The discussion revolves around finding the equation of a plane that minimizes the volume of a tetrahedron formed with the positive coordinate planes, given that the plane must pass through the point (1, 1, 1). The problem involves concepts from calculus, specifically the method of Lagrange multipliers, and the relationship between the variables defining the plane and the volume of the tetrahedron.
Some participants have provided guidance on the use of Lagrange multipliers and clarified the conditions for the plane. There are ongoing questions about the formulation of the volume in terms of the variables a, b, and c, and the implications of the results obtained. Multiple interpretations of the problem are being explored, particularly regarding the nature of the volume minimization.
Participants are grappling with the constraints of the problem, including the requirement for the plane to pass through a specific point and the implications of varying the axes of the coordinate system. There is also discussion about the absence of a maximum volume under certain conditions.
No. Your condition on the plane x/a+ y/b+ z/c= 1 is that it contains the point (1,1,1).quietrain said: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
Yes, all of this is good.so ∇V = λ∇g ?
so Va = λga
1/6 bc = λ (-1/a2)
so for Vb and Vc, i get
1/6ac = λ(-1/b2)
1/6ab = λ(-1/c2)
Well, it would have been better to show how you got that, but, yes, a= b= c= 3 is correct.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 ?
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.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?
Yes, that's fine. In fact often the best way of solving such problems is to immediately eliminate [itex]\lambda[/itex] 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.)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?
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<br /> [itex]\frac{1}{3}\frac{A^3}{(A-1)^2}[/itex]<br /> <br /> That goes to infinity as A goes to infinity and so there is NO maximum volume.[/itex]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!
[itex] <br /> oh i see. <br /> <br /> but somehow, i can't see the last point's latex. <br /> 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?)<br /> <br /> so the volume is 1/3 A<sup>3</sup>(A-1)<sup>2</sup>? <br /> but shouldn't it be, V=1/6abc = 1/6 A (2A/(A-1)) (2A/(A-1)) = 2/3 A<sup>3</sup>(A-1)<sup>2</sup>?<br /> <br /> 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?<br /> <br /> 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?[/itex]HallsofIvy said: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 [itex]\lambda[/itex] 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<br /> [itex]\frac{1}{3}\frac{A^3}{(A-1)^2}[/itex]<br /> <br /> That goes to infinity as A goes to infinity and so there is NO maximum volume.[/itex]