# Homework Help: Lagrange Multiplier-to find out the dimensions when metal used min.

1. Apr 30, 2013

### Outrageous

Lagrange Multiplier----to find out the dimensions when metal used min.

1. The problem statement, all variables and given/known data
I have a rectangular tank with a capacity of 1.0m^3. The tank is closed and the cover is made of metal half as thick as the sides and base. Find the dimensions of the tank for the total amount of metal used to build be minimum.

2. Relevant equations
Lagrange
f(x)-λ(xyz-1)=g(x,y,z,λ)

3. The attempt at a solution
the constraint is xyz=1, but what should I use for f(x)?

2. Apr 30, 2013

### LCKurtz

What are you trying to minimize? Write a formula for it. It won't be just a function of $x$ like "$f(x)$".

3. May 1, 2013

### Outrageous

Volume of the container!?
But how to use the volume of the tank to minus the volume of the solvent with capacity of 1m^3?
Assume the volume of the container is xyz, the the thickness will be t, so that will be a f(x,y,z,t) ? (x+4t)(y+4t)(z+2t) ,it will be my f(x,y,z,t)?

4. May 1, 2013

### HallsofIvy

No! Take the time to read what you wrote carefully!
"the total amount of metal used to build be minimum."

5. May 1, 2013

### Outrageous

erm.....mass?
the dimensions of the tank for the total amount of metal used to build be minimum. besides volume , I really have no idea......

6. May 1, 2013

### LCKurtz

OK, mass will work. Assume an area density of $\delta$ for everything but the top, where it is $\frac\delta 2$. So what is the formula for the mass?

7. May 1, 2013

### Ray Vickson

You are almost correct, except for omitting a constant: the volume of metal used does equal
$\text{metal volume}=f(x,y,z) = (x+4t)(y+4t)(z+3t)- xyz$ where 2t is the thickness of the bottom and sides and t is the thickness of the top (you wrote $z+2t$, which is not correct). However, since $xyz = 1$ in any allowed solution, we might as well replace the $-xyz$ term in f(x,y,z) by -1; and then we might as well drop the -1 because it is just an additive constant that will not affect the optimal solution (only the optimal final value of f). You need to minimize f.

Note that here t > 0 is a parameter, not a variable. Even though you are not told the value of t, it (presumably) remains a constant throughout the whole procedure---it is a kind of design input value. This is unlike x, y and z, which do vary during the design phase, until we finally arrive at the values we want.

Your formulation (after making the correction indicated) is a perfectly viable , and in many ways *superior*, formulation of the actual problem. However, I suspect the problem's proposer is thinking of a much simpler approximate formulation, where the actual value assigned to t is not important (provided it is small).

What amazed me is that when I solved both formulations (yours, corrected, and the small-t approximate version) I found that both have exactly the same numerical values for optimal x, y and z, irrespective of the actual value of t!

Last edited: May 1, 2013
8. May 1, 2013

### Outrageous

Thank you.

A

I tried. My answer x=z=(4y/3)., x=(4/3)^(1/3)
Then Find the dimensions of the tank for the total amount of metal used to build be minimum.
The answer should be volume of the metal used =xyz-1.
Then I substitute the answer in, I got zero.

9. May 1, 2013

### Outrageous

Thank you.

A bit dont really understand lagrange multiplier, the λ is a function or variable that can possibly make the dimension of the mass become same as the dimension of the volume?

I tried. My answer x=z=(4y/3)., x=(4/3)^(1/3)
Then Find the dimensions of the tank for the total amount of metal used to build be minimum.
The answer should be volume of the metal used =xyz-1.
Then I substitute the answer in, I got zero.

10. May 1, 2013

### Outrageous

Thank you. I really like the way you explain.

My answer for that is x=-4t, y=-4t, z=-3t.
I dont know what is wrong, since the t>0 , how can them be negative.
Sorry, what is the small t approximation version? can you tell me more please.

11. May 1, 2013

### Dick

The small t approximation is really just assuming they want you to minimize the area. So the thickness t isn't significant. And can you show more of how you worked this out? In that case λ will have dimensions. In general, it's just some constant. You haven't even said what your variables mean, which of the x,y and z correspond the height, width and length?

