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