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).
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 answer
1 answer