If the total revenue function for a computer is R(x) = 2000x - 20x^2 - x^3, find
ID: 2894077 • Letter: I
Question
If the total revenue function for a computer is R(x) = 2000x - 20x^2 - x^3, find the level of sales, x, that maximizes revenue and find the maximum revenue in dollars. x = computers R(x) = $ If the total cost function for a product is C(x) = 810 + 0-.1x^2 dollars, producing how many units, x, will result in a minimum average cost per unit? x = units Find the minimum average cost per unit. $ Analysis of daily output of a factory during a 8-hour shift shows that the hourly number of units y produced after t hours of production is y = 10t + 1/2t^2 - t^3, 0 lessthanorequalto t lessthanorequalto 8. (a) After how many hours will the hourly number of units be maximized? hr (b) What is the maximum hourly output? units/hrExplanation / Answer
Solution:(5)
Given R(x) = 2000x - 20x^2 - x^3
By taking the first derivative, which is:
.R'(x) = 2000 - 40x - 3x^2
If we set this equal to zero and solve it will give us x values of minimums or maximums.
=> 2000 - 40x - 3x^2 = 0
=> (20 - x)(100 + 3x) = 0
=> 20 - x = 0
=> x = 20 units <== this is the level of sales to produce maximum revenue.
Maximum Revenue; at x = 20;
R(x) = 2000x - 20x^2 - x^3
R(20) = 2000(20) - 20(20)^2 - (20)^3
R(20) = 40,000 - 8,000 - 8,000 = $24,000 <== MAX REVENUE