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.

  • 1
    $\begingroup$ $+- \sqrt {..} $ typo or another constraint? $\endgroup$
    – Narasimham
    Oct 29 '19 at 10:12
  • $\begingroup$ sorry, that was a typo. $\endgroup$
    – random
    Oct 29 '19 at 13:15
  • $\begingroup$ I edited to bring in $Q,$ label or else the problem gets trivial. Please roll back if not helpful. $\endgroup$
    – Narasimham
    Oct 31 '19 at 1:26

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.


$$ x+1=u\, ; y+2=v\,;$$

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 $

  • $\begingroup$ So can you find the minimum value of x+y using lagrangian multiplier? $\endgroup$
    – random
    Oct 31 '19 at 8:22
  • $\begingroup$ Yes, included that last step in terms of Q. $\endgroup$
    – Narasimham
    Oct 31 '19 at 12:37

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