Minimizing a volume of 3 variables

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Homework Statement



"Find the equation of the plane that passes through the point (1,2,3) and cuts off the smallest volume in the first octant

There are many ways to solve it, I tried two of them and actually got the correct answer, but couldn't prove if this is the min or the max volume (I guess absolute min or max?)

Homework Equations


for one attempt I used these equations:
a(x-1)+b(y-2)+c(z-3)=0
and
V=(1/3)*(A*h)

As far as the second method is concerned, I actually think this one is better and I used these two equations:

1/a+2/b+3/c=1 ( first had to find it)
V=(1/3)*(A*h)=(1/6)*(abc)

The Attempt at a Solution



I used the Lagrange Multipliers to find a,b,c

This is what I got

b=6
a=3
c=9

Now I have a few questions:
1) if I used the first/second partial derivative tests to find local min or local max, how would you check the boarders here?

2) How do I prove that the point I found is an absolute minimum or an absolute maximum?

3) "cuts off the smallest volume" means that the volume of the tetrahedron has a minimum volume? or the compliment has a min value and then the volume of the tetrahedron has a max value ?

Thanks,
Roni.
 
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1) I wouldn't be two concerned with the boarders as long as you have shown you have a local min, when the plane goes parallel to any of the axes you get an infinite volume

2) As long as you have found any local minima, i think it will be suffcient with comments as above

3) i would read it as the tetrahedron you speak of has minimum volume

so basically you can do it two (equivalent) ways
- Minimise V(a,b,c) using lagrange multipliers subject to the constraint the plane passes through (1,2,3)
- Use the constraint to find c=c(a,b), then minimise the function of 2 variables V(a,b,c(a,b))
 
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lanedance said:
1) I wouldn't be two concerned with the boarders as long as you have shown you have a local min, when the plane goes parallel to any of the axes you get an infinite volume

2) As long as you have found any local minima, i think it will be suffcient with comments as above

3) i would read it as the tetrahedron you speak of has minimum volume

so basically you can do it two (equivalent) ways
- Maximise V(a,b,c) using lagrange multipliers subject to the constraint the plane passes through (1,2,3)
- Use the constraint to find c=c(a,b), then minimise the function of 2 variables V(a,b,c(a,b))

Thank you for your answer but I actually also wanted to know how to decide whether it's a max or a min point with the Lagrange Multipliers.

here is another example:
"Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid 9x^2+36y^2+3z^2=36

I get two points, let's call them (x,y,z) and (-x,-y,-z)
I cross out the point (-x,-y,-z) because it gives me a negative volume and it's not possible (is the reasoning correct ?)

So, I am left with one point, how do I know if it's min or max ?
say I didn't have (-x,-y,-z) and I had only (x,y,z), how would you know that it's a max/min point (assuming absolute?)

I could do this with the first/ second partial derivative test and get the answer, but I want to understand how to do it with the Lagrange Multipliers method.

Thanks,
Roni.
 
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