MATH.13:5.1 - A symmetric allocation and an asymmetric optimum

Let x1,x2>=0, x1+x2=10, and minimize J=x1^2+x2^2. Swapping x1 and x2 preserves both the feasible set and J.

The cost is strictly convex on this feasible segment, and a minimizer exists because the segment is compact and the cost is continuous. Hence the minimizer is unique. The symmetry condition requires (x1,x2)=(x2,x1), so the only possible optimum is (5,5). The direct calculation J(5+h,5-h)=50+2*h^2 confirms it for every feasible h.

Now use J=x1^2+2*x2^2 under the same resource constraint. Swapping the allocations changes the cost: J(10,0)=100 and J(0,10)=200. The original exchange symmetry is gone. The optimum is (20/3,10/3), as the admissible-variation construction in MATH.10 shows.

Symmetry alone also need not make individual solutions symmetric. If the original sum-of-squares criterion is maximized on the same segment, (10,0) and (0,10) are both maxima. The swap exchanges them. The midpoint is fixed by the swap but is the minimum, so selecting it from symmetry without the uniqueness and optimization premises would answer a different question.