Lagrange multipliers not giving all the solutions

Question

I was working on a maximization problem that can be reduced to maximizing
$$f(a,b,c,d)= ab + bc + cd $$
subject to the constraint
$$a +b +c +d = 63 $$
I thought that this was a trivial and straightfoward problem for the method of lagrange multiplier, so I defined the Lagrangian as
$$\mathcal{L}(a,b,c,d,\lambda) = ab + bc + cd – \lambda (a +b +c +d – 63) $$
and setting the derivatives equal to 0, I got
$$
b – \lambda = 0 \\
a + c – \lambda = 0 \\
b +d – \lambda = 0 \\
c – \lambda = 0 \\
a +b +c +d = -63
$$
That is a linear problem that can be re-stated as finding the solution for the system $Ax = b$ with
$$ A = \begin{bmatrix}
0 & 1 & 0 & 0 & -1 \\
1 & 0 & 1 & 0 & -1 \\
0 & 1 & 0 & 1 & -1 \\
0 & 0 & 1 & 0 & -1 \\
1 & 1 & 1 & 1 & 0
\end{bmatrix}$$
and
$$
b = \begin{bmatrix}
0 & 0 & 0 & 0 & -63 \\
\end{bmatrix}^T$$
to my surprise, this system is determined with only one solution for $a,b,c$ and $d$
$$ x = \begin{bmatrix}
0 & 31.5 & 31.5 & 0 \\
\end{bmatrix}^T$$
But a direct evalution of the function to maximize and the constrains shows that, for example, the vector
$$ x = \begin{bmatrix}
31.5 & 31.5 & 0 & 0 \\
\end{bmatrix}^T$$
yields the same result for the function and is also part of the constraint surface, so it should be a critical point too, more over, there are infinetly many critical points, so way is the lagrange multipliers just giving me one result?

Answer 1

There is only one critical point but it is indefinite. This is seen for example by eliminating $d$ and studying the Hessian of the resulting unconstrained problem (see below for the details). Therefore there are no local extrema.

Your function is unbounded. For example, if we set $c=0$, then the constraint is satisfied by $d=63-a-b$, but this will not affect the value of $f$. Take a look: $$ f(a,b,0,63-a-b)=ab. $$ Clearly we can make this as large or as small as we wish, and there is no global maximum or minimum.

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IMHO it is usually easier to use a linear constraint simply to eliminate one of the variables. If we here solve $d=63-a-b-c$ from the constraint, we get the function in three variables $$ g(a,b,c)=f(a,b,c,63-a-b-c)=ab-ac-c^2+63c. $$ Its gradient is $$ \nabla g(a,b,c)=(b-c,a,63-a-2c), $$ and this vanishes only at the critical point $P_1=(a,b,c)=(0,63/2,63/2)$. The Hessian of $g$ at $P_1$ is $$ H(g,P_1)=\left(\begin{array}{ccc}0&1&-1\\ 1&0&0\\ -1&0&-2\end{array}\right). $$ Its characteristic polynomial is $$ \chi_H(x)=\det(x I_3-H)=x^3+2x^2-2x-2. $$ We could solve for the zeros using Cardano, but for our purposes it suffices to check that $\chi_H(x)$ has zeros in the intervals $\lambda_1\in(-3,-2)$, $\lambda_2\in(-1,0)$, $\lambda_3=(1,2)$. Both signs occur, so the Hessian form is indefinite. In other words, $P_1$ is a saddle point. You can use Sylvester's criterion to reach the same conclusion — $P_1$ is not a local extremum.