MATH.10:5.1 - Reallocate a fixed total, then change the criterion

Allocate ten divisible units between two uses: x1>=0, x2>=0 and x1+x2=10. The supplied cost is J=x1^2+x2^2. Take the admissible family (x1+h,x2-h), with -x1<=h<=x2.

Substitution gives

DeltaJ(h)=2*(x1-x2)*h+2*h^2.

For an interior candidate, both signs of small h are allowed. The first-order condition gives x1=x2, hence (5,5). At this candidate, the entire finite difference is 2*h^2. Every feasible allocation is (5+h,5-h) for -5<=h<=5, so this proves the unique global minimum, with cost 50.

Now change the cost to J=x1^2+2*x2^2, retaining the total and nonnegativity. The same family gives

DeltaJ(h)=2*(x1-2*x2)*h+3*h^2.

The condition becomes x1=2*x2, giving (20/3,10/3). Its difference is 3*h^2 across the whole feasible interval, so its cost 200/3 is the unique minimum. Keeping the earlier equal split would miss the change introduced by the new criterion.

If the units must instead be whole, x1 and x2 are integers. Completing the square gives J=3*(x1-20/3)^2+200/3; the admissible integer closest to 20/3 is 7. Thus (7,3) has the lowest cost, 67, in that discrete problem. The continuous stationary point helped locate the candidates; the integer comparison settles their use.

Keep the weighted cost J=x1^2+2*x2^2, make the units divisible again, and add x1<=6 while retaining the total and nonnegativity. The previous continuous minimizer (20/3,10/3) is now unavailable. At (6,4), the allowed family has -6<=h<=0 and

DeltaJ(h)=-4*h+3*h^2>=0.

This family covers every newly feasible allocation. Thus (6,4), with cost 68, is the unique global minimizer despite its nonzero derivative along the unrestricted line. The parameter domain carries the decisive information.

These are calculations for the stated cost model. Its use in an allocation decision requires the supplied cost to represent the consequence being compared.