- #1

pezzang

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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!

(PLEASE PROVIDE ME THE COMPLETE ANSWER AND EXPLANATION ON THE QUETSION.)

AGAIN THANK YOU SO MUCH!