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One tin can of Mini Ravioli 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.


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.


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
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  • $\begingroup$ Just for the record, I do have more practical applications for this than just the ravioli... All I actually want for this is the scaled Height/Width, but it would nice to finally understand the process once and for all. :-) $\endgroup$
    – ashleedawg
    Commented Apr 19, 2018 at 16:41

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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.


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 $$W=\lambda^{1/3}w\\H=\lambda^{1/3}h$$

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  • $\begingroup$ Uhhh.. I sometimes refer to myself as a "math person"... but definitely not around here. I appreciate the answer but understanding it will take me sometime. For example, I've never seen or heard of a Lambda until about 12 minutes ago. $\endgroup$
    – ashleedawg
    Commented Apr 19, 2018 at 17:02
  • $\begingroup$ As for the first method, uhhh, 300742≈(3π÷8)×(WIDTH³) ?...so 300742.6 in³=1.178097245*(WIDTH³) so 255278.2613 =(WIDTH³) so Width=63.4 (÷2×3) Height=95.1 ? ...er no, that ain't right: that would be a volume of 451113.9 in³. $\endgroup$
    – ashleedawg
    Commented Apr 19, 2018 at 17:09
  • $\begingroup$ Lambda is just a symbol i used there, you can use any letter you want. It just represents the scale that is used. In response to the second comment, where did you get 300742 from? $V$ is the volume. So this would be $4.9\,m^3$ or $174\,ft^3$. (You can choose whichever of these you want to use, based on what units you want to find $W$ in. In metres, this gives $W=1.61\dots\,m$. $\endgroup$
    – John Doe
    Commented Apr 19, 2018 at 17:20
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    $\begingroup$ @ashleedawg Ah sorry, I have just noticed you did it in $in^3$. In fact your computed $H$ and $W$ are correct. They give a volume of $$V=\pi\times\left(\frac{63.4}2\right)^2\times 95.1=300226\approx 300742\,\,\,\checkmark$$ I don't know how you got $451113\,in^3$? Note, we don't get the exact same volume only due to rounding errors, but we do get it correct to $3$ significant figures. $\endgroup$
    – John Doe
    Commented Apr 19, 2018 at 17:25
  • $\begingroup$ thanks, that was helpful. (I added an updated version of the "actual size".) - but honest, this knowledge will be re-purposed beyond ravioli. $\endgroup$
    – ashleedawg
    Commented Apr 19, 2018 at 17:51

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