# Video: Using Lagrange Multipliers to Find the Maximum Volume a Rectangular Parallelepiped Inscribed in an Ellipsoid Can Have

Find the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid (π₯Β²/πΒ²) + (π¦Β²/πΒ²) + (π§Β²/πΒ²) = 1. [A] 8πππ/(3β3) [B] πππ/(3β3) [C] 8πππ/β3 [D] 6πππ/β3 [E] 8πππ?

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### Video Transcript

Find the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid π₯ squared over π squared plus π¦ squared over π squared plus π§ squared over π squared equals one. Is it (a) eight πππ over three root three, (b) πππ over three root three, (c) eight πππ over root three, (d) six πππ over root three, or (e) eight πππ?

To find the volume of a rectangular parallelepiped, the simplest method is to take one vertex, then take the product of the lengths of its three adjacent edges. So here, for example, if these three edges have length πΌ, π½, and πΎ, then the volume of this shape π is just πΌ times π½ times πΎ. This seems easy enough. But for maximizing this value for a parallelepiped inscribed inside an ellipsoid, it is not obvious how this shape will be oriented inside the ellipsoid. So we would essentially need to solve for at least 12 variables, the π₯-, π¦-, and π§-coordinates of the first vertex and all three of its adjacent vertices.

However, we can instead use what we know about an ellipsoidβs relationship with a simpler 3D shape, a sphere. An ellipsoid is essentially just a sphere that has been scaled along three perpendicular axes. Specifically, an ellipsoid with the given equation π₯ squared over π squared plus π¦ squared over π squared plus π§ squared over π squared equals one is simply a unit sphere that has been scaled along the π₯-axis by a factor of π, along the π¦-axis by a factor of π, and along the π§-axis by a factor of π. So an ellipsoid is the result of three scalings along three perpendicular axes.

An important property of the scaling transformation is that the same scaling applied to any 3D shape will scale its volume by the same scale factor as any other 3D shape. This is slightly easier to demonstrate in two dimensions, so letβs clear a little space. Consider, for example, a unit square. It has side length one, and its area is just one unit. Now imagine stretching the square along some axis by a factor of π. Itβs easy to see that if we stretch along an axis parallel to one of its sides, its area increases by a factor of π. So the new area is π. But this in fact applies no matter which direction we stretch in.

For example, consider stretching by a factor of π along an axis parallel to the squareβs diagonal. The result is a rhombus with one unchanged diagonal length of root two and the other of π root two. The area of a rhombus is just one-half times the product of the two diagonals. So we have half times root two times π root two, which is just equal to π, the same new area as we had when we stretched along an axis parallel to one of the square sides. With matrix transformations, it is fairly straightforward to prove that this increase in area is the same along any axis and that this also generalizes to volume in three dimensions.

The upshot of this is that we can find the maximum volume of a rectangular parallelepiped that can be inscribed inside a sphere, then simply scale the parallelepiped using the same transformation as we would to transform the sphere into an ellipsoid. And this will give us the new maximum volume of the parallelepiped that can be inscribed in the ellipsoid. We can be certain that this parallelepiped will be the one with the maximum volume since it was transformed from the parallelepiped with maximum volume in the sphere. And any other parallelepiped inscribed in the ellipsoid must have changed volume by the same factor as this one. So the maximum-volume parallelepiped in the sphere corresponds with the maximum-volume parallelepiped in the ellipsoid.

This will, as we shall see, enormously simplify the problem. So we need to find the maximum volume of a parallelepiped that can be inscribed inside a unit sphere. Intuition may allow you to guess that this is a cube, and indeed it is. But weβre going to prove it using Lagrange multipliers. So consider a unit sphere centered on the origin. Its surface has equation π₯ squared plus π¦ squared plus π§ squared equals one. In order for the parallelepiped to have maximum volume, at least one of its vertices must touch the surface of the sphere. Otherwise, we could easily expand the parallelepiped along one side and increase the volume. Letβs consider then, without loss of generality, one vertex of the parallelepiped touching the sphere in the positive octant; i.e., it has coordinates π₯, π¦, π§, where π₯, π¦, and π§ are all positive.

Now, consider the edges that this vertex is adjacent to. Weβve just drawn the first part of the edges here. Since the question specifies a rectangular parallelepiped, and we will be scaling this shape with the sphere along three perpendicular axes, these edges must also all be at right angles to each other. So they will remain at right angles to each other after the scaling transformation. Since this sphere is symmetric in all three axes, we can again assume without loss of generality that the edges are parallel to each of the coordinate axes. If we now extend these edges further, we can deduce that all three edges will have maximum length. And therefore, the rectangular parallelepiped will have maximum volume when all three edges touch the sphere at their other end vertex.

