Thread: Revenue function and Marginal revenue function of P

p = 8 - 2LNx

Find the revenue function and the marginal revenue function

i got these:

revenue function R(x)= 6-2log(x)
marginal revenue (r') = -2/x

Is that correct?

and i have no idea how to start these problems based on the same numbersHeadbang)


(b) (2 pts) Find the number (in thousands) of hot dogs that can be sold during one
game such that the marginal revenue is zero.
(c) (2 pts) Find the instantaneous rate of change of the marginal revenue.
(xc) (2 pts) Use part (c) to justify that the answer in part (b) generates the maximum revenue. What is the maximum revenue?

ok now i believe revenue function will be 8x-2xlog(x)
and marginal revenue function should be 6-2log(x)

1. What is p?
2. How did you arrive at your revenue function? ie what operation did you do on p function and why?

P is the profit function

Revenue is X times P function

What is x?
Why do you say revenue is X times the profit function? Is it given in the question?
Is x the quantity?

If you want a help on this question, at least post the full question.

If p(x) is price-demand function, the revenue function is R(x) = xp(x).


"p = 8 − 2 ln x for 5 ≤ x ≤ 50, where x is the
number (in thousands) of hot dogs that can be sold during one game at a price of $p.
(a) (4 pts) Find the revenue function and the marginal revenue function when x
thousand hot dogs can be sold during one game"

Originally Posted by Craftysince1990

ok now i believe revenue function will be 8x-2xlog(x)
and marginal revenue function should be 6-2log(x)

I agree.

ok, thanks.

Could you help me start out any of these Q's?

(b) Find the number (in thousands) of hot dogs that can be sold during one
game such that the marginal revenue is zero.
(c) Find the instantaneous rate of change of the marginal revenue.
(d) Use part (c) to justify that the answer in part (b) generates the maximum revenue. What is the maximum revenue?

for (b), just solve the equation 6-2log(x) for x. I got x=e^3.
I am not sure what is 'instanteneous' rate of change, I may need to look it up first.

Got it!

(c) find second derivative of Revenue function= first derivative of the Marginal revenue function.
That is -2/x.

(d) Since x is a positive quanitity, -2/x is a negative number. That means, the Revenue function attains its maximum at the point where marginal revenue =0 (since the second derivative is negative). Then e^3 is the revenue maximumising quantity.

to find maximum revenue, substitute e^3 in the Revenue function from (a).