Show rectangular box of given volume has minimum surface area when

[jonroberts74]

Homework Statement

show rectangular box of given volume has minimum surface area when the box is a cube

[gotta show it with partial derivatives to minimize]

Homework Equations

surface area = 2(wl+hl+hw)
volume = whl

The Attempt at a Solution

so this is the one I would be minimizing

surface area = 2(wl+hl+hw)

and the partials of it, that would be

[tex]2<L+H,W+H,L+W>[/tex]

then setting those to zero and solving for critical points.

seems flawed but maybe not.

seems that it comes out to them all being of the same value

 

[Ray Vickson]

Homework Statement

show rectangular box of given volume has minimum surface area when the box is a cube

[gotta show it with partial derivatives to minimize]

Homework Equations

surface area = 2(wl+hl+hw)
volume = whl

The Attempt at a Solution

so this is the one I would be minimizing

surface area = 2(wl+hl+hw)

and the partials of it, that would be

[tex]2<L+H,W+H,L+W>[/tex]

then setting those to zero and solving for critical points.

seems flawed but maybe not.

seems that it comes out to them all being of the same value

It is very flawed indeed. You forgot about the constraint. You either need to use the constraint to eliminate one of the variables, or else use the Lagrange multiplier method.

Also: you are assuming the box is closed. If the box is open (no top) the result is not true: there are designs that are better than the cube in that case.

 

[jonroberts74]

the constraint being

[tex]wlh=k[/tex] where k is the volume

then solving for say h [tex]h=\frac{k}{lw}[/tex]

now I have [tex]\left\{\begin{array}{cc}2(L+ \frac{k}{lw}) =0\\ 2(W + \frac{k}{lw} )=0\\ 2(L+W)=0\end{array}\right.[/tex]

correct?

 

[Ray Vickson]

the constraint being

[tex]wlh=k[/tex] where k is the volume

then solving for say h [tex]h=\frac{k}{lw}[/tex]

now I have [tex]\left\{\begin{array}{cc}2(L+ \frac{k}{lw}) =0\\ 2(W + \frac{k}{lw} )=0\\ 2(L+W)=0\end{array}\right.[/tex]

correct?

No, not correct. Please take more care and try to use what you have learned already.

 

[jonroberts74]

if the volume is not the constraint then what is? It says for a rectangular box of given volume. this problem is from the chapter before lagrange multipliers and constraints are introduced

 

[Ray Vickson]

if the volume is not the constraint then what is? It says for a rectangular box of given volume. this problem is from the chapter before lagrange multipliers and constraints are introduced

What is the expression for A(L,W) = area in terms of L and W, with volume held fixed at some constant value k? Don't guess; work it out in detail.

 

[jonroberts74]

when you are asking for A(L,W) do you mean surface area or area of a given side? I apologize, I do not mean to seem like I am simply guessing.

area of a side is L x W, surface area 8(LxW) because 8 sides of the prism.

but that's under the assumption L=W which is what I have to show.

2 sides would have area LxL and 2 sides would have area WxW then the base is LxW.

So the surface area 2(LxL)+ 2(WxW) + (LxW) [with an open top]

the book doesn't give whether it is open on top or not.

if its closed then the surface area is 2(LxL)+2(WxW)+2(WxL)

and volume = (LxL)(WxW)(LxW) = k

if that's correct, taking partials

[tex]4L+1=0, 4W+1=0[/tex] [open top] this seems incorrect too though

this gives minimizer [tex](-1/4, -1/4)[/tex]

 

[Ray Vickson]

when you are asking for A(L,W) do you mean surface area or area of a given side? I apologize, I do not mean to seem like I am simply guessing.

area of a side is L x W, surface area 8(LxW) because 8 sides of the prism.

but that's under the assumption L=W which is what I have to show.

2 sides would have area LxL and 2 sides would have area WxW then the base is LxW.

So the surface area 2(LxL)+ 2(WxW) + (LxW) [with an open top]

the book doesn't give whether it is open on top or not.

if its closed then the surface area is 2(LxL)+2(WxW)+2(WxL)

and volume = (LxL)(WxW)(LxW) = k

if that's correct, taking partials

[tex]4L+1=0, 4W+1=0[/tex] [open top] this seems incorrect too though

this gives minimizer [tex](-1/4, -1/4)[/tex]

I give up. I have told you two approaches to use, and you have ignored them both.

