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minimize $3x_1^2+3x_2^2-2x_1x_2-12(x_1+x_2)+36$

s.t. $|x1+x2|=1$

Determine the maximum and minimum of this problem.

The Lagrangian can be set up as:

$L(x,\lambda)= 3x_1^2+3x_2^2-2x_1x_2-12(x_1+x_2)+36-\lambda(x1+x2-1)$

From the conditions I infer that for $\lambda =-10$ and $x_1=\frac{1}{2}$ and $x_2=\frac{1}{2}$ are solutions.

Here are my two questions.

Is there only one constraint or two because there is an absolute sign? (I tried using the constraints $x1+x2=1$ and $x1+x2= -1$ but that simply doesn't work because you then get $1-x_2+x_2+1 \neq 0$.

And thus I come to my second question, is the Lagrangian and the solutions that I inferred right?

EDIT: I thus find the following solutions:

Case 1 ($x_1+x_2=1$):

$(\frac{1}{2},\frac{1}{2})$ with $\lambda = -10$

$(-\frac{1}{2},-\frac{1}{2})$ with $\lambda = -14$

The Hessian here is:

$\begin{bmatrix} 6&-2\\-2 &6\end{bmatrix}$ so the two points above are minimum since the eigenvalues are 4 and 8 which are greater than $0$.

Case 2 ($-x_1-x_2= 1$):

$(\frac{1}{2},\frac{1}{2})$ with $\lambda = 10$

$(-\frac{1}{2},-\frac{1}{2})$ with $\lambda = 14$

$\begin{bmatrix} -6&2\\2 &-6\end{bmatrix}$ so the two points above are maximum since the eigenvalues are -4 and -8.

Steven
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    It it necessary to use Lagrange multipliers? I would think a simple two-case substitution and usual one-variable differentiation should be more than enough. Also, the absolute value separates into the two cases $x_1 + x_2 = 1$ and $x_1 + x_2 = -1$, but they need to be treated separately, not simultaneously. – Arthur Aug 05 '15 at 10:10
  • I'd like to solve it using Lagrange multipliers. But that's my point, the last two conditions are found by differentiating with lambda but if I have the two cases which you mentioned then I get $x_1+x_2-1=0$ and $x_1+x_2+1=0$. But I'm confused, these are separated and linearly indepent so I can fill in one in the other and then I get that $2 \neq 0$. If it doesn't take too much of your time can you please show me the correct way of doing it using Lagrange multipliers? – Steven Aug 05 '15 at 10:15
  • @Stephen You're misinterpreting what it means to treat those cases separately. It means that you do the whole Lagrange process twice, once with $x_1 + x_2 = 1$ as the constraint, and once with $x_1 + x_2 = -1$ as a constraint. You then compare which of the two cases gave the smallest answer. – Arthur Aug 05 '15 at 10:19
  • You're right. Thank you very much ! – Steven Aug 05 '15 at 10:23
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    Take the square of your constraint in order to consider only one case. Modify the Lagrangian accordingly. – RTJ Aug 05 '15 at 10:53

1 Answers1

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i would not go by Lagrange Multipliers i would reduce the problem to one variable, for e.g. $$x_1+x_2\geq 0$$ then we obtain $$x_1+x_2=1$$ and we get $$f(x_1,1-x_1)=8x_1^2-8x_1+27$$ therefore the searched minimum is $$25$$ for $$x_1=\frac{1}{2}$$ and also the other case is simple

Dr. Sonnhard Graubner
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