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

a) $342,491

$342.491

$389.74

b) $400

c) $320

Step-by-step explanation:

the cost function = C (x)

C (x) = 16000 + 200x + 4x^3/2

a) when we have a unit of 1000 unit, x = 1000

C (1000) = 16000 + 200 (1000) + 4 (1000) ^3/2

= 16000 + 200000 + 126491

= 342,491

Cost = $342,491

Average cost = C (1000) / 1000

= 342,491/1000

= 342.491

The average cost = $342.491

Marginal cost = derivative of the cost

C' (x) = 200 + 4 (3/2) x^1/2

= 200 + 6x^1/2

C' (1000) = 200 + 6 (1000) ^1/2

= 389.74

Marginal cost = $389.74

Marginal cost = Marginal revenue

C' (x) = C (x) / x

200 + 6x^1/2 = (16000 + 200x + 4x^3/2) / x

200 + 6x^1*2 = 16000/x + 200 + 4x^1/2

Collect like terms

6x^1*2 - 4x^1/2 = 16000/x + 200 - 200

2x^1/2 = 16000/x

2x^3/2 = 16000

x^3/2 = 16000/2

x^3/2 = 8000

x = 8000^2/3

x = 400

Therefore, the production level that will minimize the average cost is the critical value = $400

C' (x) = C (x) / x

C' (400) = 16000/400 + 200 + 4 (400) ^1/2

= 40 + 200 + 80

= 320

The minimum average cost = $320