Minimizing Volume with Lagrangian Multipliers & (2,3,4)

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Homework Help Overview

The discussion revolves around finding a linear plane that intersects the first octant and minimizes the volume beneath it, specifically passing through the point (2,3,4). The problem involves the application of Lagrangian multipliers to minimize the volume function V=(1/2)xyz while adhering to certain constraints.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formulation of the constraint for the plane based on its intercepts and the point it must pass through. There is a consideration of whether the problem can have a solution if the volume can be made arbitrarily small by adjusting the intercepts. Some participants clarify the intent of the original poster regarding the plane passing through the specified point.

Discussion Status

There is a mix of agreement on the approach suggested by one participant, with others providing additional insights into the mathematical relationships derived from the Lagrangian method. The discussion is ongoing, with participants exploring various interpretations and implications of the problem setup.

Contextual Notes

Participants are working under the assumption that the plane must pass through the point (2,3,4) and are questioning the implications of this requirement on the minimization of volume. The constraints and relationships among the variables a, b, and c are being examined in detail.

circa415
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I'm supposed to find the linear plane which cuts through first octant and results in the minimum volume underneath the plane. The plane must pass under (2,3,4). I think I need to use Lagrangian multipliers and minimize V=(1/2)xyz.. I'm not sure what the constraint would be though
 
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ok, after thinking about this some more...

x/a+y/b+z/c=1 will give me a plane where the x, y, z intercepts will be a, b, and c respectively. Therefore if the point passes through 2, 3, 4, then the constraint will have to be 2/a+3/b+4/c = 1

and then V will just be (1/2)abc

and then just apply the lagrangian method

Does that sound like a good way to approach this problem?
 
sounds good to me
 
circa415 said:
I'm supposed to find the linear plane which cuts through first octant and results in the minimum volume underneath the plane. The plane must pass under (2,3,4). I think I need to use Lagrangian multipliers and minimize V=(1/2)xyz.. I'm not sure what the constraint would be though

I don't see how this can HAVE a solution. If your purpose is to get minimum volume and your only requirement is that the plane must "pass under (2,3,4), can't you make the volume arbitrarily small by making the intercepts, x, y, z, arbitrarily close to 0?
 
I'm sure he meant that it goes THROUGH the point :)
 
yeah that's what I meant lol my mistake
 
Then the way you suggested orginally is the way to go. The volume you want to maximize would be abc, of course, and the constraint is 2/a+ 3/b+ 4/c= 1.
Taking the grad of abc give <bc, ac, ab> while taking the grad of (2/a+ 3/b+ 4/c) gives <-2/a2[/sub], -3/b2, -4/c2>.

At the values of a,b,c that maximize the volume while satisfying the constraint, we must have <bc, ac, ab>= &lambda;<-2/a2[/sub], -3/b2, -4/c2> for some &lambda;

That is, bc= -2&lambda;/a2[/sub], ac= -3&lambda;/b2[/sub], ab= -4&lambda;/c2[/sub].

Divide the first equation by the second to get
[tex]\frac{bc}{ac}= \frac{-2\lambda}{a^2}\frac{b^2}{3\lambda}[/tex]
or
[tex]\frac{b}{a}= \frac{2b^2}{3a^2}[/tex]
which is the same as 3a= 2b.

Similarly, dividing the second equation by the third gives
[tex]\frac{c}{b}= \frac{3c^2}{4b^2}[/tex]
which is the same as 4b= 3c.
That is, b= (3/2)a and c= (4/3)b= 2a.
Now put that into 2/a+ 3/b+ 4/c= 1 to solve for a, b, c.
 
This is what I did, I got -2λ/(a^2) = bc, -3λ/(b^2) = ac, -4λ/(c^2)=ab. Mutiply each by a, b, and c respectively and so: -2/a = -3/b = -4/c (take out negative and plug into constraint)
 

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