A bucket is a truncated cone (also called frustum of a cone or a conic bowl) with open top to store stuff (mostly liquids) in it:
Given constant area (side+base), what is the bucket geometry that yields maximum volume?
Let's find out by first defining some variables:
\(q:\) radius of the top, open side
\(r:\) radius of the base
\(h:\) height of the bucket
Then volume and area respectively are: $$V={\pi\over 3}h(q^2+qr+r^2)$$ $$A=\pi[(q+r)\sqrt{h^2+(q-r)^2}+r^2]$$ We can discard the constants since we want to maximize \(V\) for constant \(A\). From the second we can extract: $$h^2=\left({A-r^2\over q+r}\right)^2-(q-r)^2$$ which can be substituted into: $$W=V^2=h^2(q^2+qr+r^2)^2$$ \(W\) is a rational polynomial expression in variables \(q\) and \(r\), so it can be maximized instead of \(V\). Simplifying \(\partial W/\partial q=0\) with WolframAlpha yields: $$(q + 2 r) (2 A r^2 + 3 q^4 - A^2) = 2 r^3 (3 q^2 + 2 q r + r^2)$$ Similarly simplifying \(\partial W/\partial r=0\) yields: $$A (2 q + r) = r (q + 2 r) (3 q + 2 r)$$ Solving both for \(A, q\) (and bringing constants back) yields: $$q={1+\sqrt 7\over 2}r \approx 1.823 r$$ $$h=\root 4\of{7} r \approx 1.627 r$$ $$A=\pi(7/2+\sqrt 7)r^2 \approx 19.31 r^2$$ $$V={\pi\over 3}\root 4\of{7}(7/2+\sqrt 7)r^3 \approx 10.47 r^3$$ Note that side length (also called slant) is also \(q\), and \(V=A h/3\). There is another nice property of this optimal bucket, can you see it?
Having golden ratio not winning this time, here is a model for your 3D printer.
Also there is an article with nice colorful plots showing the unique maxima.
Let's find out by first defining some variables:
\(q:\) radius of the top, open side
\(r:\) radius of the base
\(h:\) height of the bucket
Then volume and area respectively are: $$V={\pi\over 3}h(q^2+qr+r^2)$$ $$A=\pi[(q+r)\sqrt{h^2+(q-r)^2}+r^2]$$ We can discard the constants since we want to maximize \(V\) for constant \(A\). From the second we can extract: $$h^2=\left({A-r^2\over q+r}\right)^2-(q-r)^2$$ which can be substituted into: $$W=V^2=h^2(q^2+qr+r^2)^2$$ \(W\) is a rational polynomial expression in variables \(q\) and \(r\), so it can be maximized instead of \(V\). Simplifying \(\partial W/\partial q=0\) with WolframAlpha yields: $$(q + 2 r) (2 A r^2 + 3 q^4 - A^2) = 2 r^3 (3 q^2 + 2 q r + r^2)$$ Similarly simplifying \(\partial W/\partial r=0\) yields: $$A (2 q + r) = r (q + 2 r) (3 q + 2 r)$$ Solving both for \(A, q\) (and bringing constants back) yields: $$q={1+\sqrt 7\over 2}r \approx 1.823 r$$ $$h=\root 4\of{7} r \approx 1.627 r$$ $$A=\pi(7/2+\sqrt 7)r^2 \approx 19.31 r^2$$ $$V={\pi\over 3}\root 4\of{7}(7/2+\sqrt 7)r^3 \approx 10.47 r^3$$ Note that side length (also called slant) is also \(q\), and \(V=A h/3\). There is another nice property of this optimal bucket, can you see it?
Having golden ratio not winning this time, here is a model for your 3D printer.
Also there is an article with nice colorful plots showing the unique maxima.
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