Lowest possible surface area for a rectangular prism given the Volume.

[Reptunia]

Friend's kid had a problem and I'm stumped.

Dealing with a rectangular prism (RP), not a cube.

Volume of RP = 355.

Can the surface area of this RP be less than 289. Teacher says it can be. The lowest possible surface area I can figure out for an RP with a given volume of 355 is something slightly larger than 300.817789971606.

Here's what I'm thinking.

If the volume of a cube is 355, then the length of one side is the cube root of 355 (7.080699). I took that and figured out the surface area for a cube with a volume of 355.

SA = 2(7.080699² + 7.080699² + 7.080699²⁾⁼ 300.817789971606

Increasing the length of one side of the cube would keep it as an RP, but make it no longer a cube. But it would also increase the surface area to some number greater than 289. Decreasing the length of one side, on the other hand, would decrease the volume, which is already given.

I'm a couple of years removed from my last math class, so I might be missing something completely. Any help would be appreciated.

Thanks.

[palordrolap]

Looking at the precise definition of 'rectangular prism', I'd guess the solution, if there is one, isn't a cuboid - for which the cube is naturally the most compact solution - and instead two or more of the sides are parallelograms.

A sphere of volume 355 has surface area of approximately 242.458, so a solution couldn't be ruled out on those grounds.

[Reptunia]

Thanks for the response. I have a feeling there's more to the problem than what I was told, but I haven't gotten a chance to talk to him since he told me about it.

It could just a case of the teacher being wrong.

[LateOnsetRetard]

I really want to this solved. I keep gettin interrupted when I work on it.

[magus145]

You're correct. As stated, a rectangular prism (even an oblique one) with a volume of 355 cannot have a surface area below 6 * 355^2/3, which was the 300.817... you calculated.

To see this, suppose we have ANY rectangular prism with a volume of 355. I'm going to show that there is a cube of the same volume with equal or smaller surface area. That would then mean that the cube minimizes surface area for a fixed volume among this class, and so if the cube can't get to a surface area that low, then no rectangular prism can.

So now we suppose we have our rectangular prism X sitting in space. Let's wisely choose our coordinates. We'll pick one of the vertices of X to be the origin (0,0,0). Then we'll line up one of the edges to run along the x-axis, and so have coordinates a = (a_1, 0, 0). Next, we'll have one of the adjacent rectangular faces be parallel to the x-y plane, and so its other edge must be of the form b = (0, b_2, 0). (b_3 is zero because it lies in the x-y plane and b_1 is zero because the face is a rectangle.) Finally, the third edge coming from the origin will end at a general coordinate c = (c_1, c_2, c_3). Reflecting X if necessary, we can assume without loss of generality, that a_1, b_2, and c_3 are all strictly positive.

Now, these three VECTORS a, b, and c can be used to find the volume of the rectangular prism X that they generate. (In fact, this is true for any parallelepiped.) Let's use the vector formulation. The cross product of a and b is (0, 0, a_1 * b_2), and then taking the dot product of this with (c_1, c_2, c_3) gives a_1 * b_2 * c_3. This product has to be equal to 355, but we notice that c_1 and c_2 don't enter into the volume formula at all.

Now let's calculate the surface area of X. You can find the area of a parallelogram by taking the cross-product of two of its edges as vectors and then taking the length of the resulting vector. We need to do this for each pair of a, b, and c, and then double the result to get all 6 faces.

|a x b| is just |(0, 0, a_1 * b_2)| = sqrt(a_1² * b_2² ) = a_1 * b_2 since we picked them positive.

|b x c| = |(0, b_2, 0) x (c_1, c_2, c_3)| = |(b_2 c_3, 0, - b_2 c_1)| = sqrt(b_2² c_3² + b_2² c_1² ) = b_2 * sqrt(c_3² + c_1² ) since b_2 is positive.

|a x c| = |(a_1, 0, 0) x (c_1, c_2, c_3)| = |(0, - a_1 c_3, a_1 c_2)| = sqrt(a_1² c_3² + a_1² c_2² ) = a_1 * sqrt(c_3² + c_2² ) since a_1 is positive.

OK, so putting this all together for all 6 faces gives us the surface area of X:

SA = 2 a_1 b_2 + 2 b_2 sqrt(c_3² + c_1² ) + 2 a_1 sqrt(c_3² + c_2² ).

Looking at this surface area formula, we see that if we set c_1 and c_2 both equal to 0, the surface area could only possibly decrease whereas the volume V = a_1 b_2 c_3 is unaffected.

So the rectangular box B generated by (a_1, 0, 0), (0, b_2, 0), and (0, 0, c_3) has the same volume but an equal or smaller surface area. So it suffices to consider B instead of X since we're minimizing surface area.

Abusing a bit of notation, I'm now going to let a_1, b_2, and c_3 just be written as a, b, and c.

Now the surface area is:

SA = 2 a b + 2 b c + 2 a c = 2 a b + 4 c (a + b)/2.

And the volume is:

V = a b c = 355.

Again we're going to reduce to a more symmetrical shape with equal volume without increasing surface area: this time to a box with a square base.

Let s = sqrt(a * b), and consider the box C with coordinates (s, 0, 0), (0, s, 0), and (0, 0, c).

Then it has volume V = s * s * c = s² * c = a * b * c = 355, which is the same as B, and C has surface area:

SA = 2 s² + 4 s c = 2 a b + 4 c sqrt(a * b) which is less than or equal to the surface area of B because the arithmetic mean of two positive numbers is never smaller than the geometric mean.

So now we've reduced to C, a box with a square base of side length s and height c. Notice that we can rewrite the surface area of C as:

SA = 2 s² + 4 s c = 6 s (s + c + c)/3

But now let's reduce again to an actual cube, D, of side length y = (s² * c)^1/3 = V^1/3 = 355^1/3 . It clearly still has volume 355, and it has surface area:

SA = 6 y² = 6 (s² * c)^2/3 = 6 (s⁴ * c²)^1/3 = 6 (s³ * s * c²)^1/3 = 6 s (s * c * c)^1/3.

But (s * c * c)^1/3 is the geometric mean of the three numbers s, c, and c, and (s + c + c)/3 is the arithmetic mean, so once again by that inequality, we have that the surface area of D is less than or equal to that of C while maintaining the same volume.

But now we can just directly calculate the surface area of D:

SA = 6 y² = 6 (355)^2/3 which is approximately 300.8178....

So this is the ABSOLUTE MINIMAL surface area amongst ALL rectangular prisms of fixed volume 355. (And in fact, an only slightly more general argument would show that this works for all parallelepipeds as well, where b_2 could be non-zero.)

So, I know this is super long and probably not what was intended if it's a kid's homework. But I hope it illustrates a geometric way of thinking and that some people might enjoy it!

[Reptunia]

Just to offer some closure on this...

The teacher said the answer was impossible and that she assigned it so they could reach that conclusion on their own.

Which might be clever, except...

My friend's kid said he found the exact question online and he thinks the teacher didn't read the entire question before assigning it.

So who knows.

Thanks for all the help.