Answer:
(a)[TeX]\frac{dP}{dt}=\frac{4800-1600t-240t^2}{(t^2+20)^2}[/TeX]
(b)P'(5)=-($4.54) Thousand
(c)P'(11)=-($2.10) Thousand
(d)The fifth Month
Explanation:
Given the monthly profit model:
[TeX]P(t)=\frac{240t-40t^2}{t^2+20}[/TeX]
(a)We want to derive a model that gives the Marginal Profit, P' of the book.
We differentiate
[TeX]P(t)=\frac{240t-40t^2}{t^2+20}[/TeX] using quotient rule.
[TeX]\frac{dP}{dt}=\frac{(t^2+20)(240-80t)-(240t-40t^2)(2t)}{(t^2+20)^2}[/TeX]
Simplifying
[TeX]\frac{dP}{dt}=\frac{4800-1600t-240t^2}{(t^2+20)^2}[/TeX]
We have derived a model for the marginal profit.
(b) After 5 months, at t=5
Marginal Profit=P'(5)
[TeX]\frac{dP}{dt}=\frac{4800-1600t-240t^2}{(t^2+20)^2}[/TeX]
[TeX]P^{'}(5)=\frac{4800-1600(5)-240(5)^2}{(5^2+20)^2}[/TeX]
=-($4.54) Thousand of dollars
(c)Marginal Profit 11 Months after book release
[TeX]P^{'}(11)=\frac{4800-1600(11)-240(11)^2}{(11^2+20)^2}[/TeX]
=-($2.10) Thousand of dollars
(d) Since the marginal profit at t=5 is negative, after the 5th Month, the profit starts to experience a steady decrease.