I Multivariable function optimization inconsistency

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The discussion revolves around optimizing a multivariable function, specifically minimizing the function f(r_1, r_2, q_1, q) dependent on four variables. The user initially derived critical points for r_1 and q_1, leading to a transformed function dependent on r_2 and q. However, inconsistencies arose when comparing the minima of the transformed function with the direct evaluation of f at r_1 = 0 and q_1 = 0, revealing different results. The user later identified a calculation error that contributed to the confusion regarding the existence of f at r_1 = 0. The conversation highlights the complexities of multivariable optimization and the importance of careful calculations.
RickRazor
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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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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## ...

##\ ##
 
I had made some trivial mistake in calculation. Solved it now. Thanks.
 
RickRazor said:
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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