Question

How do you find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane `x + 6y + 10z = 60`?

Answer 1

`V = x_0y_0z_0=400/3`

Explanation:

This problem can be stated as a maximum one.

Find `max f(x,y,z)=xyz` subject to
`g(x,y,z) = x + 6 y + 10 z - 60`

The lagrangian is

`L(x,y,z,lambda)= f(x,y,z)+lambda g(x,y,z)`

`L` is analytic and the stationary points include the minima/maxima points.

The stationary points are the solutions of

`grad L(x,y,z,lambda)= vec 0`

or

#{ (lambda + y z =0), (6 lambda + x z = 0), (10 lambda + x y = 0), (-60 + x + 6 y + 10 z = 0) :}#

with solutions

#( (x = 0, y= 0, z= 6, lambda = 0), (x= 0, y= 10, z= 0, lambda = 0), (x= 20, y= 10/3, z = 2, lambda= -20/3), (x= 60, y= 0, z= 0, lambda=0) )#

The solutions are in the restriction manifold so them will be qualified on

`f@g= (x y)/10 (60 - x - 6 y)`

determining the eigenvalues of

`H=grad^2 f@g = ((-y/5, 6 - x/5 - (6 y)/5),(6 - x/5 - (6 y)/5, -(6 x)/5))`

The local maxima are the stationary points in which `H -< 0` with negative eigenvalues. The maximum point is

`(x= 20, y= 10/3, z = 2)`

with eigenvalues `{-24.1702,-0.496479}` and with volume

`x_0y_0z_0=400/3`