1. If the total cost function for a product is C(x) = 200(0.02x + 6)3 dollars, where x represents the number of hundreds of units produced, producing how many units will minimize average cost? 2. A firm can produce only 3900 units per month. The monthly total cost is given by C(x) = 500 + 200x dollars, where x is the number produced. If the total venue is given by R(x) = 450x-1/100x^2 dollars, how many items, x, should the firm produce for maximum profit?
3. If the profit function for a product is P(x) = 3600x + 60x2 ? x3 ? 72,000 dollars, selling how many items, x, will produce a maximum profit?.

a. The number of units which will minimize average cost is approximately 5,130 units.

b. The firm should produce 12,500 items, x, for maximum profit.

c. The number of items, x, that will produce a maximum profit is 60 items.

Step-by-step explanation:

Note: This question is not complete as there are some signs are omitted there. The complete question is therefore provided before answering the question as follows:

1. If the total cost function for a product is C(x) = 200(0.02x + 6)^3 dollars, where x represents the number of hundreds of units produced, producing how many units will minimize average cost?

2. A firm can produce only 3900 units per month. The monthly total cost is given by C(x) = 500 + 200x dollars, where x is the number produced. If the total venue is given by R(x) = 450x-1/100x^2 dollars, how many items, x, should the firm produce for maximum profit?

3. If the profit function for a product is P(x) = 3600x + 60x2 - x3 - 72,000 dollars, selling how many items, x, will produce a maximum profit?

The explanation to the answer is now given as follows:

1. If the total cost function for a product is C(x) = 200(0.02x + 6)3 dollars, where x represents the number of hundreds of units produced, producing how many units will minimize average cost?

Given;

C(x) = 200(0.02x + 6)^3 ……………………………………….. (1)

We first simplify (0.02x + 6)^3 as follows:

(0.02x + 6)^3 = (0.02x + 6)(0.02x + 6)(0.02x + 6)

First, we have:

(0.02x + 6)(0.02x + 6) = 0.004x^2 + 0.12x + 0.12x + 36 = 0.004x^2 + 0.24x + 36

Second, we have:

(0.02x + 6)^3 = 0.004x^2 + 0.24x + 36(0.02x + 6)

(0.02x + 6)^3 = 0.00008x^3 + 0.048x^2 + 7.20x + 0.0024x^2 + 1.44x + 216

(0.02x + 6)^3 = 0.00008x^3 + 0.048x^2 + 0.0024x^2 + 7.20x + 1.44x + 216

(0.02x + 6)^3 = 0.00008x^3 + 0.0504x^2 + 8.64x + 216

Therefore, we have:

C(x) = 200(0.02x + 6)^3 = 200(0.00008x^3 + 0.0504x^2 + 8.64x + 216)

C(x) = 0.016x^3 + 10.08x^2 + 1,728x + 43,200

Therefore, the average cost (AC) can be calculated as follows:

AC(x) = C(x) / x = (0.016x^3 + 10.08x^2 + 1,728x + 43,200) / x

AC(x) = (0.016x^3 + 10.08x^2 + 1,728x + 43,200)x^(-1)

AC(x) = 0.016x^2 + 10.08x + 1,728 + 43,200x^(-1) …………………………. (2)

Taking the derivative of equation (2) with respect to x, equating to 0 and solve for x, we have:

0.032x + 10.08 - (43,300 / x^2) = 0

0.032x + 10.08 = 43,300 / x^2

X^2 * 0.32x = 43,300 – 10.08

0.32x^3 = 43,189.92

x^3 = 43,189.92 / 0.32

x^3 = 134,968.50

x = 134,968.50^(1/3)

x = 51.30

Since it is stated in the question that x represents the number of hundreds of units produced, we simply multiply by 100 as follows:

x = 51.30 * 100 = 5,130

Therefore, the number of units which will minimize average cost is approximately 5,130 units.

2. A firm can produce only 3900 units per month. The monthly total cost is given by C(x) = 500 + 200x dollars, where x is the number produced. If the total revenue is given by R(x) = 450x-1/100x^2 dollars, how many items, x, should the firm produce for maximum profit?

P(x) = R(x) - C(x) ……………. (3)

Where;

P(x) = Profit = ?

R(x) = 450x-1/100x^2

C(x) = 500 + 200x

Substituting the equations into equation (3), we have:

P(x) = 450x - 1/100x^2 - (500 + 200x)

P(x) = 450x - 0.01x^2 - 500 - 200x

P(x) = 450x - 200x - 0.01x^2 - 500

P(x) = 250x - 0.01x^2 – 500 …………………………………. (4)

Taking the derivative of equation (4) with respect to x, equating to 0 and solve for x, we have:

250 - 0.02x = 0

250 = 0.02x

x = 250 / 0.02

x = 12,500 items

Therefore, the firm should produce 12,500 items, x, for maximum profit.

3. If the profit function for a product is P(x) = 3600x + 60x2 – x^3 - 72,000 dollars, selling how many items, x, will produce a maximum profit?

Given;

P(x) = 3600x + 60x2 – x^3 - 72,000 …………………………. (5)

Taking the derivative of equation (5) with respect to x, equating to 0 and solve for x, we have:

3600 + 120x - 3x^2 = 0

Divide through by 3, we have:

1200 + 40x – x^2 = 0

1200 + 60x – 20x – x^2 = 0

60(20 + x) – x(20 + x) = 0

(60 – x)(20 + x) = 0

Therefore,

x = 60, or x = - 20

The negative value of x (i.e. x = - 20) will be will be ignored because it has no economic significance. Therefore, the number of items, x, that will produce a maximum profit is 60 items.

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.