Question
The price ‘p’ per unit is given by the relation
1 2
x p 2p 3
3
where ‘x’ is the number of units sold
then,
i. Find the revenue function R.
ii. Find the price interval for which the revenue is increasing and decreasing.
1 2
x p 2p 3
3
where ‘x’ is the number of units sold
then,
i. Find the revenue function R.
ii. Find the price interval for which the revenue is increasing and decreasing.
Answers
GPT 3.5
To find the revenue function R, we need to multiply the price per unit (p) by the number of units sold (x).
i. R = p * x
= (2p^2 - 3) * x
ii. To determine the price interval for which the revenue is increasing and decreasing, we need to find the derivative of R with respect to x and determine its sign.
dR/dx = 2p^2 - 3
To find the critical points where the derivative is zero, we set dR/dx equal to zero and solve for p:
2p^2 - 3 = 0
2p^2 = 3
p^2 = 3/2
p = ±√(3/2)
Since the revenue function R depends on both p and x, we need to consider both the sign of dR/dx and the values of p.
When p > √(3/2), the derivative dR/dx is positive, indicating that the revenue is increasing.
When p < -√(3/2), the derivative dR/dx is negative, indicating that the revenue is decreasing.
Therefore, the price interval for which the revenue is increasing is p > √(3/2), and the price interval for which the revenue is decreasing is p < -√(3/2).
i. R = p * x
= (2p^2 - 3) * x
ii. To determine the price interval for which the revenue is increasing and decreasing, we need to find the derivative of R with respect to x and determine its sign.
dR/dx = 2p^2 - 3
To find the critical points where the derivative is zero, we set dR/dx equal to zero and solve for p:
2p^2 - 3 = 0
2p^2 = 3
p^2 = 3/2
p = ±√(3/2)
Since the revenue function R depends on both p and x, we need to consider both the sign of dR/dx and the values of p.
When p > √(3/2), the derivative dR/dx is positive, indicating that the revenue is increasing.
When p < -√(3/2), the derivative dR/dx is negative, indicating that the revenue is decreasing.
Therefore, the price interval for which the revenue is increasing is p > √(3/2), and the price interval for which the revenue is decreasing is p < -√(3/2).