(1) Assume S(A)=S multiplied by A.
Graphical representation is right triangle OPQ, where O(0,0), P(0,1), Q(1/S,0), so OPQ is the feasibility region.
(a) Optimal solution:
A, B≥0⇒S≥0⇒
0<S<1⇒max(A+B)=1/S (A=1/S, B=0);
S≥1⇒max(A+B)=1 (A=0, B=1);
S=0⇒max(A+B)=1 (A any value, B=1).
(b) Infeasible: S<0.
(2) Assume S is a function of A.
(a) Optimal solution:
Range of S must include S(A)≥0 for some A≥0:
0<S(A)<1⇒max(A+B)=1/S(A) (B=0, for some A);
S(A)≥1⇒max(A+B)=1 (B=1, for some A);
S(A)=0⇒max(A+B)=1 (B=1, for some A).
(b) Infeasible: S(A)<0 for all A.
EXAMPLE: S(A)=3A-A²-2, but this would be non-linear.
1≤A≤2⇒min(S(A))=0, max(S(A))=¼; A<1 or A>2⇒S(A)<0.
(a) max(A+B)=4 at A=3/2, B=0
(b) Infeasible when A<1 or A>2.
Still investigating for (c).