Answer:
Sales will be maximized after 5 weeks .
Step-by-step explanation:
We are given that the weekly sales S are given by
[tex]S=\frac{500t}{(t+5)^2}[/tex]
[tex]S'(t)=\frac{(t+5)^2\frac{d}{dt}(500t)-500t\frac{d}{dt}((t+5)^2)}{(t+5)^4}[/tex]
[tex]S'(t)=\frac{500(t+5)^2-500t(2(t+5))}{(t+5)^4}[/tex]
[tex]S'(t)=\frac{-500(t-5)}{(t+5)^3}[/tex]
Substitute the S'(t)=0
So, [tex]S'(t)=\frac{-500(t-5)}{(t+5)^3}=0[/tex]
[tex]-500(t-5)=0[/tex]
[tex]t=5[/tex]
[tex]S'(t)=\frac{-500(t-5)}{(t+5)^3}\\S''(t)=\frac{(t+5)^3\frac{d}{dt}(-500(t-5))-(-500(t-5))\frac{d}{dt}((t+5)^3)}{(t+5)^6}\\S''(t)=\frac{(t+5)^3(-500)-(-500(t-5))(3(t+5)^2)}{(t+5)^6}[/tex]
Substitute t = 5 in S''(t)
[tex]S''(t)=\frac{(5+5)^3(-500)-(-500(5-5))(3(5+5)^2)}{(5+5)^6}\\S''(t)=-0.5[/tex]
Since S''(t) at t =5 is less than 0
So, maximum
So, Sales will be maximized at t = 5
Hence Sales will be maximized after 5 weeks .
