# Max Vol. Q.

## Main Question or Discussion Point

"Hi, I have a question on max vol. q. Its invloved with multivariable calculus.

Q) Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid 9x^2+36y^2 + 4z^2 = 36.

What i did was i found the three x,y and z-intersection points.
(2,0,0), (0,1,0), and (0,0,3)

Then, I just assumed the following equation:

(2x)/4 + y +(3z)/9 = 1. <- I substituted value of x,y and z for the intersection value.
And if i simplify it, i get: z = 3 - 3x/4 - 3y

To find volume,

V = xyz

so,
V = xy(3 - 3x/4 - 3y)
and fsubx = 3y-3xy-3y^2 = 0
fsuby = 3x - (3x^2)/2 -6xy = 0.

If i do the calculation

i get: 6y - 3x -3y^2 +3(x^2)/2 = 0.

and x = 2/3 and y = 1/3.

And if i sub 2/3 and 1/3 for x and y in the original equation, i get
V = 2/9.

IS THIS THE RIGHT WAY TO DO IT? I ASSUMED THE BEGINNING PART OF THE PROBLEM SOLVING SO I MIGHT BE COMPLETELY WRONG. ANY OF YOU MATH EXPERT, PLEASE HELP ME OUT!!!! THANK YOU SO MUCH AND HAVE A NICE DAY!!!

AGAIN THANK YOU SO MUCH!!!

HallsofIvy
Homework Helper
What you are calling the "intersection points" are the vertices of the ellipsoid. There is no point on the ellipsoid having those x y z values as coordinates.

Let (x,y,z) be the vertex of the rectangular box in the first octant. Then the box has dimensions 2x by 2y by 2z and its volume is
V= 8xyz. That's the function you want to maximize. The vertices all lie on the ellipsoid so we must have 9x2+ 35y2+ 4z2= 36.

That is, the problem is to maximize 8xyz subject to the condition
9x2+ 35y2+ 4z2= 36.

One way of doing this is to solve the subsidiary equation for one of the variables and put that into the "object" function V= 8xyz to reduce to two variables. Then take the partial derivatives with respect to those two variables and set them equal to 0.

A better way (if you haven't already learned it, you should soon) is to use the "Lagrange multiplier" method. Since we must remain in a given set (the subsidiary equation), we do not require that the derivative (gradient) be 0 but that it be perpendicular to the set.

Specifically, here, we would take partial derivatives of V= 8xyz:
Vx= 8yz, Vy= 8xz, Vz= 8xy. The partial derivatives of the subsidiary equation (and so the perpendicular to it) are 18x, 70y, and 8z. The x, y, z that maximize the volume of this rectangle satisfy 8yz= 18 lambda x, 8xz= 70lambda y, 8xy= 8lambda z for some number lambda. Those three equations, together with 9x2+ 35y2+ 4z2= 36 can be solved for x, y, z.