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Question Number 49678 by Rio Michael last updated on 09/Dec/18

Commented by Rio Michael last updated on 09/Dec/18

$$AclosedCylindercanismadefromafixamountAofthin$$$$metalsheet.Findtherelationbetweenitsradiusandheight,$$$$ifitistocontainthemaximumpossiblevolume$$$$(Note:Wastageofmaterialignored).$$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Dec/18

$$v=\pi r^{2}h$$$$s=2\pi rh+2\pi r^2$$$$h=\frac{s-2\pi r^2}{2\pi r}$$$$v=\pi r^2\times \left(\frac{s-2\pi r^2}{2\pi r}\right)$$$$v=\frac{r}{2}(s-2\pi r^2)=\frac{rs}{2}-\pi r^3$$$$\frac{dv}{dr}=\frac{s}{2}-3\pi r^2$$$$formax/min\frac{dv}{dr}=0$$$$3\pi r^2=\frac{s}{2}{r}^2=\frac{s}{6\pi }r=\sqrt{\frac{s}{6\pi }}$$$$\frac{d^{2}v}{{dr}^2}=-6\pi r=-6\pi \times \sqrt{\frac{s}{6\pi }}=-\sqrt{6\pi s}$$$$\frac{d^{2}v}{{dr}^2}<0atr=\sqrt{\frac{s}{6\pi }}$$$$somaxvolumewhenr=\sqrt{\frac{s}{6\pi }}$$$$h=\frac{s-2\pi r^2}{2\pi r}=(\frac{s}{2\pi r}-r)$$$$h=(\frac{s}{2\pi \sqrt{\frac{s}{6\pi }}}-\sqrt{\frac{s}{6\pi }})$$$$h=(\frac{1}{2}\times \frac{\sqrt{s}}{\sqrt{\pi }}\times \sqrt{6}-\sqrt{\frac{s}{6\pi }})$$$$$$

Commented by Rio Michael last updated on 09/Dec/18

$$thankyousir$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Dec/18

$$mostwelcome...$$

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