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).

(b) how can the profit function be found?
○ a. find the derivative of ( r(x)-c(x) ).
○ b. find ( r(x)-c(x) ).
○ c. find the derivative of ( r(x) ).
○ d. find ( r(x)-c(x) ).
the profit function is ( p(x)= )
(c) the marginal profit when ( x = 100 ) can be found by evaluating
the marginal profit when ( x = 100 ) is

Explanation:

(a) Marginal Revenue

Step1: Recall the formula for marginal revenue

Marginal revenue \(R'(x)\) is the derivative of the revenue function \(R(x)\). Given \(R(x)=20\ln(5x + 1)\), use the chain - rule \((\ln(u))'=\frac{u'}{u}\), where \(u = 5x+1\) and \(u'=5\).

$$ R'(x)=\frac{20\times5}{5x + 1}=\frac{100}{5x+1} $$

(b) Profit function \(P(x)\)

Step1: Recall the formula for profit function

Profit function \(P(x)=R(x)-C(x)\). Given \(R(x)=20\ln(5x + 1)\) and \(C(x)=\frac{x}{5}\), then

$$ P(x)=20\ln(5x + 1)-\frac{x}{5} $$

(c) Marginal profit when \(x = 100\)

Step1: Find the marginal profit function \(P'(x)\)

Differentiate \(P(x)=20\ln(5x + 1)-\frac{x}{5}\) using the sum - rule \((u - v)'=u'-v'\).
For \(u = 20\ln(5x + 1)\), \(u'=\frac{20\times5}{5x+1}=\frac{100}{5x + 1}\) (by the chain - rule), and for \(v=\frac{x}{5}\), \(v'=\frac{1}{5}\).
So \(P'(x)=\frac{100}{5x+1}-\frac{1}{5}\)

Step2: Evaluate \(P'(x)\) at \(x = 100\)

Substitute \(x = 100\) into \(P'(x)\):

$$ P'(100)=\frac{100}{5\times100+1}-\frac{1}{5}=\frac{100}{501}-\frac{1}{5}=\frac{500 - 501}{2505}=-\frac{1}{2505}\approx - 0.0004 $$

Answer:

(a) The marginal revenue function is \(R'(x)=\frac{100}{5x + 1}\)
(b) The profit function is \(P(x)=20\ln(5x + 1)-\frac{x}{5}\)
(c) The marginal profit when \(x = 100\) is \(-\frac{1}{2505}\approx - 0.0004\)
(d) When \(x = 100\) items are produced and sold, the profit is decreasing at a rate of approximately \(0.0004\) per item.