Find plane eqn that minimize Volume

[quietrain]

Homework Statement

find plane that minimize Volume V if plane is constrain to pass through point 1,1,1

Homework Equations

plane equation = x/a +y/b + z/c = 1 , a,b,c > 0
with the positive coordinate planes and the plane, they form a tetrahedon of volume V = 1/6 abc

The Attempt at a Solution

how do i even start? :x
am i suppose to use the method of lagrangian multipliers?

thanks!

 

[Dick]

Lagrange multipliers sounds like a good approach to me. Why don't you just try it?

 

[quietrain]

er ok, so is this right?

i want to minimize V(a,b,c)=1/6 abc
my g(a,b,c) = x/a +y/b +z/c = 1

so ∇V = λ∇g ?
so V_a = λg_a
1/6 bc = λ (-1/a²)

so for V_b and V_c, i get
1/6ac = λ(-1/b²)
1/6ab = λ(-1/c²)

so since the plane must pass through (1,1,1), i get g(a,b,c) = 1/a + 1/b +1/c = 1

so solving i get a=b=c=3?

so the plane is x/3 + y/3 +z/3 = 1 ?


but i have a few questions

1) shouldn't V be a function of x,y,z ? and not a,b,c? and so g should also be of x,y,z?

2) i didn't find the value of λ when i solve the simultaneous equations, is this ok? and even if i find it, what does it symbolize?

3) lastly, how do i know that the value a=b=c=3 gives me the minimized volume? why not it be the maximized?

thanks!

 

[HallsofIvy]

er ok, so is this right?

i want to minimize V(a,b,c)=1/6 abc
my g(a,b,c) = x/a +y/b +z/c = 1

No. Your condition on the plane x/a+ y/b+ z/c= 1 is that it contains the point (1,1,1).
Your condition is that g(a,b,c)= 1/a+ 1/b+ 1/c= 1

so ∇V = λ∇g ?
so V_a = λg_a
1/6 bc = λ (-1/a²)

so for V_b and V_c, i get
1/6ac = λ(-1/b²)
1/6ab = λ(-1/c²)

Yes, all of this is good.

so since the plane must pass through (1,1,1), i get g(a,b,c) = 1/a + 1/b +1/c = 1

so solving i get a=b=c=3?

so the plane is x/3 + y/3 +z/3 = 1 ?

Well, it would have been better to show how you got that, but, yes, a= b= c= 3 is correct.

but i have a few questions

1) shouldn't V be a function of x,y,z ? and not a,b,c? and so g should also be of x,y,z?

No. (x, y, z) would be a point on the plane. You want to determine which plane which means you want to find a, b, and c.

2) i didn't find the value of λ when i solve the simultaneous equations, is this ok? and even if i find it, what does it symbolize?

Yes, that's fine. In fact often the best way of solving such problems is to immediately eliminate $\lambda$ by dividing your equations. Your condition is that the two gradients are parallel- their individual lengths are not relevant. (lambda is the ratio of those two lengths.)

3) lastly, how do i know that the value a=b=c=3 gives me the minimized volume? why not it be the maximized?

thanks!

Suppose a= A, some fixed large number and that b and c are the same. It is easy to show that [itex]b= c= 2A/(A-1)[/itex and so the volume would be
$\frac{1}{3}\frac{A^3}{(A-1)^2}$

That goes to infinity as A goes to infinity and so there is NO maximum volume.

 

[quietrain]

No. Your condition on the plane x/a+ y/b+ z/c= 1 is that it contains the point (1,1,1).
Your condition is that g(a,b,c)= 1/a+ 1/b+ 1/c= 1 Yes, all of this is good. Well, it would have been better to show how you got that, but, yes, a= b= c= 3 is correct.


No. (x, y, z) would be a point on the plane. You want to determine which plane which means you want to find a, b, and c. Yes, that's fine. In fact often the best way of solving such problems is to immediately eliminate $\lambda$ by dividing your equations. Your condition is that the two gradients are parallel- their individual lengths are not relevant. (lambda is the ratio of those two lengths.) Suppose a= A, some fixed large number and that b and c are the same. It is easy to show that [itex]b= c= 2A/(A-1)[/itex and so the volume would be
$\frac{1}{3}\frac{A^3}{(A-1)^2}$

That goes to infinity as A goes to infinity and so there is NO maximum volume.

oh i see.

but somehow, i can't see the last point's latex.
the code seems to suggest that b=c=2A/(A-1) if i sub a=A into the g(a,b,c) ? (but why can we let b=c?)

so the volume is 1/3 A³(A-1)²?
but shouldn't it be, V=1/6abc = 1/6 A (2A/(A-1)) (2A/(A-1)) = 2/3 A³(A-1)²?

so you mean that if i can show that V=1/6abc, as long as when a or b or c tends to infinity, if it causes V to tend to infinity, then there is no maximum volume?

also, i was thinking, what if now the axes are not the positive x=0,y=0,z=0 planes making the volume with the plane, but rather, some other planes like x=5. then i just need to find the equation of the volume of the new tetrahedron and do the same right?