Critical points, minima and maxima of a constrained optimisation problem

Question

$f(x_1,x_2,x_3) = x_1 x_2 + x_2 x_3 + x_3 x_1$ with the constraint $x_1+ x_2 + x_3 = 1$.

Now I test the critical points by taking $\nabla f = 0$, hence at $(x_2 + x_3, x_1 + x_3, x_2 + x_1) =0$. This only occurs when $x_1 = x_2 = x_3 = 0$, hence there are no critical points with the constraint. Therefore we must check the boundary. The max should exist at $x_1=x_2=x_3= \frac13$, and the minimum I will work out after classes that I must now attend.

Answer 1

Since both the function and the constraint equation are unchanged (invariant) by "exchanges" among the three variables, there is symmetry about the line $ \ x_1 \ = \ x_2 \ = \ x_3 \ $ We would expect then that there is a critical point where that line intersects the plane $ \ x_1 \ + \ x_2 \ + \ x_3 \ = \ 1 \ $ , giving the result found in the other answers.

It should be said that the function has no global minimum under the single constraint. For $ \ x_1 x_2 + x_2 x_3 + x_3 x_1 \ = \ c \ $ , the level surfaces for positive $ \ c \ $ are two separate objects in the first octant and the one directly opposite through the origin. But for negative $ \ c \ $ , the separate surfaces "merge" into a single hyperboloid of one sheet, which always intersects the constraint plane. An example is graphed below:

two views of the hyperboloid $ \ x_1 x_2 + x_2 x_3 + x_3 x_1 \ = \ -10 \ $ and the constraint plane $ \ x_1 \ + \ x_2 \ + \ x_3 \ = \ 1 \ $ ; there is a circle of intersection for all level surfaces $ \ c \ < \ \frac{1}{3} \ $ ; this circle increases in diameter with decreasing values of $ \ c \ $

If there is a further restriction that the variables must be non-negative, then the maximum of our function has been found in the interior of the triangle in the first octant on the constraint plane. We still need to check the boundaries of this triangle; owing to the "triangular" symmetry mentioned, we can just check one vertex and one edge of the triangle.

At $ \ (1, \ 0, \ 0 ) \ \ , \ f(x_1, \ x_2, \ x_3) \ $ is zero, so the value is the same at all three vertices. In the plane $ \ x_3 \ = \ 0 \ $ , the "edge" of the triangle is $ \ x_1 \ + \ x_2 \ = \ 1 \ \Rightarrow \ x_2 \ = \ 1 \ - \ x_1 \ $ . The function becomes $ \ f(x_1, \ x_2, \ x_3) \ = \ x_1 \ ( 1 \ - \ x_1 ) \ $ , for which it is straightforward to locate a local maximum at $ \ (\frac{1}{2}, \ \frac{1}{2}, \ 0) \ $ , where the value of the function is $ \ \frac{1}{4} \ $ . The same value occurs at the midpoint of all three edges then; however, we see that these are not global extrema.

Answer 2

Let $g(x_1,x_2,x_3)=x_1+x_2+x_3$. Using Lagrange multipliers, setting $f_{x_{1}}=\lambda g_{x_{1}}, f_{x_{2}}=\lambda g_{x_{2}}, f_{x_{3}}=\lambda g_{x_{3}}$

gives $x_2+x_3=\lambda, x_1+x_3=\lambda, x_1+x_2=\lambda$ and therefore $x_1=x_2=x_3$.

Then substituting into $x_1+x_2+x_3=1$ gives $x_1=x_2=x_3=\frac{1}{3}$.

Answer 3

Or look at: $$g(x_1,x_2) := f(x_1,x_2, 1 - x_1 - x_2) = x_1x_2 + (1 - x_1 - x_2)(x_1 + x_2)$$ Now we find the critical points of $g$. Simplifying, we get: $$g(x_1, x_2) = x_1x_2 + x_1 + x_2 - x_1^2 - x_1x_2 - x_2x_1 - x_2^2$$ And cancelling: $$g(x_1,x_2) = x_1 - x_1^2 + x_2 - x_2^2 - x_1x_2$$ Now, taking the derivative: $$\nabla g = (1 - 2x_1 - x_2, 1 - 2x_2 - x_1) = \mathbf{0}$$ It's not hard to see that $x_1 = x_2 = 1/3$ really is a solution. Since the equations are linearly independent, it is the only solution. Going back all the way, this gives $x_3 = 1/3$ too.