Find revenue, maximum revenue?

Question

A manufacturer of tablet computers, after extensive research established the following price-demand, and cost functions:

$p(x)= 360-20x$

$c(x)= 300+95x$

where $p(x)$ is the wholesale price in dollars at which $x$ million tablet computers can be sold. The cost $c(x)$ is in millions of dollars. The domain of each function is $1 < x < 15$.

A) Find the revenue function $R(x)$ and sketch the graph of $R(x)$.

B) Find the output which will produce the maximum revenue. What is the maximum revenue? What is the wholesale price, per tablet computer, that produces the maximum revenue?

C) For what outputs will a loss occur? For what outputs will a profit occur?

Answer 1

First of all we know that Total Revenue = The number of Units you Sell * The price of each unit.

So we write: Revenue = Price * Quantity Or $R(X) = P(x)*x$

So yep, your revenue function is just $R(X) = 360x -20x^2$

Just as a point of interest: Note that revenue and profit are two different things. We say revenue is the total turnover (or price times quantity as I just said) and profit is Total Revenue - Total Cost. We'll get to this later.

Second, your maximum revenue: This occurs at the point where reveue is no longer growing clearly if the revenue is no longer going up, then it's at a maximum value.

To find this, we look for the function is no longer growing, in fact we look for where the function has stopped moving at all, because then we can be sure it is not going down either.

So we take the derivative of the revenue function and set it equal to 0.

$R'(x) = 360 - 40x = 0 \implies x = 360/40 = 9$ So the maximum revenue is achieved when you sell 9 units. We call this derivative function the marginal revenue, and what we just did is called taking first order conditions.

I'll leave the last part to you.

As a hint, note that Total Profit = Total Revenue - Total Cost. We usually denote profit with a pi sign:

$\pi(x) = TR(x) - TC(x)$ To find total cost, do exactly the same as we did above with total revenue.

Answer 2

This is just an addition to my comment. As André Nicolas and I showed you, you can obtain your revenue function $R(x)$ by multiplying price $p(x)$ by the quantity $x$

$$R(x)=p(x)\cdot x=(360-20x)x=360x-20x^2$$

Graph:

b) you should be able to solve with some elementary calculus and for c) you should try to find $x$'s for which Cost $>$ Revenue

Edit:

For b), setting $$p(x)=c(x)$$

will not give you the maximum revenue. You can even see this by looking at the graph.

The break-even point $p(x)=c(x)$ occurs after maximum revenue has been reached.