Cylinder dimensions based on volume and (height:width) ratio

Question

One has dimensions 4.4" tall × 2.9" wide.

Since 1" = 2.54cm and $V = \pi r^{2}h$, that means:

                     Imperial     Metric
        Height         4.4"      11.2 cm
        Width/Depth    2.9"       7.4 cm
        Volume        29.1 in³  476.3 cm³

I have consumed $10348$ cans of Mini Ravioli.
∴ I have consumed ~174 ft³ (~4.9 m³) of Mini Ravioli.

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If all that Mini Ravioli had come in One Big Can — having the same $W:H$ proportion as the original ($\approx 2:3$) — what would the can dimensions be?

_{Scaling Volume like this has always confused me. Obviously I can't just multiply the width & height by the quantity, or else the new volume would be > 32 trillion inches³. I suspect the solution is related to this answer but I can't quite wrap my head around it.}

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Edit: _^(Solution)

It would be, relative to average adult height, this big:

7.9 feet tall × 5.4 feet wide
2.4 metres tall × 1.6 metres wide

Answer 1

The proportion between W and H is $2:3$. So $3W=2H$ $$V=\pi r^2 h=\pi\times\left(\frac{W}{2}\right)^2\times H=\frac\pi4\times W^2\times\frac32W=\frac{3\pi}{8}W^3$$ So given the volume, you can work out the width of a can with this proportion.

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Alternative method: You already know the volume, width and height for a smaller version of this cylinder. Now you want width and height for a cylinder with a different volume, but the same proportions, i.e. a scaled up version of the small cylinder. Say the small cylinder has volume $v$, height $h$, width $w$ and the large one has volume $V$, height $H$, width $W$. We have that $V=\lambda v$ for some number $\lambda$ which you can work out. $V$ is a volume, $W$ is a length, so if $V$ has scaled up by $\lambda$, then $W$ would have scaled up by $\lambda^{1/3}$. This is true because volume is proportional to length cubed. It can also be shown to be true by the method above. So we have that $$W=\lambda^{1/3}w\\H=\lambda^{1/3}h$$