I have to solve this problem without Lagrange Multipliers but cannot do this without lagrange multipliers method.Can someone help me?

Question

What is the volume of the largest rectangular parallelepiped which can be inscribed in the ellipsoid x²/9+y²/16+z²/36=1? Show that the answer you get gives you the largest volume.(DO NOT USE LAGRANGE MULTIPLIERS)

Answer 1

Let (x, y,z) is vertex of ellipsoid in 1st octant. Then volume V = 2x·2y·2z = 8xyz.

Now z = 6(1 - x²/9 - y²/16)^1/2_. So, volume V(x,y) = 48xy(1- x²/9 - y²/16)^1/2_.

∂V/∂x = 48y(1 - x²/9 - y²/16)^1/2 - 16/3x²y(1 - x²/9 - y²/16)^-1/2 = 0;

∂V/dy = 48x(1 - x²/9 - y²/16)^1/2 - 3xy²(1 - x²/9 - y²/16)^-1/2 = 0

48y(1 - x²/9 - y²/16)^1/2 = 16/3x²y(1 - x²/9 - y²/16)^-1/2

48x(1 - x²/9 - y²/16)^1/2 = 3xy²(1 - x²/9 - y²/16)^-1/2_.

By division we get: y/x = 16/9x/y or y² = 16/9x² and y = 4/3x;

48x(1 - x²/9 - x²/9) = 3x·16/9x²; 9x(1 - 2x²/9) = x³; x = 0 or 9 - 2x² = x²; x² = 3, x = √3. then y = 4√ 3/3

and z = 6(1 - 1/3 - 1/3)^1/2 = 2√3.

Then V = 8√3·4√3/3·2√3 = 64√3.

If x = 0 then V = 0 < 64√3