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 <u>$5</u>.
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 <u>$5</u>.
