By solving for #z#,
#x+2y+3z=6 Leftrightarrow z= 2-1/3x-2/3y#
Let #(x,y,z)# be the vertex on the plane, where #x,y,z>0#.
The volume #V# of the rectangular box can be expressed as
#V=xyz=xy(2-1/3x-2/3y)=2xy-1/3x^2y-2/3xy^2#
Second Partial Test
Let us find critical points.
#V_x=2y-2/3xy-2/3y^2=2/3y(3-x-y)=0 Rightarrow x+y=3#
#V_y=2x-1/3x^2-4/3xy=1/3x(6-x-4y)=0 Rightarrow x+4y=6#
#Rightarrow (x,y)=(2,1)# is the only critical point.
Let us find second partials.
#V_{x x}=-2/3y Rightarrow V_{x x}(2,1)=-2/3#
#V_{y y}=-4/3x Rightarrow V_{yy}(2,1)=-8/3#
#V_{xy}=2-2/3x-4/3y Rightarrow V_{x y}(2,1)=-2/3#
By Second Partial Test,
#D(2,1)=(-2/3)(-8/3)-(-2/3)^2=4/3>0#
and
#V_{x x}(2,1)=-2/3<0#,
we may conclude that
#V(2,1)=(2)(1)(2-2/3-2/3)=4/3#
is the largest volume.
Lagrange Multiplier
Let #g(x,y,z)=x+2y+3z#
#grad V=lambda grad g Rightarrow (yz,xz,xy)=lambda(1,2,3)#
#Rightarrow {(yz=1/2xz Rightarrow y=x/2),(yz=1/3xy Rightarrow z=x/3):}#
#Rightarrow g(x,x/2,x/3)=x+2(x/2)+3(x/3)=3x=6#
#Rightarrow x=2#, #y=2/2=1#, #z=2/3#
Hence,
#V(2,1,2/3)=2cdot1cdot2/3=4/3#
is the largest volume.
I hope that this was helpful.
Here's another source for the same solution, with some similar problems: http://people.whitman.edu/~hundledr/courses/M225/Exam2qSOL.pdf
