MATH SOLVE

3 months ago

Q:
# For each cost function (given in dollars), find (a) the cost,average cost, and marginal cost at a production level of 1000units; (b) the production level that will minimize the averagecost; and c) the minimum average cost.C(x)= 16,000x + 200x+ 4x3/2

Accepted Solution

A:

Answer:a) $342,491 $342.491 $389.74b) $400c) $320Step-by-step explanation:the cost function = C(x)C(x) = 16000 + 200x + 4x^3/2a) when we have a unit of 1000 unit, x= 1000C(1000) = 16000 + 200(1000) + 4(1000)^3/2= 16000 + 200000 + 126491= 342,491Cost = $342,491Average cost= C(1000) / 1000= 342,491/1000= 342.491The average cost = $342.491Marginal cost = derivative of the costC'(x) = 200 + 4(3/2) x^1/2 = 200 + 6x^1/2C'(1000) = 200 + 6(1000)^1/2 = 389.74Marginal cost = $389.74Marginal cost = Marginal revenue C'(x) = C(x) / x200 + 6x^1/2 = (16000 + 200x + 4x^3/2) / x200 + 6x^1*2 = 16000/x + 200 +4x^1/2Collect like terms 6x^1*2 - 4x^1/2 = 16000/x + 200 -2002x^1/2 = 16000/x2x^3/2 = 16000x^3/2 = 16000/2x^3/2 = 8000x = 8000^2/3x = 400Therefore, the production level that will minimize the average cost is the critical value = $400C'(x) = C(x) / xC'(400) = 16000/400 + 200 + 4(400)^1/2= 40 + 200 + 80= 320The minimum average cost = $320