## 28 Reputation

17 years, 276 days

## maybe not in that case?...

Hi, I have no time to prove the identity you mentioned here, but even if it were true, it might not work in the case as the matrix K is a singular one. I am eager to see the proof first of all.

Removed.

## I like the solution, but it would be dif...

I like the solution that Roman posted but I must admit it would be rather difficult for me to think of. Thereby, I paste my rather heuristic one: ```Obj := (1/4*a[1]+3/4*a[2]+7/8*a[3]+a[4]+15/8*a[5]+2*a[6] +19/8*a[7]+21/8*a[8]-1/4*b[1]-3/4*b[2]-7/8*b[3]-b[4]-15/8*b[5] -2*b[6]-19/8*b[7]-21/8*b[8])^2+(1/4*b[1]+3/4*b[2]+7/8*b[3]+b[4] +15/8*b[5]+2*b[6]+19/8*b[7]+21/8*b[8]-1/4*a[1]-3/4*a[2]-7/8*a[3] -a[4]-15/8*a[5]-2*a[6]-19/8*a[7]-21/8*a[8])^2;``` ```#noting symetry, lets incorporate variable c=a-b into the model rep:={seq(a[i]=c[i]+b[i],i=1..8)}: New_obj:=subs(rep,Obj);``` ```#as New_obj satisfies the equation #New_Obj = 2*(nObj)^2 implying thus that #maximum of New_Obj and nObj considering resp. constraints #are accomplished whenever nObj is max or min #we need only to adjust constraints for c #de facto it is sufficient to find maximum of nObj only #and taking -1*nObj yelds min nObj:=1/4*c[1]+3/4*c[2]+7/8*c[3]+c[4]+15/8*c[5]+2*c[6]+19/8*c[7]+21/8*c[8]; nC1:=add(c[i],i=1..8)=0; nC2:=seq(c[i]^2=1,i=1..8); with(Optimization): NLPSolve(nObj, {nC1, nC2}, maximize=true): extremum_of_nObj:=%[1]; ext_at_point=%%[2]; clearly New_Obj is nonnegative and the question is if it can be zero 'solve' provides the answer here_nObj_is_zero=solve({nObj, nC1, nC2}, {seq(c[i],i=1..8)}); ```

## This is weird!...

Maybe I'm missing something, but it appears to me as: Obj:=(1/4*a[1]+3/4*a[2]+7/8*a[3]+15/8*a[5]+2*a[6]+19/8*a[7]+21/8*a[8]+1/4*b[1]+ 3/4*b[2]+7/8*b[3]+15/8*b[5]+2*b[6]+19/8*b[7]+21/8*b[8])^2= 1/4*(a[1]+b[1])+3/4*(a[2]+b[2])+7/8*(a[3]+b[3])+...+21/8*(a[8]+b[8]) what in connection with CON3 is equal to 1/4+3/4+7/8+...+21/8, which is constant and there is nothing to minimize! What is it about, then?

## For me its a LP...

As I noticed, the obj. function is a square of a linear combination of a's and b's.

## making [a=1, b=2, c=3, and so on]...

Why did none of you mention the 'Equate' command?? it solves the first problem offhand, doesn't it? Happy New Year....
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