# Is there a reason that the maximal volumes of rectangular prisms with multiple open faces but constant surface area follows this pattern?

Suppose a rectangular prism has a surface area of $12 \text{ m}^2$. The maximal volume of this prism is well known. If the side lengths of the prism are $x$, $y$, and $z$, then the surface area is given by $$A=2xy+2xz+2yz = 12$$ and the volume is given by $$V=xyz$$
One can use Lagrange multipliers to show that $x=y=z$; that is, the prism is a cube. With the given surface area of $12 \text{ m}^2$, one can easily show that the side length is $\sqrt{2} \text{ m}$ and the maximal volume is $2^{3/2} \text{ m}^3$. This is a standard optimization exercise.

$\hskip2in$

A similar classic optimization problem involves maximizing the volume of a box with one face removed. If we assume that the face removed is an "$xy$" face, the problem becomes optimizing the same volume function under the modified constant $$A = xy+2xz+2zy = 12$$ Using similar techniques as before, one can show $x=y=2z$, and the dimensions of the box are $2\times 2\times 1$ for a volume of $4 \text{ m}^3$. Note that this value is greater than the previous value by a factor of $\sqrt{2}$.

$\hskip2in$

One may continue to extend the problem by considering a rectangular prism with two faces removed under a surface area constraint. If the removed faces are on opposite sides of the prism, then the volume can be increased without bound by enlarging the removed faces and shortening the remaining side to keep the surface area constant. However, if the removed sides share an edge, the volume is limited again. Using the same surface area $A = 12 \text{ m}^2$, The optimal dimensions are $\sqrt{2} \times \sqrt{2} \times 2\sqrt{2}$ with the removed edge having length $2\sqrt{2}$. This gives a volume of $4\sqrt{2} \text{ m}^3$. Again, the maximal volume (if we ignore the infinite case) is greater than the previous maximum by a factor of $\sqrt{2}$.

$\hskip2in$

We can keep going and consider a rectangular prism with three faces removed. Although there are 120 ways to choose three faces to remove, they all fall into two patterns: either all the remaining faces share a vertex, or the three faces form an arch shape. If the remaining sides form an arch, then the volume can be increased without bound by making the arch very tall and wide, but not very deep so that the surface area remains constant. If the remaining sides share a vertex, then the optimal dimensions are $2 \times 2 \times 2$, for a volume of $8 \text{ m}^3$. Again, this is larger than the previous volume by a factor of $\sqrt{2}$.

$\hskip2in$

Continue to the case with four faces removed from the prism. Despite 360 different ways to remove the faces, these also only produce two patterns: either the remaining sides are opposite to one another, or they are joined like pages of a book. If the sides are opposite to one another, they can trivially be pulled apart indefinitely, so the volume can be increased without bound. In the case of the book pages, the volume can also be increased without bound by making the pages wider and shortening the binding between the pages. So with four sides removed, the volume of the prism is never bounded.

It is trivial that the volume is unbounded when five or six faces are removed from the prism.

So for the cases where the volume can be bounded, removing one side of the prism increases the maximum possible volume by a factor of $\sqrt{2}$.

Is there a reason that this occurs? Is there a reason that the factor is the same each time? Is there anything special about the value $\sqrt{2}$?

• Interesting. I haven't tried to answer but here's a project: try asking answer same question in the plane and in four space. That may suggest an approach. – Ethan Bolker May 16 '16 at 15:00
• See my answer for a proper proof and explanation of your observation. – user21820 May 31 '16 at 15:45

