I was trying to solve the problem where you need to find the max and min value of
$$ P = x+y$$
knowing that
$$x-3\sqrt {x+1} -3\sqrt{y+2}+y= Q $$
For the maximum value, I used lagrange multiplier to find it and the result was correct.
For the minimum value, I use the lagrange multiplier to find the maximum value of P' = -x-y and the maximum value of $P'' = \frac{1}{x+y}$ Both turned out to have the same value of x and y as when I tried to find the maximum value of P. This is clearly wrong but I don't know why and how to find the correct answer. Can somebody help me?
Thank you.
Some clarity is to be brought in to see what we are doing.
To illustrate a simpler case if in a rectangle you want to extremise (x+y) subject to constant area A , then by Lagrange Multiplier you obtain
$$x= \sqrt A \quad =y\quad and \quad (x+y)= 2 \sqrt {A}$$
The constraint function is required to be labelled compulsorily and object functioned optionally.
I shall solve this problem with another label Q avoiding P altogether for sake of clarity. The final maximization result/answer should be seen as a function of Q alone.
Rewrite
$$ x+1=u\, ; y+2=v\,;$$
You want to extremise
$$(u+v) \tag1 $$
subject to constraint
$$ u -3 \sqrt {u}+ v -3 \sqrt {v} = Q+3 \tag2 $$
Using Lagrange Multiplier
$$ u- 3 \sqrt{u} -(Q+3)/2 =0 \tag3$$
with solution (two solutions appear due to quadratic)
$$ u=3-\frac{Q}{2} \pm \frac32 \sqrt{3-2Q}= v \tag 4 $$
and $(u+v)_{max}=(6-Q) \pm 3\sqrt{3-2Q} \tag5 $
or $(x+y)_{max}=(3-Q) \pm 3\sqrt{3-2Q}. \tag6 $