The demand for a new computer game can be modeled by p (x )equals 48.5 minus 5 ln x, for 0less than or equalsxless than or equals800, where p(x) is the price consumers will pay, in dollars, and x is the number of games sold, in thousands. Recall that total revenue is given by R (x)= x * p (x). Complete parts (a) through (c) below.

a. Find R(x).

b. Find the marginal revenue, R'(x)

c. Is there any price at which revenue will be maximized? Why or why not? Choose the correct answer below.

a. No, because p(x)R'(x). Therefore, the revenue cannot be maximized.

b. No, because the marginal revenue, R'(x), has no maximum. Therefore, revenue will always increase

c. Yes, because the revenue will be maximized when R(x)-R'(x). The price to the consumer would be $:_

d. Yes, because the revenue will be maximized when R'(x)=0. The price to the consumer would be $:_

Question

22-03-2021
Business

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The demand for a new computer game can be modeled by p (x )equals 48.5 minus 5 ln x, for 0less than or equalsxless than or equals800, where p(x) is the price consumers will pay, in dollars, and x is the number of games sold, in thousands. Recall that total revenue is given by R (x)= x * p (x). Complete parts (a) through (c) below.

a. Find R(x).

b. Find the marginal revenue, R'(x)

c. Is there any price at which revenue will be maximized? Why or why not? Choose the correct answer below.

a. No, because p(x)R'(x). Therefore, the revenue cannot be maximized.

b. No, because the marginal revenue, R'(x), has no maximum. Therefore, revenue will always increase

c. Yes, because the revenue will be maximized when R(x)-R'(x). The price to the consumer would be $:_

d. Yes, because the revenue will be maximized when R'(x)=0. The price to the consumer would be $:_

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amcool amcool · Answer 1 · 2021-03-26T17:35:03+01:00

Answer:

a. R(x) = 48.5x - x5inx

b. R'(x) = 43.5 - 5lnx

c. The correct option is d. Yes, because the revenue will be maximized when R'(x)=0. The price to the consumer would be $5.

Explanation:

Given:

p(x) = 48.5 - 5lnx, for 0 ≤ x ≤ 800 ............. (1)

a. Find R(x).

R (x) = x * p (x) ........................ (2)

Substitute equation (1) into (2), we have:

R(x) = x(48.5 - 5lnx)

R(x) = 48.5x - x5inx .............................. (3)

Equation (1) is the R(x) required to be found.

b. Find the marginal revenue, R'(x)

The derivative of equation (3) is taken to obtain R'(x) as follows:

R'(x) = 48.5 - 5lnx - x5(1/x)

R'(x) = 48.5 - 5lnx - 5

R'(x) = 48.5 - 5 - 5lnx

R'(x) = 43.5 - 5lnx ...................... (4)

Equation (4) is the marginal revenue, R'(x), required to found.

c. Is there any price at which revenue will be maximized? Why or why not? Choose the correct answer below.

Profit is maximized when R'(x) = 0

Equating equation (4) to 0 and solve for x, we have:

43.5 - 5lnx = 0

43.5 = 5lnx

5lnx = 43.5

lnx = 43.5 / 5

x = e^(43.5 / 5)

x = 6,002.91

Substituting x = 6,002.91 into equation (1), we have:

p(x) = 48.5 - (5 * ln(6,002.91))

p(x) = 48.5 - (5 * 8.70)

p(x) = 48.5 - 43.5

p(x) = $5

Therefore, the correct option is d. Yes, because the revenue will be maximized when R'(x)=0. The price to the consumer would be $5.