# 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 flatten, where ##x,y,z>0##.

The work ##V## of the athwart box can be developed as

##V=xyz=xy(2-1/3x-2/3y)=2xy-1/3x^2y-2/3xy^2##

Second Partial Test

Let us experience crucial purposes.

##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 merely crucial purpose.

Let us experience relieve 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 Relieve 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 close that

##V(2,1)=(2)(1)(2-2/3-2/3)=4/3##

is the largest work.

Lagrange Multiplier

Let ##g(x,y,z)=x+2y+3z##

##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 work.

I anticipation that this was advantageous.

Here's another commencement for the selfselfsame discontinuance, delay some homogeneous problems: http://people.whitman.edu/~hundledr/courses/M225/Exam2qSOL.pdf