 

[jonroberts74]

I did not ignore them, I am trying to figure them out, thanks anyway.

 

[jonroberts74]

the constraint being

[tex]wlh=k[/tex] where k is the volume

then solving for say h [tex]h=\frac{k}{lw}[/tex]

now I have [tex]\left\{\begin{array}{cc}2(L+ \frac{k}{lw}) =0\\ 2(W + \frac{k}{lw} )=0\\ 2(L+W)=0\end{array}\right.[/tex]

correct?

It was an odd numbered question so Igave in and checked the back of the book, the answer it gave said to do this.

so I don't know what to think because you told me this is the incorrect route

 

[Ray Vickson]

It was an odd numbered question so Igave in and checked the back of the book, the answer it gave said to do this.

so I don't know what to think because you told me this is the incorrect route

I advised you to be careful. Your derivatives are all wrong. The route is OK, but you made serious errors along the way.

 

[ehild]

First write up the area in terms of W and L (substitute k/(WL) for H). Simplify and find the partial derivatives with respect to W and L after.

ehild

 

[jonroberts74]

okay,

I'll try this way

a new function F

[tex]F(x,y,z,\lambda) = 2(xy+xz+yz) - \lambda(xyz-V)[/tex]

taking partials

[tex]\left\{\begin{array}{cc}F_{x}=2y+2z\\F_{y} = 2x+2z\\F_{z}=2x+2y\\ F_{\lambda}=-xyz+V\end{array}\right.[/tex]

I'll solve for z

[tex]z=-x, z=-y \Rightarrow -x=-y \Rightarrow x=y[/tex]

[tex]x(x)y=V[/tex] and [tex]x(y)y=V[/tex]

I do appreciate the help, I've said it previously. I have a lot of difficulty reading english. So it may seem like I am not trying but I am.

 

[ehild]

okay,

I'll try this way

a new function F

[tex]F(x,y,z,\lambda) = 2(xy+xz+yz) - \lambda(xyz-V)[/tex]

taking partials

[tex]\left\{\begin{array}{cc}F_{x}=2y+2z\\F_{y} = 2x+2z\\F_{z}=2x+2y\\ F_{\lambda}=-xyz+V\end{array}\right.[/tex]

The partials are wrong again. You ignored λ(xyz-V).

ehild.

 

[jonroberts74]

The partials are wrong again. You ignored λ(xyz-V).

ehild.

I did, thank you.

[tex]\left\{ \begin{array}{cc} 2y+2z - \lambda yz \\ 2x+2z - \lambda xz \\ 2y+2x - \lambda zx \\ -xyz +V \end{array}\right.[/tex]
[tex]z=\frac{V}{xy}[/tex]

[tex]\left\{ \begin{array}{cc} 2y+2\frac{v}{xy} - \lambda y\frac{v}{xy} \\ 2x+2\frac{v}{xy} - \lambda x\frac{v}{xy} \\ 2y+2x - \lambda \frac{v}{xy} x \\ -xy\frac{v}{xy} +V \end{array}\right.[/tex]

 

[ehild]

I did, thank you.

[tex]\left\{ \begin{array}{cc} 2y+2z - \lambda yz \\ 2x+2z - \lambda xz \\ 2y+2x - \lambda zx \\ -xyz +V \end{array}\right.[/tex]
[tex]z=\frac{V}{xy}[/tex]

The third equation is wrong. You have to differentiate with respect to z.
Do not proceed by substituting z=V/(xy). Isolate y from the first equation and x from the second one in terms of z and λ. Substitute for x and y in the third equation. Solve for z in terms of λ. With the expression for z, get x and y...

ehild

 

[jonroberts74]

[tex]\left\{ \begin{array}{cc} 2y+2z - \lambda yz \\ 2x+2z - \lambda xz \\ 2y+2x - \lambda yx \\ -xyz +V \end{array}\right.[/tex]

[tex]y(2-\lambda z) = -2z \Rightarrow y= \frac{-2z}{2-\lambda z} [/tex]

[tex]x(2-\lambda z) = -2z \Rightarrow x = \frac{-2z}{2-\lambda z}[/tex]

[tex]2\frac{-2z}{2-\lambda z}+ 2\frac{-2z}{2-\lambda z} = \lambda \frac{-2z}{2-\lambda z}\frac{-2z}{2-\lambda z} [/tex]

I got to run to class, will finish this after

 

[vela]

The third equation is wrong. You have to differentiate with respect to z.
Do not proceed by substituting z=V/(xy). Isolate y from the first equation and x from the second one in terms of z and λ. Substitute for x and y in the third equation. Solve for z in terms of λ. With the expression for z, get x and y...

