Answer
a. $2$ dollars.
b. $1.96$ dollars.
c. $0$, see explanations.
Work Step by Step
a. Given the revenue equation $r(x)=20000(1-\frac{1}{x})$, we have $r'(x)=20000(\frac{1}{x^2})$ and the marginal revenue when 100 machines are produced is $r'(100)=20000(\frac{1}{100^2})=2$ dollars.
b. Let $x=101$; the increase in revenue that will result from increasing production from 100 machines a
week to 101 machines a week is:
$\Delta r=r'(101)(101-100)=20000(\frac{1}{101^2})\approx1.96$ dollars.
c. We can find $\lim_{x\to\infty}r'(x)=\lim_{x\to\infty}20000(\frac{1}{x^2})=0$ and this means that the marginal revenue will decrease as the number of machines produced, $x$, increases and it will approach zero when the $x$ value is very large.
