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# Math Problem:Minimizing production Costs:?

The total monthly cost (in dollars) incurred by cannon precision instruments for manufacturing x units of the model MI digital camera is given by the function

C(x)= .0025x^2 +80x +10,000

a. Find the average cost function C.

b. Find the level of production that results in the smallest average production cost.

C. find the level of production for which the average cost is equal to the marginal cost.

Please please help! I have been stuck for days and i just dont get it.

### 1 Answer

- MichaelLv 72 years ago
a. Find the average cost function A

( using A since C is the total cost function)

Average is the total divided by the quantity

A = (.0025x^2 +80x +10,000) / x

A = .0025x + 80 + 10000/x <––––––

b. Find the level of production that results in the smallest average production cost.

Set the derivative of A to zero

A' = .0025 - 10000/x²

.0025 - 10000/x² = 0

.0025 = 10000/x²

x² = 10000/.0025

x² = 4000000

x ± 2000

x must be positive

x = 2000 <––––––

c. find the level of production for which the average cost is equal to the marginal cost.

The marginal cost is the derivative of the total cost

C' = .005x + 80

Set A equal to C'

.0025x + 80 + 10000/x = 0.005x + 80

-.0025x +10000/x = 0

-.0025x² + 10000 = 0

.0025x² = 10000

x² = 4000000

x = ±2000

x must be positive

x = 2000 <––––––