Find-among-all-right-circular-cylinders-of-fixed-volume-V-that-one-with-smallest-surface-area-counting-the-areas-of-the-faces-at-top-and-bottom-

Question Number 144200 by bramlexs22 last updated on 23/Jun/21

Find, among all right circular

cylinders of fixed volume V

that one with smallest surface area

(counting the areas of the faces

at top and bottom)

Answered by MJS_new last updated on 23/Jun/21

S = 2 π r^{2} + 2 π r h

V = π r^{2} h \Rightarrow h = \frac{V}{π r^{2}}

\Rightarrow

S = 2 π r^{2} + \frac{2 V}{r}

\frac{d S}{d r} = 4 π r - \frac{2 V}{r^{2}} = 0 \Rightarrow r = \sqrt[3]{\frac{V}{2 π}} \land h = \sqrt[3]{\frac{4 V}{π}}

or simply h = 2 r

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