# Equilibrium Revenue

• Oct 21st 2007, 07:34 PM
aphan19
Equilibrium Revenue
Please Help. I have a midterm, and this is an example question of what might be on it.

The manager of a CD store has found that if the price of CD is p(x)=80-x/6, then x CDs will be sold. Find an expression for the total revenue from the sale of x CDs (hint: revenue= demand x price). Use your expression to determine the maximum revenue.
• Oct 21st 2007, 07:55 PM
coe236
Revenue is price*demand, so the price(x) here is 80-x/6 and the demand is x, so the expression would be x*(80-x/6)= 80x - x^2/6. Max revenue, in terms of microeconomics, is usually at equilibrium or the midpoint on the demand curve. If thats given, then you just plug it into the expression
• Oct 21st 2007, 07:55 PM
Jhevon
Quote:

Originally Posted by aphan19
Please Help. I have a midterm, and this is an example question of what might be on it.

The manager of a CD store has found that if the price of CD is p(x)=80-x/6, then x CDs will be sold. Find an expression for the total revenue from the sale of x CDs (hint: revenue= demand x price). Use your expression to determine the maximum revenue.

is the formula: $p(x) = \frac {80 - x}6$ ? if so, you should use parentheses to indicate that.

Revenue = demand * Price, that is, it is the number of items sold times the price each item is sold for.

thus the revenue function is given by:

$R(x) = x \cdot \frac {80 - x}6 = \frac {40}3x - \frac 16 x^2$

this is a parabola, it's maximum occurs at its vertex

the vertex of a parabola, $f(x) = ax^2 + bx + c$ is the point $\left( \frac {-b}{2a}, f \left( \frac {-b}{2a}\right)\right)$

the answer to your last question is the y-coordinate of that formula

can you find it?

(you could also complete the square to get the vertex, which method do you prefer?)
• Oct 21st 2007, 08:05 PM
aphan19
It is not (80-x)/6, it is actually 80-(x/6). Sorry, for not adding the parenthesis in my problem before. If the problem is actually 80-(x/6), now what do I do to solve it?
• Oct 21st 2007, 08:15 PM
Jhevon
Quote:

Originally Posted by aphan19
It is not (80-x)/6, it is actually 80-(x/6). Sorry, for not adding the parenthesis in my problem before. If the problem is actually 80-(x/6), now what do I do to solve it?

do exactly what we said before: multiply through by x, and solve for the vertex of the resulting parabola by whatever method you feel comfortable with