What are the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 125 cm^3?

Mar 21, 2018

Radius of can =$3.4139 c m$, height of can=$3.4139 c m$

Explanation:

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= $\pi {r}^{2} h = 125 c {m}^{3}$......$\left[1\right]$

Surface area of can = $2 \pi r h + \pi {r}^{2}$

From ......$\left[1\right] ,$h=$\frac{125}{\pi {r}^{2}}$ and substituting this value for $h$ in .....$\left[1\right]$

Area=$\left[2 \pi r\right] \frac{125}{\pi {r}^{2}} + \pi {r}^{2}$, =$\left[\frac{250}{r} + \pi {r}^{2}\right]$......$\left[2\right]$

Differentiating ....$\left[2\right]$ with respect to $A$ [area], $\frac{\mathrm{dA}}{\mathrm{dx}}$ = -$\frac{250}{r} ^ 2 + 2 \pi r$ For max/min $\frac{\mathrm{dA}}{\mathrm{dx}} = 0$, i.e. $2 \pi r = \frac{250}{r} ^ 2$

solving this for $r$, $r = \frac{\sqrt[3]{250}}{2 \pi}$, which is $3.4139$ to four dec places.

Substituting this value for $r$ in.....$\left[1\right]$ will give the value of $h$.

The second derivative ${d}^{2} \frac{A}{\mathrm{dx}} ^ 2$= $\frac{250}{r} ^ 3 + 2 \pi$ 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 required to minimise the surface area of the can and thus it's mass, subject to the assumptions made.