A cola company requires a can in the shape of a right circular cylinder of height \(10~\rm{cm}\) and radius \(3~\rm{cm}\). If the manufacturer of the cans claims a percentage error of no more than \(0.2\%\) in the height and no more than \(0.1\%\) in the radius, what is the approximate maximum variation in the volume of the can?
845
The relative error in the radius \(R\) is \(\dfrac{|\Delta R|}{R}=\dfrac{|dR|}{R}=0.001\), and the relative error in the height \(h\) is \(\dfrac{\left\vert \Delta h\right\vert }{h}=\dfrac{|dh|}{h}=0.002\). The relative error in the volume \(V\) is \[ \begin{eqnarray*} \frac{|\Delta V|}{V} &\approx &\frac{|dV|}{V}=\frac{\left\vert 2\pi Rh\, dR+\pi R^{2} dh\right\vert }{\pi R^{2}h}=\left\vert 2\dfrac{dR}{R}+\dfrac{dh}{h}\right\vert\\[4pt] &=&\left\vert 2\dfrac{\Delta R}{R}+\dfrac{\Delta h}{h} \right\vert \leq 2\frac{\left\vert \Delta R\right\vert }{R}+\frac{\left\vert \Delta h\right\vert }{h} \\ &=&2(0.001)+0.002=0.004 \end{eqnarray*} \]
The maximum variation in the volume is approximately \(0.4\%\), so the actual volume of the container varies as follows: \[ \begin{equation*} V=\pi R^{2}h\pm 0.004( \pi R^{2}h) =\pi R^2 h ( 1\pm 0.004) =90\pi ( 1\pm 0.004) \rm{cm}^{3} \end{equation*} \]
The volume \(V\) is between \(89.64\pi \approx 281.612\;\rm{cm}^{3}\) and \(90.36\pi \approx 283.874\rm{cm}^{3}\).