I think it would be simpler to solve for ##\lambda## in each of the three equations and set the results equal to each other. It's pretty straightforward from there to show that x=y=z.

 

[Ray Vickson]

I did, thank you.

[tex]\left\{ \begin{array}{cc} 2y+2z - \lambda yz \\ 2x+2z - \lambda xz \\ 2y+2x - \lambda zx \\ -xyz +V \end{array}\right.[/tex]
[tex]z=\frac{V}{xy}[/tex]

[tex]\left\{ \begin{array}{cc} 2y+2\frac{v}{xy} - \lambda y\frac{v}{xy} \\ 2x+2\frac{v}{xy} - \lambda x\frac{v}{xy} \\ 2y+2x - \lambda \frac{v}{xy} x \\ -xy\frac{v}{xy} +V \end{array}\right.[/tex]

The equation ##F_x = 0## can be written as ##Y + Z = \lambda/2##, where ##Y = 1/y## and ##Z = 1/z## (obtained from ##F_x/(yz) = 0##). Similarly for the others, so altogether (with ##X = 1/x## as well) we have:
[tex] X+Y = \lambda/2\\
Y + Z = \lambda/2 \\
Z + X = \lambda/2[/tex]

You can either solve for ##X,Y,Z## in terms of ##\lambda## or else just use the equations to show that ##X = Y = Z##. This corresponds to a cube.

For the open-top case the same method can be used.

 

[jonroberts74]

I believe this is what I was trying to get at before. As always, we complicate something confusing ourselves--especially new material. Thanks everyone

 

[ehild]

Yes, the simplest way would have been without Lagrange multiplier, but your partial derivatives were not correct.

The area was A=2(xy+yz+xz) with the constraint V=xyz.

z=V/(xy), substitute into A:

A=2(xy+V/x+V/y)

Take the partial derivatives with respect to x and y and equate them with 0. Proceed

ehild

 

[jonroberts74]

Yes, the simplest way would have been without Lagrange multiplier, but your partial derivatives were not correct.

The area was A=2(xy+yz+xz) with the constraint V=xyz.

z=V/(xy), substitute into A:

A=2(xy+V/x+V/y)

Take the partial derivatives with respect to x and y and equate them with 0. Proceed

ehild

A=2(xy+V/x+V/y) gives partials

[tex]A_{x} = 2y - \frac{2V}{x^2}; A_{y} = 2x - \frac{2V}{y^2}[/tex]

now [tex]x = \frac{V}{y^2}; y = \frac{V}{x^2}[/tex]

I can see from when I solved for z that what I get for x and y, that y^2 is equal to yz had I solved for x in the constraint. and similar for initially solving for y in the constraint. I see the equality but I am having difficulty composing my thoughts

 

[ehild]

A=2(xy+V/x+V/y) gives partials

[tex]A_{x} = 2y - \frac{2V}{x^2}; A_{y} = 2x - \frac{2V}{y^2}[/tex]

now [tex]x = \frac{V}{y^2}; y = \frac{V}{x^2}[/tex]

I can see from when I solved for z that what I get for x and y, that y^2 is equal to yz had I solved for x in the constraint. and similar for initially solving for y in the constraint. I see the equality but I am having difficulty composing my thoughts

Substitute ##x = \frac{V}{y^2}## into the equation for A_x: ##y - \frac{V}{x^2}=0## and solve for y...

ehild

 

[jonroberts74]

got it! thanks so much and thank you for your patience, everyone. much appreciated

 

[ehild]

now [tex]x = \frac{V}{y^2}; y = \frac{V}{x^2}[/tex]

I can see from when I solved for z that what I get for x and y, that y^2 is equal to yz had I solved for x in the constraint. and similar for initially solving for y in the constraint. I see the equality but I am having difficulty composing my thoughts

Show your thoughts mathematically . Your method is nice and simple, but it is difficult to follow when you write it in words. So you substituted ##x = \frac{V}{y^2}## into the constraint equation V=xyz and you got ##V=\frac{V}{y^2}yz## which simplifies to ##1=\frac{z}{y}## that is, y=z. Doing the same with ##y = \frac{V}{x^2}## you got x=z. So x=y=z .ehild