The revenue function for a lawnmower shop is given by R(x) = x • p(x) dollars where x is the number of units sold and p(x) = 200 - 0.2x is the unit price. Find the maximum revenue.

Rmax = 50,000 dollars

Step-by-step explanation: from the question,

x= number of units sold

p(x) = price function = 200 - 0.2x

Revenue = price * quantity

Revenue = x ( 200 - 0.2x)

Revenue = 200x - 0.2x².

Let R = revenue, R = 200x - 0.2x²

To get the maximum revenue (Rmax), we need the maximum quantity (Xmax) that will produce that.

And to get maximum quantity, we equate the first derivative (dR/dx) of the revenue (R) with respect to quantity (x) to zero..

If R = 200x - 0.2x²

dR/dx = 200 - 0.4x = 0

200 - 0.4x = 0

200 = 0.4x

x = 200/ 0.4

x = 500

Hence Xmax = 500 units

Substitute Xmax into the revenue function to get maximum revenue,

R = 200x - 0.2x², but x = 500

Rmax = 200(500) - 0.2 (500)²

Rmax = 100,000 - 50,000

Rmax = 50,000 dollars

my guess would be b and d, but i didn’t actually get out the pen and paper. lol