Applied Mathematics - Lagrangian multiplication or other technique?

TheSetMiner

I have been stuck on this question from my college applied maths course for about a week.

Find the maximum and minimum volumes of a rectangular box whose surface area equals 4000 square cm and whose edge length (sum of lengths of all edges) is 180 cm. Hint: It can be found that the box is not a cube, so if A,B, and C are the lengths of the sides, you may want to let A represent a side with A not equal to B and A not equal to C.

I suspect that a lagrangian multiplication technique is used to get a value for lambda and mu, two additional variables but I am very unsure how to implement this. I have tried to follow a youtube example but ended up with 3 equations and 5 unknowns and no idea how to solve for them. Can anyone help guide me in the right direction or provide a solution?

Thanks to anyone who can offer me any help at all.

bnt

When I saw this kind of problem before, it didn't involve anything like that, it fell under the heading of "Optimisation problems". You derive a formula for the volume in terms of one of the sides (say x), then use calculus to find the maximum and/or minimum values for x. Khan Academy has a series on this, here.

TheSetMiner

bnt wrote: »

When I saw this kind of problem before, it didn't involve anything like that, it fell under the heading of "Optimisation problems". You derive a formula for the volume in terms of one of the sides (say x), then use calculus to find the maximum and/or minimum values for x. Khan Academy has a series on this, here.

Hi, thanks for replying.

Does this technique work for 3 variables though? namely a f(A, B, C)

bnt

As I recall the volume optimisation problem, you don't actually have three independent variables, because the problem has those defined constraints that link them together.

a rectangular box whose surface area equals 4000 square cm and whose edge length (sum of lengths of all edges) is 180 cm

The edge length would be expressed as 4A+4B+4C = 180 and the surface area would be 2AB+2AC+2BC=4000. If I'm right, you could treat those as simultaneous equations and try to simplify to a two-variable equation for the volume (the 2nd equation), then find the minima/maxima by calculus.

(Note that I'm working from general principles here, not a specific algorithm to solve this particular problem - those videos I mentioned would go in to more depth.)

ZorbaTehZ

The standard way to solve these is Lagrange multipliers. The objective function here is F(A,B,C,λ1,λ2)=ABC - λ1(2AB - 2AC - 2BC - 4000) - λ2(4A - 4B - 4C - 180). Then the idea is to set the partial derivatives with respect to A,B,C,λ1,λ2 equal to zero and solve these.

You end up with 5 equations, and dont need λ1,λ2, so the first 3 can be combined to obtain one equation. Then you need to consider the cases separately A=-C and A!=-C, this will allow you to obtain 1 value for B in the first case (using the second constraint equation) in the first case and 2 values for B in terms of C in the second case. Then you should be able to simplify it further, but it will probably be long.

I dont really have enough time to do it out fully. In total there are 6 critical points, but only 2 are real. So that you can check your answers, the correct values via mathematica are {A -> 9 + √481, B -> -27 + 2√481, C -> -9 - √481} and {A -> 9 - √481, B -> -27 - 2√481, C -> -9 + √481}, you can check yourself which gives minimum and which gives maximum.

TheSetMiner

(x^2)/4 + (y^2)/9 + z^2 = 1

Quick Question! How does one go about solving for the minimum/maximum distances of an ellipse of this equation from the origin to a point on its surface? Intuitively I know D^2 = X^2 + Y^2 + Z^2 (for a unit circle, at least?) but this doesn't seem to help here at all!

ZorbaTehZ

That is an ellipsoid and you put it in the form x^2/a^2+y^2/b^2+z^2/c^2=1 then take max(a,b,c) and min(a,b,c). Just think about what it looks like in R^3 to see why.

TheSetMiner

ZorbaTehZ wrote: »

That is an ellipsoid and you put it in the form x^2/a^2+y^2/b^2+z^2/c^2=1 then take max(a,b,c) and min(a,b,c). Just think about what it looks like in R^3 to see why.

But how does lagrange multiplication tie into this? thanks!

TheSetMiner

Ah wolfram alphas 3d graph has definitely clarified the max(a, b, c) step!

Blackpanther95

This is a Lagrange MULTIPLIERs question (not Lagrange multiplication...as this sounds less cool). One gradient vector is a scalar multiple of the other. Set your f, and g(the constraint), then differentiate both. Solve the simultaenous equations (linear if your lecturers nice), sub into the constraint...there's your answer.

