $a = 2q + 10$ looks correct. Just solve the equation for $q$ and the interpretation is clear. That is, the quantity produced is $q = (a - 10)/2$. If you want to know the price, then just substitute $q$ in for $D(p)$ like so: $a - p = D(p) = q = (a - 10)/2$.</p>

CC BY-SA 3.0
2015-01-19T23:26:39.153Z
I calculated the value of $a$: $a = 2q + 10$. But how am I supposed to interpret this? In this case, $a$ is just the $b$ in a standard 7th grade $y = mx + b$ formula, so it corresponds to the y intercept in Cartesian coordinates. But $q$ can't take negative values
and presumably must be integers. Therefore, for $a < 30$ he wouldn't produce, but for $a \geq 30$ he would. Does that make any sense?

Economics confuses me sometimes to the point where I forget how to add and subtract because I can't tell what I should be performing the operations on.
Stan Shunpike
CC BY-SA 3.0
2015-01-19T23:29:10.713Z
You don't need to give me the correct answer. Simply telling me if my deductions are true based on the premises I gave would be enough. That would leave it to me to decide if my premises are correct.
Stan Shunpike
CC BY-SA 3.0
2015-01-19T23:04:59.063Z
2015-01-19T23:59:23.440Z
2015-01-19T23:59:23.440Z
jmbejara
<p>I was given a word problem. No formulas. So I set up the following equations: </p>

<p>Demand Function 
\begin{equation}
D(p) = a - p
\end{equation} 
Cost Function: 
\begin{equation}
c(q) = 9 + 10q
\end{equation} 
Marginal Cost 
\begin{equation}
MC = 10
\end{equation} </p>

<p>I am supposed to find how much the man in the problem would produce for any value of $a$. But, without a production function, I am unsure how to determine how much he will produce. </p>

<p>Specifically, suppose I write </p>

<p>\begin{equation}
R = pD(p) 
\end{equation}  </p>

<p>I can't then differentiate R by q to find marginal revenue. </p>

<p>So how do I know how much he would produce? </p>

<supply-and-demand><production-function>
How do you find marginal revenue if you don't have a production function?
Stan Shunpike
# How do you find marginal revenue if you don't have a production function?

(Asked by: username_0 on 2015-01-19)

I was given a word problem. No formulas. So I set up the following equations:

Demand Function
\begin{equation}
D(p) = a - p
\end{equation}
Cost Function:
\begin{equation}
c(q) = 9 + 10q
\end{equation}
Marginal Cost
\begin{equation}
MC = 10
\end{equation}

I am supposed to find how much the man in the problem would produce for any value of $a$. But, without a production function, I am unsure how to determine how much he will produce.

Specifically, suppose I write

\begin{equation}
R = pD(p)
\end{equation}

I can't then differentiate R by q to find marginal revenue.

So how do I know how much he would produce?

**Tags:** supply-and-demand, production-function

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## 1 answers

### Answer by username_1 on 2015-01-19. Score: 2

Here are some hints. Either solve the profit function in terms of $p$, substituting out all references to $q$ (recall that $D(p) = q$), or solve the profit function in terms of $q$, substituting out all references to $p$. Then you'll find that the profit function is concave and the differentiated with respect to $p$ in the first case or $q$ in the second case will give you the profit maximizing solution.

EDIT: Yeah, $a = 2q + 10$ looks correct. Just solve the equation for $q$ and the interpretation is clear. That is, the quantity produced is $q = (a - 10)/2$. If you want to know the price, then just substitute $q$ in for $D(p)$ like so: $a - p = D(p) = q = (a - 10)/2$.

> **Comments:**
> **username_0**: I calculated the value of $a$: $a = 2q + 10$. But how am I supposed to interpret this? In this case, $a$ is just the $b$ in a standard 7th grade $y = mx + b$ formula, so it corresponds to the y intercept in Cartesian coordinates. But $q$ can't take negative values and presumably must be integers. Therefore, for $a < 30$ he wouldn't produce, but for $a \geq 30$ he would. Does that make any sense?  Economics confuses me sometimes to the point where I forget how to add and subtract because I can't tell what I should be performing the operations on.  
> **username_0**: You don't need to give me the correct answer. Simply telling me if my deductions are true based on the premises I gave would be enough. That would leave it to me to decide if my premises are correct.  

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