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Max volume of a box

mark

I box with edges parallel to the coordinate axes lies with one vertex at the origin and opposite vertex in the first octant and on the plane

2x+y+z=12,

as shown in the figure below. What is the maximum possible volume of such a box?

mearing

The formula for the box is

V=xyz

\nabla f= \langle yz,xz,xy \rangle \space \text{and} \space \nabla g =\langle 2,1,1 \rangle

So:

yz= \lambda2 \space \space xz=\lambda \space \space xy=\lambda

Reducing to a single variable:

2\frac {y}{2}+y+y=12 \space\space \text{and} \space\space 2\frac{z}{2}+z+z=12

So:

x=2 \space, \space y=4 \space , \space \text{and} \space z=4

Plugging into the volume equation:

V=(2)(4)(4)

V=32

fcarrill

So I solved this by doing the Lagrange method
So the box is going to be my f(x)=xyz
then the equation given is going to be my g(x)=2x+y+z=12
So my \nabla f=<yz,xz,xy>
and my \nabla g= <2,1,1>$

So by using \nabla f = \lambda \nabla g
I have x:yz= \lambda 2
y:xz= \lambda 1 = (y=2x)
z:xy= \lambda 1 = ( z=2x)
if I plug that in to my g(x) equation then I have 2x+2x+2x=12
6x=12
x=2
and since y=2x and z=2x then i have x=2,y=4,z=4
and then my V =32