TheSetMiner

Blackpanther95 wrote: »

This is a Lagrange MULTIPLIERs question (not Lagrange multiplication...as this sounds less cool). One gradient vector is a scalar multiple of the other. Set your f, and g(the constraint), then differentiate both. Solve the simultaenous equations (linear if your lecturers nice), sub into the constraint...there's your answer.

i think finding the f and g is my problem. I only have one single equation - the ellipsoid.

ZorbaTehZ

TheSetMiner wrote: »

i think finding the f and g is my problem. I only have one single equation - the ellipsoid.

f is the equation you want to minimise/maximise. g is the constraint equation. So if you want the maximum distance from the origin to a point on the ellipsoid then you should take f=equation expressing distance from (0,0,0) to (x,y,z) right? Then set F=f-a.g where a is a variable and do as I indicated before or as BlackPanther did above.

TheSetMiner

thanks

MathsManiac

ZorbaTehZ wrote: »

The standard way to solve these is Lagrange multipliers. The objective function here is F(A,B,C,λ1,λ2)=ABC - λ1(2AB - 2AC - 2BC - 4000) - λ2(4A - 4B - 4C - 180). Then the idea is to set the partial derivatives with respect to A,B,C,λ1,λ2 equal to zero and solve these.

You end up with 5 equations, and dont need λ1,λ2, so the first 3 can be combined to obtain one equation. Then you need to consider the cases separately A=-C and A!=-C, this will allow you to obtain 1 value for B in the first case (using the second constraint equation) in the first case and 2 values for B in terms of C in the second case. Then you should be able to simplify it further, but it will probably be long.

I dont really have enough time to do it out fully. In total there are 6 critical points, but only 2 are real. So that you can check your answers, the correct values via mathematica are {A -> 9 + √481, B -> -27 + 2√481, C -> -9 - √481} and {A -> 9 - √481, B -> -27 - 2√481, C -> -9 + √481}, you can check yourself which gives minimum and which gives maximum.

It seems to me that this basically means that there are no solutions to the problem posed, since all three dimensions of an actual box would have to be positive.

MathsManiac

...which means that the two constraints are incompatible: there is no rectangular box whose surface area is 4000cm^2 and whose edge-lengths sum to 180 cm.

(This is comfortably provable, by the way: for the surface area of the box to be 4000, at least one face must be at least 4000/6 in area. The edges of this face alone would have to sum to at least 4*sqrt(4000/6) and the same for the opposite face, so we have already exceeded 180 cm before even considering the remaining four edges of the solid.)

Perhaps the intention behind the question was that the sum of the three edge-length dimensions should be 180, rather than the sum of the 12 actual edge lengths. Or maybe it was intended to be an open box.

Blackpanther95

So I did your problem out...is it possible you took down the question wrong, or left out something you thought was unimportant. Usually when you do Lagrange multipliers you have only-one constraint. Here you have two. So if we use the surface area as a constraint, we see that lambda=200/9 ...and if we solve for this then the equations are pretty much inconsistent...anyone get this problem

(I add all the equations together that you get from the Lagrange multipliers and then used the constraints (they appear pretty handily))

MathsManiac

Blackpanther95 wrote: »

So I did your problem out...is it possible you took down the question wrong, or left out something you thought was unimportant. Usually when you do Lagrange multipliers you have only-one constraint. Here you have two. So if we use the surface area as a constraint, we see that lambda=200/9 ...and if we solve for this then the equations are pretty much inconsistent...anyone get this problem

(I add all the equations together that you get from the Lagrange multipliers and then used the constraints (they appear pretty handily))

I think we've already established that the problem has no solution. (Post #5, 14, & 15.)

Blackpanther95

MathsManiac wrote: »

I think we've already established that the problem has no solution. (Post #5, 14, & 15.)

Yes...I wonder what the real question though was...

MathsManiac

Blackpanther95 wrote: »

Yes...I wonder what the real question though was...

Well, if the intention had been that the three edge-dimensions sum to 180, then there would be a solution. Instead of using Lagrange multipliers, you can just use the two constraints to eliminate two of the dimensions and express the volume in terms of one only. I think you get

V = A^3 - 180A^2 + 2000A

Leading, I think, to {A,B,C} = {60 - (20/3)sqrt(66), 60 - (20/3)sqrt(66), 60 + (40/3)sqrt(66)}.

(Numerically: 5.84 * 5.84 * 168.32 = 5740.65)