Respuesta :

[tex]\bf a)\\\\ V=\stackrel{\textit{volume of cone}}{\cfrac{\pi r^2\cdot \boxed{3r}}{3}}+\stackrel{\textit{volume of hemisphere}}{\cfrac{2\pi r^3}{3}}\implies V=\cfrac{3\pi r^3}{3}+\cfrac{2\pi r^3}{3} \\\\\\V=\cfrac{5\pi r^3}{3} \\\\\\ S=\stackrel{\textit{lateral area of cone}}{\pi r\sqrt{r^2+\boxed{3^2r^2}}}+\stackrel{\textit{area of hemisphere}}{2\pi r^2}\implies S=\pi r\sqrt{r^2+9r^2}+2\pi r^2[/tex]

[tex]\bf S=\pi r\sqrt{10r^2}+2\pi r^2\implies S=\pi r^2\sqrt{10}+2\pi r^2 \\\\\\ S=\pi r^2(2+\sqrt{10})\\\\\\ b)\\\\ \stackrel{A=kS}{A=k\cdot \pi r^2(2+\sqrt{10})}\qquad \qquad \stackrel{C=kV}{C=k\cdot \cfrac{5\pi r^3}{3}}[/tex]

[tex]\bf c)\\\\ A\ge C\implies k\cdot \pi r^2(2+\sqrt{10})\ge k\cdot \cfrac{5\pi r^3}{3} \\\\\\ k\pi r^2(2+\sqrt{10})\ge \cfrac{5k\pi r^2r}{3}\implies \cfrac{\underline{k\pi r^2}(2+\sqrt{10})}{\underline{k\pi r^2}}\ge\cfrac{5r}{3} \\\\\\ 3(2+\sqrt{10})\ge 5r\implies 6+3\sqrt{10}\ge 5r\implies \cfrac{6+3\sqrt{10}}{5}\ge r[/tex]

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