Extremal problem

The task is to calculate $a'$ of a square (you cut out) if the volume of a cube is maximum.

(you cut out white squares and put together grey squares so you get a cube without a cover/cap)

I don't know what I'm doing wrong, but when I try to calculate critical points I always get $0$, and there isn't even a $\max$ but a $\min$!

$V={a'}^3$ $\implies V'=3{a'}^2$

$\Rightarrow \text{critical points}: 0=3{a'}^2$ $\implies a'=0 (?!)$

The solution is $\dfrac{a}{6}$.

• From the picture it may seem that it is a=3a' but it's not. It's just a sketch. Commented Nov 23, 2013 at 22:38
• The flattened square is missing one side of the cube. Commented Nov 23, 2013 at 22:40
• Yes, it's supposed to. You have a square and you cut out it's borders (in shape of a square- white squares on the picture), so you get a cube without a cap/cover. Commented Nov 23, 2013 at 22:47
• There isn't going to be any extremum if the container must be a cube(that's why you're getting $0$ as critical point since there is no variation, the solution is just a point is the solution space). A maximum will only occur if the container is a cuboid, that is, variations are allowed in the dimensions of the container. Commented Nov 23, 2013 at 23:09
• apparently this is a task for Einstein! It's extra hard to solve it (correctly) and the one who succeedes to solve this is a genius! Commented Nov 23, 2013 at 23:11

$$3a' = a\implies a' = \frac 13 a$$
$$V = a'^3\implies V = \left(\frac 13 a\right)^3 = \frac 1{27} a^3$$
$$V' = \frac 19a^2$$
• No, Anja97: $V'(a) = \frac 19a^2$ Be clear about what you need to find. You say you need to find $a$: that's impossible. We can find the derivative of $V'$ in terms of $a$, but it's not $a/6$. Commented Nov 23, 2013 at 23:02
• You have not provided enough information to determine the maximum volume in terms of $a$, there is no maximum to volume, as posted. As $a$ increases without bound, so does $V$. Did you, perhaps, leave out information about any constraints on $a$? For example, constraints on surface area? Commented Nov 23, 2013 at 23:11