Unexpected behavior from MaxValue and Maximize

Question

consider the following function

α = 1/Sqrt[(1 - η^2)^2 + 4 D0^2 η^2];
p1 = Plot[{Evaluate[Table[α, {D0, 0, 1, .1}]], 1/(1 + η^2)}, {η, 0, 2}, 
       PlotRange -> {{0, 2}, {0, 3}}]

i now wanted to calculate the function that goes through all the maxima of this function for $\eta \in(0,1)$.

Using

MaxValue[α, D0]

gives

while

Maximize[α, D0]

yields

which are both not what I'm looking for. Using the straight forward approach

max = α /. Solve[D[α, η] == 0]

p2 = Plot[max[[1]], {η, 0, 1}, PlotRange -> {{0, 2}, {0, 3}}]
Show[p1, p2]

it works. I guess i didn't quite get how MaxValue works. Can someone teach me, what I did wrong?

Answer 1

Forgive me if this is entirely not what you were asking but your final line:

I guess i didn't quite get how MaxValue works. Can someone teach me, what I did wrong?

suggests that perhaps an example of MaxValue doing something, anything, useful might help.

mv = MaxValue[α, η];

p3 = Plot[Evaluate[Table[mv, {D0, 0, 1, .1}]], {η, 0, 2}, 
  PlotRange -> {{0, 2}, {0, 3}}]

Show[p1, p3]

So MaxValue quite handily gives is the maximum value that is reached for each value of D0.

Likewise MaxValue[α, D0] gives us the maximum value that is reached for each value of η, which is the uppermost line in the plot:

mv2 = MaxValue[α, D0]

p4 = Plot[Evaluate[Table[mv2, {D0, 0, 1, .1}]], {η, 0, 2}, 
  PlotRange -> {{0, 2}, {0, 3}}]

Show[p1, p4]