Answer
a) The marginal revenue is $2$ dollars.
b) The increase in revenue is $1.96$ dollars.
c) $\lim_{x\to\infty}r'(x)=0$
Work Step by Step
$$r(x)=20000\Big(1-\frac{1}{x}\Big)=20000-\frac{20000}{x}$$
a) The marginal revenue is the derivative of the revenue from selling $x$ washing machines $r(x)$. So we would call the marginal revenue here $r'(x)$.
$$r'(x)=0-\Big(-\frac{20000}{x^2}\Big)=\frac{20000}{x^2}$$
The marginal revenue when $100$ washing machines are produced is $$r'(100)=\frac{20000}{100^2}=2(dollars)$$
b) The marginal revenue when $101$ washing machines are produced is $$r'(101)=\frac{20000}{101^2}\approx1.961(dollars)$$
This number, or the marginal revenue for $101$ machines, represents the change in the revenue as we increase the production from $100$ machines to $101$ machines.
In other words, increasing production to $101$ machines a week will increase the revenue by $1.961$ dollars.
c) $$\lim_{x\to\infty}r'(x)=\lim_{x\to\infty}\frac{20000}{x^2}=0$$
As $x\to\infty$, $20000/x^2$ will approach $0$.
This shows us that even if we have the capacity to increase production infinitely, the revenue will not increase continuously. There will come a point when it we will stop gaining revenue anymore.
