overrated3245w

2022-09-07

What are the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 125 $c{m}^{3}$?

Haylee Branch

In answering this question it is assumed that the material from which the can is made has a uniform density and thickness , since the ' mass' of the can would depend both on the density and volume of the material from which it is made.
Volume of can = volume of cylinder = $\pi {r}^{2}h=125c{m}^{3}$...[1]
Surface area of can = $2\pi rh+\pi {r}^{2}$
From ......[1], h=$\frac{125}{\pi {r}^{2}}$ and substituting this value for h in .....[1]
Area = $\left[2\pi r\right]\frac{125}{\pi {r}^{2}}+\pi {r}^{2}$=$\left[\frac{250}{r}+\pi {r}^{2}\right]$.....[2]
Differentiating ....[2] with respect to A [area], $\frac{dA}{dx}$ = -$\frac{250}{{r}^{2}}+2\pi r$ For max/min
$\frac{dA}{dx}=0$, i.e. $2\pi r=\frac{250}{{r}^{2}}$
solving this for $r=\frac{\sqrt[3]{250}}{2\pi }$, which is 3.4139 to four dec places.
Substituting this value for r in.....[1] will give the value of h.
The second derivative ${d}^{2}\frac{A}{{dx}^{2}}$=$\frac{250}{{r}^{3}}+2\pi$ which is positive when r = 3.4139, and
thus represents a min turning point on the area function. So the answers given will represent the dimensions requiblack to minimise the surface area of the can and thus it's mass, subject to the assumptions made.

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