First of all we will say
Budget"30=9C+3S or S=10-3C"
Therefore
"U = S\u2022C+10(S+C) = S\u2022C+10S+10C = C(10-3C)+10(10-3C)+10C"
Simplify to
"U = 10C-3C\u00b2+100-30C+10C = 100-10C-3C\u00b2"
this means
"C\u22650 and from budget: C[MAX]=30\/9\u22483.33 ; at S=0"
"S\u22650 and from budget: S[MAX]=30\/3=10.00 ; at C=0"
"3.33\u2265C\u22650"
"0\u2264S\u226410"
Now maximize U(C)→MAX
100-10C-3C²→MAX & 10/3≥C≥0
∂U(C)/∂C= -10-3C < 0
Maximum will be a lower boundary since its downward sloping for all allowed ranges.
Therefore utility-maximizing solution backet is:C=0 & S=10 &U[MAX]=100.
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