Volume máximo de um cilindro

Sem utilizar derivadas:

[latex]\\\mathrm{V_c=2\pi r^2_cx=2\pi x\left ( R^2-x^2 \right )=2\pi x(R+x)(R-x)}\\\\ \mathrm{Sejam\ a\ e\ b\ constantes,logo,V_c=\frac{2\pi }{ab}\left [ ax(R+x)b(R-x) \right ]}\\\\ \mathrm{Manipulando\ igualdade\ anterior:\sqrt[3]{\mathrm{V_c}}=\sqrt[3]{\mathrm{\frac{2\pi }{ab}}}\sqrt[3]{\mathrm{\left [ ax(R+x)b(R-x) \right ]}}}\\\\ \mathrm{Da\ Desigualdade\ das\ M\acute{e}dias:\ M_G\leq M_A,logo:}\\\\ \mathrm{\sqrt[3]{\mathrm{V_c}}=\sqrt[3]{\mathrm{\frac{2\pi }{ab}}}\sqrt[3]{\mathrm{\left [ ax(R+x)b(R-x) \right ]}}\leq \sqrt[3]{\mathrm{\frac{2\pi }{ab}}}\left [ \frac{x+a(R+x)+b(R-x)}{3} \right ]}\\\\ \mathrm{\sqrt[3]{\mathrm{\left [ ax(R+x)b(R-x) \right ]}}\leq \left [ \frac{R(a+b)+(a-b+1)x}{3} \right ]}\\\\ \mathrm{Impondo\ a-b=1\ para\ anularmos\ o\ termos\ "(a-b+1)x"\ e\ sabendo\ }\\\\ \mathrm{M_G=M_A\ se\ x=a(R+x)=b(R-x),vem:a=\frac{x}{R+x}\ e\ b=\frac{x}{R-x}}\\\\ \mathrm{\therefore\ \frac{x}{R+x}-\frac{x}{R-x}=1\to x=\frac{R}{\sqrt{3}}\ \therefore\ r_c=\sqrt{R^2-x^2}=\sqrt{2}, pois\ R=\sqrt{3 }} [/latex]