Answer:
a) 1/2
b) 250
Step-by-step explanation:
The start of the question doesn't matter entirely, although is interesting to read. What we are trying to do is find the value for [tex]p[/tex] such that [tex]1000p(1-p)[/tex] is maximized. Once we have that [tex]p[/tex], we can easily find the answer to part b.
Finding the value that maximizes [tex]1000p(1-p)[/tex] is the same as finding the value that maximizes [tex]p(1-p)[/tex], just on a smaller scale. So, we really want to maximize [tex]p(1-p)[/tex]. To do this, we will do a trick called completing the square.
[tex]p(1-p)=p-p^2=-p^2+p=-(p^2-p)=-(p^2-p+1/4)-(-1/4)=-(p-1/2)^2+1/4[/tex].
Because there is a negative sign in front of the big squared term, combined with the fact that a square is always positive, means we need to find the value of [tex]p[/tex] such that the inner part of the square term is equal to [tex]0[/tex].
[tex]p-1/2=0\\p=1/2[/tex].
So, the answer to part a is [tex]\boxed{1/2}[/tex].
We can then plug [tex]1/2[/tex] into the equation for p to find the answer to part b.
[tex]1000(1/2)(1-1/2)=1000(1/2)(1/2)=1000*1/4=250[/tex].
So, the answer to part b is [tex]\boxed{250}[/tex].
And we're done!
