[tex]\displaystyle\bf\\\frac{a}{b-15}=\frac{a+14}{b} \Leftrightarrow ab=(a+14)(b-15)=ab-15a+14b-210.\\ab=ab-15a+14b-210~\Big |-ab \implies -15a+14b-210=0.\\-15a=210-14b=\mathcal{M}_{14} \implies a=\mathcal{M}_{14} \implies \exists n\in\mathbb{N}~a.i.~\boxed{\bf a=14n}.\\-15\cdot14n=210-14b \Leftrightarrow 14b=210+15\cdot14n |:14 \implies\\\boxed{\bf b=15+15n=15(n+1)}.\\[/tex]
[tex]\displaystyle\bf\\am~redus~problema~la~a~afla~valoarea~maxima~a~~\frac{15(n+1)}{14n}.\\\frac{15(n+1)}{14n}=\frac{15n+15}{14n}=\frac{15n}{14n}+\frac{15}{14n}=\frac{15}{14}+\frac{15}{14n},~valoarea~maxima\\este~pentru~valoarea~maxima~a~fractiei~\frac{15}{14n}~care~se~atinge~pentru\\\boxed{\bf n=1} \implies \boxed{\bf~a=14~si~b=30}.[/tex]
