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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

Respuesta :

Answer:

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