​Rite-Cut riding lawnmowers obey the demand equation p=-1/10X + 1800 . The cost of producing x lawnmowers is given by the function C(x)=80x+6000

Question

a) Express the revenue R as a function of x.

​b) Express the profit P as a function of x.

​c) Find the value of x that maximizes profit. What is the maximum​ profit?

​d) What price should be charged in order to maximize​ profit?

Answer 1

a). Revenue = (price)(quantity)

So, R(x) = (-1/10 x + 800)(x) = (-1/10)x² + 800x

b). Profit = Revenue - cost

P(x) = R(x) - C(x) = (-1/10)x² + 720x - 6000

c), The graph of y = P(x) ix a parabola opening downward. P(x) is maximized when x is the x-coordinate of the vertex. So, maximum profit when x = -(720) / (2(-1/10)) = 144.

d). Profit is maximized when p = -(1/10)(144) + 800 = $785.60