Given k natural numbers {a1,…,ak}⊂N with 1≤a1<a2<…<ak and gcd(a1,…,ak)=1, let be R(a1,…,ak)={λ1a1+⋯+λkak| λi∈N,i=1÷k} and ¯R(a1,…,ak)=N∖R(a1,…,ak). It is easy to see that |¯R(a1,…,ak)|<∞. The Frobenius Problem related to the set {a1,…,ak} consists on the computation of f(a1,…,ak)=max, also called the \textit{Frobenius number}, and the cardinal | \overline{R}(a_1, \ldots ,a_k)|. The solution of the Frobenius Problem is the explicit computation of the set \overline{R} (a_1,\ldots ,a_k). In some cases it is known a sharp upper bound for the Frobenius number. When k=3 this bound is known to be F(N)=\max\limits_{\substack{0 \lt a \lt b \lt N \\ \mathrm{gcd}(a,b,N)=1}} f(a,b,N)= \begin{cases} 2(\lfloor N/2 \rfloor -1)^2-1 & \textrm{if } N \equiv 0 (\mod 2),\\ 2 \lfloor N/2 \rfloor (\lfloor N/2 \rfloor -1) -1 & \textrm{if } N \equiv 1 (\mod 2).\\ \end{cases} This bound is given in [Dixmier1990]. In this work we give a geometrical proof of this bound which allows us to give the solution of the Frobenius problem for all the sets \{\alpha ,\beta ,N\} such that f(\alpha ,\beta ,N)=F(N).