Is there a reason that this occurs? Is there a reason that the factor is the same each time? Is there anything special about the value $\sqrt{2}? Here's the start of a discussion, and a conjecture. In the plane, the unit square has area$1$and perimeter$4$. If you use three sides to make three quarters of a rectangle the maximum area is$2$, with sides$1$,$2$and$1$. (The two flanking sides have to be the same, by symmetry). If you use two sides meeting at a corner, each of length$2$, then the rectangle is a square with area$4$. So at each of the three steps the maximum area has doubled. We can solve the extreme case in$n$dimensions. The unit cube there has$2n$faces. If we want the maximum solid with$n$faces (meeting at a corner) then to keep the total surface area ($n-1$-dimensional volume) constant, each will have to be an$(n-1)$-cube with twice the volume of the original faces. The edge length for those face cubes in$2^{1/(n-1)}$so the volume of the new big cube is$2^{n/(n-1)}$. That agrees with the two examples computed directly, for$n=2$(above) and$n=3$(in the question). Here's what I think happens in between: each time you remove a face the maximum volume increases by a factor of$2^{1/(n-1)}$, as in the original question for$n=3$. Conjecture. If you remove$d$faces of the unit cube in dimension$n$(so that completing the rectangular parallelepiped has bounded volume) then the maximum volume achievable is $$2^{d/(n-1)}.$$ It should be possible to prove this using the pattern of partial variation; no need for Lagrange multipliers. Note: the factor$2^{1/(n-1)}$decreases to$1$as$n$grows. That is probably because in high dimensions most of the volume of a cube is near the boundary. Here's an analogous observation that supports the conjecture. It's a continuous version, for the unit$n$-ball$B$and its bounding ($n-1)$-sphere$S$with area (i.e.$(n-1)$-volume)$a$. (There's a formula for$a$we won't need.) Suppose we imagine removing a fraction of$S$, leaving behind a figure$T$with area$f \times a$($ 0 < f \le 1$). Then expand$T$so that its area becomes$a$. That means increasing its radius by a factor of$(1/f)^{1/(n-1)}$. The corresponding sphere the increases in volume by a factor of $$((1/f)^{1/(n-1)})^n = (1/f)^{n/(n-1)}.$$ When you keep just half this is our old friend$2^{n/(n-1)}$. • See my answer. We don't need anything beyond elementary algebra, not even calculus! – user21820 May 31 '16 at 7:18 Chop the cube in half. You get half the volume and half the surface area, with one side missing. If you make the shape$\sqrt{2}$bigger, the area is multiplied by$2$, to its original 12. The volume grows by$\sqrt{2}^3$. So the volume was halved by the axe, but grew back by$2\sqrt{2}$, and overall grew by$\sqrt{2}$. Chop the shape in half again. You lose a second side, get half the surface area, half the volume. Regrow to get the same area, and the volume grows overall by another$\sqrt{2}$. • Your answer has the right idea, but you just need to rigorously express it. Namely, take any box with one face missing. Double it and fit them together, forming a cuboid with double the volume and surface area. Applying this procedure to all boxes of a given surface area results in all cuboids of double that surface area. Clearly then the maximal box will result in in a maximal cuboid. Same for the rest. – user21820 May 31 '16 at 6:55 Michael's answer provides an elegant geometric approach. Here is an algebraic approach to observe the$\sqrt{2}$factor popping out each time. Let$x,y,z$be the (non-negative) lengths of the sides of the cuboid (rectangular prism). As you stated, the basic question reduces to maximizing$xyz$given$2xy+2yz+2zx$. Instead of using lagrange multipliers, one should for such symmetric expressions use AM-GM:$8(xyz)^2 = (2xy)(2yz)(2zx) \le \frac13(2xy+2yz+2zx)^3$with equality at$2xy = 2yz = 2zx$. To solve the variant where you don't count the area of one face, you get:$4(xyz)^2 = (xy)(2yz)(2zx) \le \frac13(xy+2yz+2zx)^3$with equality at$xy = 2yz = 2zx$. Not counting two non-parallel faces gives:$2(xyz)^2 = (xy)(yz)(2zx) \le \frac13(xy+yz+2zx)^3$with equality at$xy = yz = 2zx$. Not counting three non-parallel faces gives:$(xyz)^2 = (xy)(yz)(zx) \le \frac13(xy+yz+zx)^3$with equality at$xy = yz = zx$. Throughout we can see that given a fixed surface area, the constant in the left-hand expression each time divides by$2$for the precise reason that a$2$drops out from the right-hand expression for the surface area. And that causes the maximum$xyz$to change by a factor of$\sqrt{2}$precisely since the left-hand expression is that constant times$(xyz)^2$. Note that the AM-GM inequality cannot apply beyond three non-parallel faces, and you should expect it not to, since there would be 'missing' terms in the right-hand expression and so it's not able to 'majorize' the left-hand expression. (These intuitive concepts come from the ideas behind homogenous inequalities.) One can now easily extend this analysis to higher-dimensional analogues. It simply boils down to the product of the coefficients on the faces, of which we take the$(d-1)$-th root because that is precisely the power 'required' to homogenize the inequality, since the faces are one dimension lower than the whole. • +1 I think the last sentence is the heart of the reason for the OP's$\sqrt{2}\$.. It's what the end of my answer stresses. Your use of the AM-GM inequality is a nice way to prove the general statement. – Ethan Bolker May 31 '16 at 13:22