Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. Assume that revenue, R(x), and cost, C(x), of producing x units are in dollars.
R(x) = 60x-.5x^2 and C(x) = 4x+20

The maximum profit at x =5.6 units is

Pmₐₓ = 136.8 per unit cost

Step-by-step explanation:

Explanation:-

Given Revenue function  R(x) = 60x-.5x^2 and

Given the cost function is C(x) = 4x+20

Profit = R(x) - C(x)

profit = 60x-.5x^2 - (4x+20)

         = 60x - 5x^2 -4x -20

    P    =  56x - 5x^2 -20  … ( i )

Now differentiating equation (1) with respective to 'x'

  P¹     = 56 - 5(2x) -0

  P¹     = 56 -10x  

The derivative of profit function is equating zero

56 -10x  =0

56 =10x

x = 5.6

The maximum profit at x =5.6

        Pmₐₓ    =  56x - 5x^2 -20  

                    = 56(5.6) -5(5.6)^2-20

                   =  136.8

Final answer:-

The maximum profit at x =5.6 units is

Pmₐₓ = 136.8 per unit cost