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.
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).
To find the revenue function R, we need to multiply the price p by the number of units sold x.
i. Revenue function R: R = x * p
Given that p = 2x^2 + 3, we can substitute this value into the revenue function:
R = x * (2x^2 + 3)
R = 2x^3 + 3x
ii. To determine the price interval for which the revenue is increasing and decreasing, we need to examine the slope of the revenue function.
Take the derivative of the revenue function with respect to x:
dR/dx = 6x^2 + 3
For the derivative to be positive, the quadratic term (6x^2) must be positive. This implies that x^2 > 0, which is always true.
Therefore, dR/dx will be positive for all values of x.
In conclusion, the revenue is always increasing and never decreasing (as long as the number of units sold is positive).