Question

if total revenue received from the sale of x items is given by ( r(x)=20ln(5x + 1) ), while the total cost to produce x items is ( c(x)=\frac{x}{5} ). find the following
(a) the marginal revenue
(b) the profit function ( p(x) )
(c) the marginal profit when ( x = 100 )
(d) interpret the results of part (c).

(a) how can the marginal revenue be found?
○ a. find ( rleft(\frac{x}{2}
ight) ).
○ b. find ( r(x)-c(x) ).
○ c. find the derivative of ( r(x) ).
○ d. find the derivative of ( r(x)-c(x) )
the marginal revenue when x items are produced is
(b) how can the profit function be found?
○ a. find the derivative of ( r(x)-c(x) ).
○ b. find ( r^{prime}(x)-c(x) ).

Explanation:

Step1: Recall the definition of marginal revenue

Marginal revenue is the derivative of the revenue function. So for \(R(x) = 20\ln(5x + 1)\), we use the chain - rule. The chain - rule states that if \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). Let \(u = 5x+1\), then \(R(x)=20\ln(u)\). The derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\), and the derivative of \(u = 5x + 1\) with respect to \(x\) is \(5\).

Step2: Calculate \(R^\prime(x)\)

By the chain - rule, \(R^\prime(x)=\frac{20}{5x + 1}\cdot5=\frac{100}{5x+1}\)

Answer:

C. Find the derivative of \(R(x)\)
The marginal revenue when \(x\) items are produced is \(\frac{100}{5x + 1}\)