# Is there any possibility to visually examine a Lagrange problem with Mathematica?

I have the function: $$f(a_1, a_2) = a_1^2 \sigma_1^2 + a_2^2 \sigma_2^2$$, which is subjected to the constraint: $$a_1 + a_2 = 1$$. I have obtained that: $$a_i = \frac{\sigma_i^{-2}}{\sum_{j = 1}^2 \sigma_j^{-2}},$$ optimizes $$f$$. I have two questions:

1. Is there a possibility to verify that the above $$a_i$$ maximizes or minimizes the $$f$$ using Mathematica, in particular, by visualization?

2. If the answer to the above question is yes, how the values of the function $$f$$ changes if we alter the exponents in $$a_i$$ from $$-2$$ to $$-1$$ or from $$-2$$ to $$-3$$?

• Have you checked Minimize? For example Minimize[{a1^2 s1^2 + a2^2 s2^2 + a3^2 s3^2, a1 + a2 + a3 == 1}, {a1, a2, a3}, Reals] // FullSimplify Sep 26 at 19:42
• @Domen Thanks for your comment. I have tried that. In fact, I don't want to solve the optimization problem with Mathematica; I'm looking if there is any way of visualization with Mathematica in order to see if my obtained result, $a_i$, minimizes the function or not, then I would like to change the exponents in $a_i$ to see if the values of function increases or decreases. But I don't know how to do this.
– ARez
Sep 26 at 19:49
• @Domen I'm thinking that if I remove one of the variables and make the problem as: $f(a_1, a_2) = a_1^2 \sigma_1^2 + a_2^2 \sigma_2^2$, perhaps I could investigate the effects of changing the exponents in my formula for $a_i$ on the function $f$ by visualization.
– ARez
Sep 26 at 20:23

Due to the constrain, you can write your function as a function of only 2 variables:

f[a1_,a2_]= a1^2 s1^2 + a2^2 s2^2 + (1 - a1 - a2)^2 s3^2;


To find all critical points (max., min or saddle) the following derivatives must be zero:

e1 = D[f[a1, a2], a1] == 0
e2 = D[f[a1, a2], a2] == 0
sol = Solve[{e1, e2}, {a1, a2}]


This results in:

and a3 we get by:

(1 - a1 - a2) /. sol // Simplify


Note that by dividing by s1^2s2^2s3^2 we get your solution.

To analyze what sort of critical point we have (max, min or saddle) we need the second derivatives (called "Hessian matrix"):

hes = D[f[a1, a2], {{a1, a2}, 2}]
(*{{2 s1^2 + 2 s3^2, 2 s3^2}, {2 s3^2, 2 s2^2 + 2 s3^2}}*)


We have an extremum if the determinant is >0:

Det[hes]
(*4 s1^2 s2^2 + 4 s1^2 s3^2 + 4 s2^2 s3^2*)


This shows that we indeed have an extremum. We will have a minimum if both second derivatives are >0. That is

hes[[1,1]]>0
hes[[2,2]]>0


As both terms consist of squares we have a minimum.

• Many thanks for writing an answer. I modified my question in order to become simpler. I have a question: Now, we know that $a_i = \frac{\sigma_i^{-2}}{\sum_{j = 1}^2 \sigma_j^{-2}}$ minimizes the function $f$. Is there any way, in particular visualization, to see how the values of $f$ changes if we alter the exponents in $a_i$. In other words, when one changes the exponents from $-1$ to $-3$, how $f$ reaches its minimum at $-2$.
– ARez
Sep 26 at 20:46
• Changing the question after getting an answer makes for confusion. Would be more clear if asked as a new question. Sep 26 at 22:25
• @DanielLichtblau Sorry, I didn't know the culture here. Should I open a new question?
– ARez
Sep 26 at 22:33
• @ARez yes, you should open a new question as opposed to changing your original question (especially when it has been answered). Sep 26 at 22:48