+9466 volume of candle = \(\frac13\pi r^2h\) Since h = r , we can replace h with r .
volume of candle = \(\frac13\pi r^2r\)
volume of candle = \(\frac13\pi r^3\)
surface area of candle = \(\pi r^2+\pi r\sqrt{r^2+h^2}\) Since h = r , we can replace h with r .
surface area of candle = \(\pi r^2+\pi r\sqrt{r^2+r^2}\)
surface area of candle = \(\pi r^2+\pi r\sqrt{2r^2}\)
surface area of candle = \(\pi r^2+\pi r\cdot r\sqrt{2}\)
surface area of candle = \(\pi r^2+\pi r^2\sqrt{2}\)
ratio of the volume of candle to its surface area = \(\frac{\text{volume of candle}}{\text{surface area of candle}}\)
\(\frac{\text{volume of candle}}{\text{surface area of candle}}\,=\,\frac{\frac13\pi r^3}{\pi r^2+\pi r^2\sqrt2}\\~\\ =\,\frac{\frac13r^3}{r^2+r^2\sqrt2} \\~\\ =\,\frac{\frac13r}{1+\sqrt2} \\~\\ =\,\frac{r}{3(1+\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1+\sqrt2)(1-\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1-2)} \\~\\ =\,\frac{r(1-\sqrt2)}{-3} \)
+9466 volume of candle = \(\frac13\pi r^2h\) Since h = r , we can replace h with r .
volume of candle = \(\frac13\pi r^2r\)
volume of candle = \(\frac13\pi r^3\)
surface area of candle = \(\pi r^2+\pi r\sqrt{r^2+h^2}\) Since h = r , we can replace h with r .
surface area of candle = \(\pi r^2+\pi r\sqrt{r^2+r^2}\)
surface area of candle = \(\pi r^2+\pi r\sqrt{2r^2}\)
surface area of candle = \(\pi r^2+\pi r\cdot r\sqrt{2}\)
surface area of candle = \(\pi r^2+\pi r^2\sqrt{2}\)
ratio of the volume of candle to its surface area = \(\frac{\text{volume of candle}}{\text{surface area of candle}}\)
\(\frac{\text{volume of candle}}{\text{surface area of candle}}\,=\,\frac{\frac13\pi r^3}{\pi r^2+\pi r^2\sqrt2}\\~\\ =\,\frac{\frac13r^3}{r^2+r^2\sqrt2} \\~\\ =\,\frac{\frac13r}{1+\sqrt2} \\~\\ =\,\frac{r}{3(1+\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1+\sqrt2)(1-\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1-2)} \\~\\ =\,\frac{r(1-\sqrt2)}{-3} \)
