Question

The weekly average cost C ˉ (x) in dollars of manufacturing x units of a certain camera is given by C ˉ (x)=0.004x+85+16000x −1 (x>0) (a) What is the marginal average cost function? (b) What is the marginal average cost of producing 1500 cameras weekly? Write out the answer giving a clear interpretation of this answer. (Do not use the word marginal in your sentence.) (c) How many cameras should they produce weekly, to minimize the average cost? What is the minimum average cost? i. Find the critical point. ii. Determine if the critical point is a relative minimum or maximum by either making a sign chart for the derivative or using the second derivative. iii. Write a sentence answer to the question.

Answer 1

(a) The marginal average cost function is Cˉ ′ (x) = 0.004 - 16000/x^2.

(b) The marginal average cost of producing 1500 cameras weekly is $0.03. This means that for each additional camera produced beyond 1500, the average cost will decrease by approximately $0.03.

(c) To minimize the average cost, the company should produce approximately 28 cameras weekly. The minimum average cost is $85.14.

i. The critical point is x ≈ 28.3.

ii. The critical point is a relative minimum.

iii. To minimize the average cost, 28 cameras should be produced weekly.

Step-by-step explanation:

(a) Marginal average cost function

The marginal average cost function is the derivative of the average cost function.

Cˉ ′ (x) = dCˉ (x)dx=d(0.004x+85+16000x −1)dx=0.004−16000x−2

b) Marginal average cost of producing 1500 cameras weekly

The marginal average cost of producing 1500 cameras weekly is:

Cˉ ′ (1500) = 0.004 - 16000/1500^2

≈ -0.031.

This means that for each additional camera produced beyond 1500, the average cost will decrease by approximately $0.03.

(c) Minimizing the average cost

To minimize the average cost, we need to find the critical point of the average cost function. This is the point where the marginal average cost is equal to zero.

Cˉ ′ (x) = 0.004 - 16000/x^2

Cˉ ′ (x) = 0

Solving for x, we get x ≈ 80/√5 ≈ 28.3.

This means that to minimize the average cost, the company should produce approximately 28 cameras weekly.

Second derivative test

To determine whether the critical point is a relative minimum or maximum, we can use the second derivative test.

Cˉ ′ ′ (x) = 32000/x^3 > 0 for all x > 0

Since the second derivative is positive for all x > 0, we know that the critical point is a relative minimum. This means that the average cost is minimized at x = 28.3.

Minimum average cost

The minimum average cost is Cˉ (28.3) ≈ 85.14. This means that the average cost of producing 28 cameras weekly is approximately $85.14.

i. The critical point is x ≈ 28.3.

ii. The critical point is a relative minimum. This can be determined by making a sign chart for the derivative or using the second derivative. The second derivative test is easier in this case:

Cˉ ′ ′ (x) = 32000/x^3 > 0 for all x > 0

Since the second derivative is positive for all x > 0, we know that the critical point is a relative minimum.

iii. To minimize the average cost, the company should produce approximately 28 cameras weekly.