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Disclaimer: In the definition (Stewart Calculus, 7E): "Method of Lagrange Multipliers" part (b)- Evaluate $f$ at all extreme points $(x,y,z)$ from step a. The largest of these values is the maximum and the smallest of these values are the minimum.

It is impossible to infer whether an extreme point is a maximum or minimum from this definition when there is only one extreme point in the constraint. Can it be either proven or inferred that it is a maximum or minimum value using another method? Wolfram Alpha Lagrange Calculator said it was a minima (this is an even number exercise, ie no answer in back of book).

Second Disclaimer: I will show my work as to how I only got one value for verification, otherwise if not verified and there is an error the question could be ended there depending on the nature of the error. Then I would like you to tell me how/if it can be proven or inferred that a extreme value in Lagrange's method is a max or min when Stewart's definition doesn't apply and there is only one point.

Function: $f(x,y,z) = x^2 + y^2 + z^2$

Constraint: $x + y + z = 12$

$\nabla f$ = $\lambda \nabla g $:

$2x = \lambda$

$2y = \lambda$

$2z = \lambda$

Multiplying Equation 1 by yz, Equation 2 by xz, Equation 3 by xy. Gives the following:

$2xyz = \lambda yz$

$2xyz = \lambda xz$

$2xyz = \lambda xy$

Which implies:

$\lambda yz = \lambda xz = \lambda xy \to x=y=z$

We will say that $x=y=z=a$ so that plugging into the constraint gives:

$3a = 12 \to a = 4 \to x=y=z=4$

We have our only extreme value at $f(4,4,4)= 48

How is this a minimum?

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  • $\begingroup$ This isn't really a direct answer, but from the variable symmetry in the equations, you can tell that the minimum should be of the from (x,x,x) $\endgroup$ – SquirtleSquad May 20 '16 at 23:52
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In this case, simple geometric considerations tell you that you have a minimum: the objective function is the square of the distance from the origin and the constraint is the equation of a plane.

More generally, one can examine the bordered Hessian to determine the nature of the stationary points found via Lagrange multipliers. In this case, $Hf = 2I_3$ and $\nabla g=(1,1,1)$, so the bordered Hessian is $$\pmatrix{0&1&1&1 \\ 1&2&0&0 \\ 1&0&2&0 \\ 1&0&0&2}.$$ There is one constraint, so we examine the determinants of the entire matrix and the $3\times3$ principal minor. They are $-12$ and $-4$, respectively, which means that the stationary point gives a (constrained) local minimum.

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  • $\begingroup$ interesting information thank you $\endgroup$ – K. Gibson May 23 '16 at 23:20
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Well you can always compare it to other values. For example, take $x=1, y=1, z=10$, $x+y+z=12$, and $f(x)=1+1+100=102>48$

So 48 has to be a minimum.

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