Since the sphere is symmetric about all three axes and these lines are parallel to the three axes, the vertices of the parallelepiped on the other end of these edges must be the same distance along their respective parallel axes, but in the negative direction from the original point. So this point must have coordinates negative π₯, π¦, π§. This point must have coordinates π₯, negative π¦, π§. And this point must have coordinates π₯, π¦, negative π§. This in turn means that this side is of length two π₯, this side is of length two π¦, and this side is of length two π§, which means that the volume of the full rectangular parallelepiped must be π equals two π₯ times two π¦ times two π§ equals eight π₯π¦π§.

This is the function to be maximized. And our constraint is that our original chosen point at π₯, π¦, π§ must lie on the sphere, meaning the values of π₯, π¦, and π§ satisfy the equation of the sphere π₯ squared plus π¦ squared plus π§ squared equals one. Putting this into the conventional form for a constraint function, we have π equals π₯ squared plus π¦ squared plus π§ squared minus one equals zero. Recall that the Lagrangian function is given by πΏ of π±, π equals π of π± plus ππ of π±, where π is the function to be maximized, π is the constraint function, and π is the scalar known as the Lagrange multiplier. The π± vector is the tuple of the variables of π, in this case π₯, π¦, π§.

So, in our case, we have πΏ of π₯, π¦, π§, π equals eight π₯π¦π§ plus π times π₯ squared plus π¦ squared plus π§ squared minus one. To find the maximum or minimum of π subject to the constraint π, we take partial derivatives of the Lagrangian πΏ with respect to π₯, π¦, π§, and π; set these equal to zero; then solve the resulting system of equations for π₯, π¦, and π§. We will not need to solve explicitly for π. Letβs first clear some space and move our function up here.

So we now need to take partial derivatives of πΏ with respect to π₯, π¦, π§, and π. ππΏ by ππ₯ equals eight π¦π§ plus two ππ₯, ππΏ by ππ¦ equals eight π₯π§ plus two ππ¦, ππΏ by ππ§ equals eight π₯π¦ plus two ππ§, and ππΏ by ππ equals π₯ squared plus π¦ squared plus π§ squared minus one, which weβll notice is just the same as the constraint function π. We now set all of these equal to zero and solve this system of equations for π₯, π¦, and π§.

From the symmetry of this system of equations, you may be able to see where this is going, but letβs continue. We can solve these equations in any order we like. Letβs start with equation one, and weβre going to need a little more space. From equation one, we can rearrange to get π§ equals negative ππ₯ over four π¦. Substituting this expression for π§ into equation two, we get eight π₯ times negative ππ₯ over four π¦ plus two ππ¦ equals zero. Dividing by two and taking π as a common factor before simplifying, we get π times π¦ minus π₯ squared over π¦ equals zero, which means that either π equals zero or π¦ minus π₯ squared over π¦ equals zero. If we substitute this expression for π§ into the third equation instead, we get eight π₯π¦ plus two π times negative ππ₯ over four π¦ equals zero.

Rearranging and dividing through by π₯, which we can do since π₯ equal to zero would be the trivial result with a parallelepiped of zero volume, we get 16π¦ squared equals π squared. π¦ is also not equal to zero, so π is not equal to zero. Therefore, from our previous equation, we must have π¦ minus π₯ squared over π¦ equals zero. Rearranging this equation, we get π₯ squared equals π¦ squared, which is also equal to π squared over 16. If we now use equation four and substitute in π¦ squared for π₯ squared and π§ squared negative ππ₯ over four π¦, we get π₯ squared plus π₯ squared plus negative ππ₯ over four root π squared over 16 all squared is equal to one.

This all simplifies down to three π₯ squared equals one, which then means that π₯ squared equals π¦ squared equals one-third. From our equation for π§, we can get π§ squared equals π squared π₯ squared over 16π¦ squared, which simplifies to π₯ squared, so π§ squared is also equal to one-third. So we have now solved this system of equations for π₯, π¦, and π§, and π₯ equals π¦ equals π§ equals one over root three. And this is the positive root three because we initially stated that π₯, π¦, and π§ are all positive. Remember that each side of the parallelepiped was of length two π₯, two π¦, and two π§, respectively. So each side of this parallelepiped is the same length, two over root three.

So this is indeed a cube. And its volume is given by eight π₯π¦π§, which is just eight times one over root three times one over root three times one over root three, which gives us eight over three root three, the maximum volume of a rectangular parallelepiped that can be inscribed inside a unit sphere. We now need to stretch this unit sphere and the cube inside it along the π₯-axis by a factor of π, along the π¦-axis by a factor of π, and along the π§-axis by a factor of π. This scaling will multiply the volume of the cube by π, π, and π. So this gives the volume of the largest rectangular parallelepiped that can be inscribed in this ellipsoid as eight πππ over three root three. Going back to our possible answers, we can see that this matches with (a) eight πππ over three root three.