Multivariable function optimization inconsistency

  • #1
RickRazor
17
3
TL;DR Summary
Missing conceptual detail in optimization problems
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I have a function dependent on 4 variables ##f(r_1,r_2,q_1,q)##. I'm looking to minimize this function in the domain ##0\leq r_1 \leq r_2 \leq 1## with respect to the variables ##r_1, r_2## and ##q_1##.

To find the minima, I first solved ##\frac{\partial f}{\partial r_1}=0## and ##\frac{\partial f}{\partial q_1}=0##, giving ##r_1^*(r_2,q)## and ##q_1^*(r_2,q)##. Now I have the function of the form ##f(r_1^*(r_2,q),r_2,q_1^*(r_2,q),q).##

Now I solved ##\frac{\partial f(r_1^*(r_2,q),r_2,q_1^*(r_2,q),q)}{\partial r_2}=0## for ##r_2^*(q)##.

So, the final function is of the form ##f(r_1^*(r_2^*(q),q),r_2^*(q),q_1^*(r_2^*(q),q),q)## which is fine. Now I see later that ##r_1^*(r_2^*(q),q)=0## and ##q_1^*(r_2^*(q),q)=0##.

So, if I directly look for the function ##f(0,r_2,0,q)## and it's minima wrt ##r_2##, it's giving a different result, i.e. I have

##\min_{r_2} f(0,r_2,0,q) \neq f(r_1^*(r_2^*(q),q),r_2^*(q),q_1^*(r_2^*(q),q),q)## even though ##r_1^*(r_2^*(q),q)=0## and ##q_1^*(r_2^*(q),q)=0##. Why is this the case? Are there other simple examples?

The function is
##f(r_1,r_2,q_1,q)=3r_1+r_2+q_1^2/r_1+2(q-q_1)^2/(r_2-r_1)## and

##f(0,r_2,0,q)=r_2+2(q-q_1)^2/r_2##

##r_2^*(q)=\sqrt{\frac{2}{3}}q##,
##r_1(r_2^*(q),q)=0## and ##q_1(r_2^*(q),q)=0##.
 
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Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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4,371
Hi,

I have diffculty following the steps; perhaps you can post them ?

And I don't see how ##r_1(r_2^*(q),q)=0## can come out: ##f## does not exist for ##r_1 = 0## ...

##\ ##
 
  • #3
RickRazor
17
3
I had made some trivial mistake in calculation. Solved it now. Thanks.
 
  • #4
scottdave
Science Advisor
Homework Helper
Insights Author
Gold Member
1,892
883
I had made some trivial mistake in calculation. Solved it now. Thanks.
I'm glad you solved it.

Did it have to do with having r1 in the denominator in one instance then r2 in the denominator in the second instance?
 

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