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

[Outrageous]

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

Homework Statement

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.

Homework Equations

Lagrange
f(x)-λ(xyz-1)=g(x,y,z,λ)

The Attempt at a Solution

the constraint is xyz=1, but what should I use for f(x)?

Please guide , thank you.

 

[LCKurtz]

Homework Statement

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.

Homework Equations

Lagrange
f(x)-λ(xyz-1)=g(x,y,z,λ)

The Attempt at a Solution

the constraint is xyz=1, but what should I use for f(x)?

Please guide , thank you.

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

 

[Outrageous]

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

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)?

 

[HallsofIvy]

Volume of the container!?

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

∠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)?

 

[Outrageous]

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

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...

 

[LCKurtz]

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...

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?

 

[Ray Vickson]

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...

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!

 

[Outrageous]

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?

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.

 

[Outrageous]

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?

Thank you.

A bit don't 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.

 

[Outrageous]

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.

Thank you. I really like the way you explain.

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!

My answer for that is x=-4t, y=-4t, z=-3t.
I don't know what is wrong, since the t>0 , how can them be negative.
can you please tell me your answer? I will try again , please.
Sorry, what is the small t approximation version? can you tell me more please.

 

[Dick]

Thank you. I really like the way you explain.
My answer for that is x=-4t, y=-4t, z=-3t.
I don't know what is wrong, since the t>0 , how can them be negative.
can you please tell me your answer? I will try again , please.
Sorry, what is the small t approximation version? can you tell me more please.

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?

 

[Ray Vickson]

Thank you. I really like the way you explain.



My answer for that is x=-4t, y=-4t, z=-3t.
I don't know what is wrong, since the t>0 , how can them be negative.
can you please tell me your answer? I will try again , please.
Sorry, what is the small t approximation version? can you tell me more please.

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:
$$\text{metal volume} = \left( 4t + \frac{1}{3} 6^{2/3}\right)^2 \left( 3t + \frac{1}{4} 6^{2/3} \right) -1.$$

 

[LCKurtz]

Thank you.

A bit don't 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?

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 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.

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.

 

[Outrageous]

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:
$$\text{metal volume} = \left( 4t + \frac{1}{3} 6^{2/3}\right)^2 \left( 3t + \frac{1}{4} 6^{2/3} \right) -1.$$

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

 

[Outrageous]

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.

Volume of metal used is xyz-1 ?

 

[LCKurtz]

Volume of metal used is xyz-1 ?

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.

 

[Outrageous]

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 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.

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.

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?

 

[Ray Vickson]

Volume of metal used is xyz-1 ?

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!

 

[LCKurtz]

The objective function is ∫2xy+∫2yz+∫(xyz/2), where ∫ is the area density.

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

The constraint function is xyz=0.

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

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?

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.

 

[Outrageous]

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

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?

 

[Outrageous]

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!

because I mistook xyz as the volume of the container.

 

[LCKurtz]

because I mistook xyz as the volume of the container.

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

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?

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

 

[Outrageous]

xyz is the capacity of the container , not the volume of the container.
In m^3 , so answer should be in volume
]

A bit don't 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?
[/B]

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

 

[LCKurtz]

xyz is the capacity of the container , not the volume of the container.

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.

 

[Outrageous]

Thank you guys for spending time to explain.