An automobile manufacturer sells cars in America, Europe, and Asia, charging a different price in each of the three markets. The price function for cars sold in America is p = 29 − 0.2x (for 0 ≤ x ≤ 145),

a. The company's profit function P(x, y, z) = 24x +12y +16z – 0.2x^2 – 0.1y^2 – 0.1z^2 – 26

b. The number of cars that must be sold in each market are 60 in America, also 60 in Europe and 80 in Asia.

Step-by-step explanation:

Note: This question is not complete and it has some errors in the figures used. The correct complete question is therefore provided before answering the question as follows:

An automobile manufacturer sells cars in America, Europe, and Asia, charging a different price in each of the three markets. The price function for cars sold in America is p = 27 − 0.2x  (for 0 ≤ x ≤ 135), the price function for cars sold in Europe is q = 15 − 0.1y  (for 0 ≤ y ≤ 150), and the price function for cars sold in Asia is r = 19 − 0.1z (for 0 ≤ z ≤ 190), all in thousands of dollars, where x, y, and z are the numbers of cars sold in America, Europe, and Asia, respectively. The company's cost function is C = 26 + 3(x + y + z) thousand dollars.

(a) Find the company's profit function P(x, y, z). [Hint: The profit will be revenue from America plus revenue from Europe plus revenue from Asia minus costs, where each revenue is price times quantity.]

P(x, y, z) =

(b) Find how many cars should be sold in each market to maximize profit. [Hint: Set the three partials Px, Py, and Pz equal to zero and solve. Assuming that the maximum exists, it must occur at this point.]

The explanation to the answers is now given as follows:

(a) Find the company's profit function P(x, y, z). [Hint: The profit will be revenue from America plus revenue from Europe plus revenue from Asia minus costs, where each revenue is price times quantity.]

p = price function for cars sold in America = p = 27 − 0.2x (for 0 ≤ x ≤ 135)

q = price function for cars sold in Europe = 15 − 0.1y (for 0 ≤ y ≤ 150)

r = price function for cars sold in Asia = 19 − 0.1z (for 0 ≤ z ≤ 190)

C = company's cost function = 26 + 3(x + y + z) = 26 + 3x + 3y + 3z

P(x, y, z) = profit function

x, y, and z are the numbers of cars sold in America, Europe, and Asia, respectively

Therefore, we have:

TR = Total revenue = px + qy + rz …………………………….. (1)

Substituting the relevant values into equation (1), we have:

TR = (27 − 0.2x)x + (15 − 0.1y)y + (19 − 0.1z)z

TR = 27x – 0.2x^2 + 15y – 0.1y^2 + 19z – 0.1z^2

Also,

P(x, y, z) = TR – C ……………………………………… (2)

Substituting the relevant values into equation (2), we have:

P(x, y, z) = 27x – 0.2x^2 + 15y – 0.1y^2 + 19z – 0.1z^2 – (26 + 3x + 3y + 3z)

P(x, y, z) = 27x – 0.2x^2 + 15y – 0.1y^2 + 19z – 0.1z^2 – 26 – 3x – 3y – 3z

P(x, y, z) = 27x – 3x +15y – 3y +19z – 3z – 0.2x^2 – 0.1y^2 – 0.1z^2 – 26

P(x, y, z) = 24x +12y +16z – 0.2x^2 – 0.1y^2 – 0.1z^2 – 26 ……………….. (3)

Therefore, the company's profit function P(x, y, z) = 24x +12y +16z – 0.2x^2 – 0.1y^2 – 0.1z^2 – 26.

(b) Find how many cars should be sold in each market to maximize profit. [Hint: Set the three partials Px, Py, and Pz equal to zero and solve. Assuming that the maximum exists, it must occur at this point.]

As indicated in the question, the number cars that should be sold in each market to maximize profit can be calculated by deriving the three partials Px, Py, and Pz from the profit function P(x, y, z), i.e. equation (3) in part a above, and set equal to zero and then solve as follows:

In America

From equation (3), we have:

Px = 24 – 0.4x = 0

Therefore, we have:

24 – 0.4x = 0

24 = 0.4x

x = 24 / 0.4

x = 60

In Europe

From equation (3), we have:

Py = 12 – 0.2y = 0

Therefore, we have:

12 – 0.2y = 0

12 = 0.2y

y = 12 / 0.2

y = 60

In Asia

From equation (3), we have:

Pz = 16 - 0.2z = 0

Therefore, we have:

16 - 0.2z = 0

16 = 0.2z

z = 16 / 0.2

z = 80

Based on the above calculations, the number of cars that must be sold in each market are 60 in America, also 60 in Europe and 80 in Asia.

d.192

step-by-step explanation:

12*16=192

b.an increase in a causes b to decrease