Most Efficient Bucket

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: $$V=\frac{\pi }{3}h(q^2+qr+r^2)$$ $$A=\pi [(q+r)\sqrt{h^2+(q-r{)}^2}+r^2]$$ We can discard the constants since we want to maximize $V$ for constant $A$. From the second we can extract: $$h^2={\left(\frac{A-r^2}{q+r}\right)}^2-(q-r{)}^2$$ 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: $$(q+2r)(2A{r}^2+3q^4-A^2)=2r^3(3q^2+2qr+r^2)$$ Similarly simplifying $\partial W/\partial r=0$ yields: $$A(2q+r)=r(q+2r)(3q+2r)$$ Solving both for $A,q$ (and bringing constants back) yields: $$q=\frac{1+\sqrt{7}}{2}r\approx 1.823r$$ $$h=\sqrt[4]{7}r\approx 1.627r$$ $$A=\pi (7/2+\sqrt{7})r^2\approx 19.31r^2$$ $$V=\frac{\pi }{3}\sqrt[4]{7}(7/2+\sqrt{7})r^3\approx 10.47r^3$$ Note that side length (also called slant) is also $q$, and $V=Ah/3$. 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.