The price p (in dollars) and the quantity x sold of a certain product satisfy the demand equation x=-5p+200. What is the maximizes revenue?
R=$
To find the revenue, we need to multiply the price (p) by the quantity sold (x):
R = p*x
Given x = -5p + 200, we can substitute this into the revenue equation:
R = p*(-5p + 200)
R = -5p^2 + 200p
To find the maximum revenue, we need to find the value of p that maximizes the revenue. This can be done by taking the derivative of the revenue function with respect to p and setting it equal to zero:
dR/dp = -10p + 200
Setting dR/dp = 0 gives:
-10p + 200 = 0
10p = 200
p = 20
Therefore, the price that maximizes revenue is $20. To find the maximum revenue, we plug this price back into the revenue function:
R = -5(20)^2 + 200(20)
R = -5(400) + 4000
R = -2000 + 4000
R = $2000
Therefore, the maximum revenue is $2000.