What is a boundary point when solving for a max/min using Lagrange Multipliers? After you solve the required system of equation and get the critical maxima and minima, when do you have to check for boundary points and how do you identify them?

e.g. Optimise (1+a)(1+b)(1+c) given constraint a+b+c=1, with a,b,c all non-negative.

After using the Lagrange multiplier equating the respective partial derivatives, I get (a,b,c)=(1/3, 1/3, 1/3). Clearly there must be both a maximum and minimum, and I assume this is the maximum. Where is the minimum? (0,0,1) optimises best for the minimum, and I assume using 0 is a boundary point but why? And what effect does the restriction to non-negative reals have?


Your example serves perfectly to explain the necessary procedure.

You are given a function $f(x,y,z):=(1+x)(1+y)(1+z)$ in ${\mathbb R}^3$, as well as a compact set $S\subset{\mathbb R}^3$, and you are told to determine $\max f(S)$ and $\min f(S)$.

Differential calculus is a help in this task insofar as putting suitable derivatives to zero brings interior stationary points of $f$ in the different dimensional strata of $S$ to the fore. The given simplex $S$ is a union $S=S_0\cup S_1\cup S_2$, whereby $S_0$ consists of the three vertices, $S_1$ of the three edges (without their endpoints), and $S_2$ of the interior points of the triangle $S$.

If the global maximum of $f$ on $S$ happens to lie on $S_2$ it will be detected by Lagrange's method, applied with the condition $x+y+z=1$. If the maximum happens to lie on one of the edges it will be detected by using Lagrange's method with two conditions, or simpler: by a parametrization of these edges (three separate problems!). If the maximum happens to lie at one of the vertices it will be taken care of by evaluating $f$ at these vertices.

In all we obtain a (hopefully finite) candidate list $\{p_1,p_2,\ldots, p_N\}$. The global maximum of $f$ on the set $S$ will be the largest of the values $f(p_k)$ $(1\leq k\leq N)$. Note that we don't need to compute any second derivatives.

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At first - about elementary way. $$(1+a) + (1+b) + (1+c) = 4.$$ Using AM-GM, one can get: $$(1+a)(1+b)(1+c)\le \left(\dfrac{1+a+1+b+1+c}3\right),$$ so $\left(\dfrac13,\dfrac13,\dfrac13\right)$ is maximum. Note that the issue conditions are significant in this case.

Partitial derivatives of Lagrange multipliers method for $$f(a,b,c,\lambda) = (1+a)(1+b)(1+c)+\lambda(a+b+c-1)$$ can give $$\begin{cases} (1+b)(1+c) + \lambda = 0\\ (1+a)(1+c) + \lambda = 0\\ (1+a)(1+b) + \lambda = 0\\ a+b+c = 1 \end{cases}$$

$$\begin{cases} (b-a)(1+c) = 0\\ (1+a)(c-b) = 0\\ a+b+c =1, \end{cases}$$ and one can get that $\left(\dfrac13,\dfrac13,\dfrac13\right)$ is multiple root for maximum.

So the function has not a global minima, and boundary conditions work.

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First of all, if the non negativity condition is not given (if a,b,c can be any real numbers), then there is no minimum. Indeed, let c=0, a be a large negative number, b be a large positive number such that a+b=1. Hence (1+a)(1+b)(1+c) tends to $-\infty$.

When it is solved by the Lagrange multipliers method, four (not one) constraints must be considered.

Optimize $(1+a)(1+b)(1+c)$ subject to $a+b+c=1, a,b,c\geq0$.

Then the Kuhn-Tucker conditions must be checked by considering various cases...

Another approach (to imagine better): let's look at the 2-variable function:

Optimize $z=(1+x)(1+y)$ subject to $x+y=1, x,y\geq0$.

Substitute $y=1-x$ into the objective function: $z=(1+x)(1+1-x)=-x^2+x+2.$

Equivalent problem: Optimize $z=-x^2+x+2$ subject to $x\geq0$.

According to the Extreme Point Theorem, the extreme values of the function occur either at the border or the critical point(s).

Border: x=0. Critical point(s): $z'_x=0 \Rightarrow -2x+1=0 \Rightarrow x=\frac{1}{2}.$

Evaluation: $z(0)=2 - min$; $z(\frac{1}{2})=\frac{9}{4} - max.$

Or referring to the initial two variable objective function $z=(1+x)(1+y):$

$z(0,1)=2 - min; z(\frac{1}{2},\frac{1}{2})=\frac{9}{4} - max$.

Note: Now it can be generalized to the 3-variable function.

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