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Question Number 121848 by danielasebhofoh last updated on 12/Nov/20

Answered by TANMAY PANACEA last updated on 12/Nov/20

$$\pi r^{2}h+\frac{1}{3}\pi r^2\left(\frac{3}{4}r\right)=V$$$$curvedsurfacearea=S=\pi rl+2\pi rh$$$$l=\sqrt{r^2+{\left(\frac{3r}{4}\right)}^2}=\frac{5r}{2}$$$$S=\pi r\left(\frac{5r}{2}\right)+2\pi rh$$$$now\pi r^{2}h=V-\frac{\pi r^3}{4}\to h=\frac{V}{\pi r^2}-\frac{r}{4}$$$$S=\frac{5\pi r^2}{2}+2\pi r(\frac{V}{\pi r^2}-\frac{r}{4})=\frac{5\pi r^2}{2}+\frac{2V}{r}-\frac{\pi r^2}{2}=2\pi r^2+\frac{2V}{r}$$$$\frac{dS}{dr}=4\pi r-\frac{2V}{r^2}$$$$formax/min\frac{dS}{dr}=0\to 4\pi r-\frac{2V}{1}\times \frac{1}{r^2}=0$$$$4\pi r^3=2V\to r={\left(\frac{V}{2\pi }\right)}^{\frac{1}{3}}$$$$$$$$h=\frac{V}{\pi r^2}-\frac{r}{4}=\frac{V}{\pi {\left(\frac{V}{2\pi }\right)}^{\frac{2}{3}}}-\frac{1}{4}{\left(\frac{V}{2\pi }\right)}^{\frac{1}{3}}$$$$h=\frac{V^{\frac{1}{3}}}{\pi }\times 2^{\frac{2}{3}}\times {\pi }^{\frac{2}{3}}-\frac{1}{4}\times \frac{V^{\frac{1}{3}}}{{\pi }^{\frac{1}{3}}}\times \frac{1}{2^{\frac{1}{3}}}$$$$h={\left(\frac{V}{\pi }\right)}^{\frac{1}{3}}\times 2^{\frac{2}{3}}-{\left(\frac{V}{\pi }\right)}^{\frac{1}{3}}\times \frac{1}{4}\times \frac{2^{\frac{2}{3}}}{2}$$$$={\left(\frac{V}{\pi }\right)}^{\frac{1}{3}}\times 2^{\frac{2}{3}}\times (1-\frac{1}{8})={\left(\frac{4V}{\pi }\right)}^{\frac{1}{3}}\times \frac{7}{8}$$$$$$$$\mathit{S}=\frac{5\pi r^2}{2}+2\pi rh$$$$=\frac{5\pi }{2}\times {\left(\frac{V}{2\pi }\right)}^{\frac{2}{3}}+2\pi {\left(\frac{V}{2\pi }\right)}^{\frac{1}{3}}\times \frac{7}{8}\times {\left(\frac{4V}{\pi }\right)}^{\frac{1}{3}}$$$$={\left(\frac{V}{2\pi }\right)}^{\frac{2}{3}}\left(\frac{5\pi }{2}\right)+2\pi {\left(\frac{V}{2\pi }\right)}^{\frac{1}{3}}\times \frac{7}{8}\times 2\times {\left(\frac{V}{2\pi }\right)}^{\frac{1}{3}}$$$$={\left(\frac{V}{2\pi }\right)}^{\frac{2}{3}}(\frac{5\pi }{2}+\frac{7\pi }{2})=6\pi \times {\left(\frac{V}{2\pi }\right)}^{\frac{2}{3}}$$$$$$

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