Finding maximum surface area of a box with fixed diagonal - issue with Lagrange multipliers

Question

Consider a box with dimensions $x_1, x_2, x_3$ whose diagonal is $1$ (i.e. $x_1 ^2+ x_2^2+ x_3^2 = 1$ ). Its surface area is $2(x_1x_2 + x_2x_3 + x_1x_3)$. I am trying to find such a box with maximal surface area. When I use Lagrange multipliers I get that there is no maximum, which is strange because this problem is assigned in a textbook. I am not sure if my reasoning is mistaken, or if there is really no solution.

So the problem is $$\min f(x) = -(x_1x_2 + x_2x_3 + x_1x_3)$$ $$\text{s.t. } \hspace{0.1cm }h(x) = x_1 ^2+ x_2^2+ x_3^2 - 1 = 0$$ with the implicit assumption that $x_1, x_2, x_3 > 0$. Assuming there is a local minimum, by the Lagrange multiplier theorem, there exists a real number $-\dfrac {\lambda}{2}$ such that $$\nabla f(x) - \dfrac {\lambda}{2} \nabla h(x) = 0.$$

Now $$\nabla f(x) = -\begin{bmatrix}x_2 + x_3\\ x_1 + x_3\\ x_1 + x_2 \end{bmatrix} \text{and } \nabla h(x) = \begin{bmatrix}2x_1\\ 2x_2 \\ 2x_3 \end{bmatrix}$$

This gives the system of equations $$ \begin{align*} \lambda x_1 + x_2 + x_3 & = 0\\ x_1 + \lambda x_2 + x_3 & = 0\\ x_1 + x_2 + \lambda x_3 & = 0 \end{align*}$$

Adding all of these equations together yields $$(1 + \lambda)(x_1 + x_2 + x_3) = 0$$ If $\lambda \not = -1$, then we have $x_1 + x_2 + x_3 = 0$. However, this is not possible under the implicit assumption that $x_1, x_2, x_3 > 0$, so we conclude that $\lambda = -1$.

But if $\lambda = -1$, then the above system becomes $$\begin{align} - x_1 + x_2 + x_3 & = 0\\ x_1 - x_2 + x_3 & = 0\\ x_1 + x_2 - x_3 & = 0 \end{align}$$

whose only solution is $(0, 0, 0)$, which is unacceptable.

Answer 1

Forming the lagrangian

$$ L = x y + y z + z x +\lambda(x^2+y^2+z^2-1) $$

the stationary points are the solutions for

$$ \nabla L = 0 = \cases{2 \lambda x+y+z\\ 2 \lambda y+x+z\\ 2 \lambda z+x+y\\ x^2+y^2+z^2-1} $$

now to solve this system of equations, first we will try to solve

$$ \cases{2 \lambda x+y+z=0\\ 2 \lambda y+x+z=0\\ 2 \lambda z+x+y=0} $$

but the unique solution gives $x=y=z=0$ which is unacceptable. To have a non trivial solution we need that

$$ \det\left( \begin{array}{ccc} 2 \lambda & 1 & 1 \\ 1 & 2 \lambda & 1 \\ 1 & 1 & 2 \lambda \\ \end{array} \right) = 2 (\lambda +1) (2 \lambda -1)^2 = 0\ \ \ \ \ \ \ \ \ (1) $$

so this operation determines the feasible $\lambda$. Now taking $x,y$ for instance from

$$ \cases{2 \lambda x+y+z=0\\ 2 \lambda y+x+z=0} $$

and substituting into $x^2+y^2+z^2-1$ we get

$$ \frac{2 z^2}{(2 \lambda +1)^2}+z^2-1=0 $$

and solving for $z$ after substituting $\lambda$ with the solutions from $(1)$ we will have all the stationary points from the lagrangian: after that we should choose the feasible solutions.

Answer 2

As pointed out by "peter a g" in the comments, the issue is: The equation $$(1 + \lambda)(x_1 + x_2 + x_3) = 0$$

should instead be $$(2 + \lambda)(x_1 + x_2 + x_3) = 0$$ This leads us to $\lambda = -2$, and solving the corresponding system gives $x_1 = x_2 = x_3$. Plugging this into $x_1^2 + x_2^2 + x_3^2 = 1$ yields $x _1 = x_2 = x_3 = \dfrac {1}{\sqrt 3}.$

Since the constraint set $C$ is compact, $f(x) = x_1x_2 + x_2 x_3 + x_1 x_3$ (restricted to $C$) achieves a global maximum on $C$. To finish, we would have to find all solutions of the system obtained using the Lagrange multipleir method, and check all of them.