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