12. May 1, 2013

### Ray Vickson

Show your work, so we can see where you went wrong.

I get
$$x = y = \frac{1}{3} 6^{2/3} \doteq 1.100642, \: z = \frac{1}{4} 6^{2/3} \doteq 0.825482$$
This is for ANY value of t > 0!.

Note, however, that the volume of metal used depends on the thickness, so if the sides and bottom have thickness 2t and the top has thickness t, we have:
$$\text{metal volume} = \left( 4t + \frac{1}{3} 6^{2/3}\right)^2 \left( 3t + \frac{1}{4} 6^{2/3} \right) -1.$$

13. May 1, 2013

### LCKurtz

You didn't show your work, but I assume you calculated the mass by taking $\delta$ times the surface area $S$ for the bottom and sides, and $\delta /2$ times the area of the top. Is that how you got your answers below? If not, I need to see what you did. But if so, read on.

I don't see how you get zero. I probably labeled the variables differently. I got the the dimensions of the bottom are both (4/3)^(1/3) and the height is (3/4)(4/3)^(1/3). So if you labeled the bottom dimensions x and z and the height y, we have the same answer. And if you plug them in the formula for $S$ you definitely won't get 0.

[Edit--Added after I saw Ray's last post] These numbers agree with Ray's solution.

14. May 1, 2013

### Outrageous

Only x, y z vary. Then t is constant.

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15. May 1, 2013

### Outrageous

Volume of metal used is xyz-1 ?

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16. May 2, 2013

### LCKurtz

The volume of the box is xyz and it is given to be 1. That is not the same as the volume of metal in the sides of the box.

You expect us to read sideways??? You never wrote your objective function separately. That is what you are trying to minimize. The point is that using an area density, the volume of a sheet of metal is density times area. It is the part of your formula with the with the $\delta$'s, and it is the first thing you should have written down.

You were asked for the dimensions. You have them. And you get the volume of the box is xyz=1. If you want to calculate the minimum mass, put the values of x,y, and z in your objective function.

17. May 2, 2013

### Outrageous

The objective function is ∫2xy+∫2yz+∫(xyz/2), where ∫ is the area density.
The constraint function is xyz=0.
So I am trying to minimize the mass.
Area density is mass per area. Volume is should be (mass) per density. Then volume is (area density times area) per density? How do you get the density times area? ∫'s is differentiate the surface rea? With respect to?

18. May 2, 2013

### Ray Vickson

Why would you say that? You quoted previous posts that had precise expressions given for the metal volume. Then you just ignored everything they said!

19. May 2, 2013

### LCKurtz

Don't use an integral sign for density, that is very confusing. The objective function is$$M = \delta(2yz+2xz+\frac 3 2 xy)$$depending on how you label the sides

No. The volume of the box is given to be 1. xyz=1.

If you have a thin sheet of metal with area mass 2kg per square meter and you have a piece that is 10 sq meters, how much does it weigh? Think about that.

20. May 3, 2013

### Outrageous

Subtitute x=z=4y/3=(4/3)^(1/3) into M , then the answer gained will be the minimum mass.
one more question,
Find the dimensions of the tank for the total amount of metal used to build be minimum.
Then my answer should be in mass or volume?

21. May 3, 2013

### Outrageous

because I mistook xyz as the volume of the container.

22. May 3, 2013

### LCKurtz

Mistook??? xyz is the volume of the container. length x width x height.

The dimensions are length, width, and height. What units are they in in this problem?

23. May 3, 2013

### Outrageous

xyz is the capacity of the container , not the volume of the container.
In m^3 , so answer should be in volume
]
We can simply add xyz-1 into the function because xyz-1=0 , so it doesn't matter what is the usage of the λ. Isn't ?

Really thank you

24. May 3, 2013

### LCKurtz

I don't know what you are thinking. The capacity of a container and its volume are the same thing.

I think it is time for me to abandon this thread. I have explained how to solve the problem and I can hardly do more without working it for you.

You need to understand that the approach RGVickson has suggested is a different approach. Perhaps that is confusing you. Use his method or mine, but don't mix them.

Last edited: May 3, 2013
25. May 3, 2013

### Outrageous

Thank you guys for spending time to explain.