Find the volume of the largest rectangular box in the first octant with three faces in the coordinate plane and one vertex in the plane x + 2y + 3z = 6 a. using the Second Partial Test . b. using Lagrange multipliers. How do I do this?
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
