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#?

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Answer 1 · 2016-06-25T23:27:21

#V = x_0y_0z_0=400/3#

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#