# Homework Help: Show rectangular box of given volume has minimum surface area when

1. Jul 15, 2014

### jonroberts74

1. The problem statement, all variables and given/known data

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

[gotta show it with partial derivatives to minimize]

2. Relevant equations

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

3. 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

$$2<L+H,W+H,L+W>$$

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

2. Jul 15, 2014

### Ray Vickson

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.

3. Jul 15, 2014

### jonroberts74

the constraint being

$$wlh=k$$ where k is the volume

then solving for say h $$h=\frac{k}{lw}$$

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

correct?

4. Jul 15, 2014

### Ray Vickson

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

5. Jul 15, 2014

### 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

6. Jul 15, 2014

### Ray Vickson

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.

7. Jul 15, 2014

### 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 thats 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 thats correct, taking partials

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

this gives minimizer $$(-1/4, -1/4)$$

Last edited: Jul 15, 2014
8. Jul 15, 2014

### Ray Vickson

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

9. Jul 15, 2014

### jonroberts74

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

10. Jul 15, 2014

### jonroberts74

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

11. Jul 15, 2014

### Ray Vickson

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

12. Jul 15, 2014

### 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

13. Jul 15, 2014

### jonroberts74

okay,

I'll try this way

a new function F

$$F(x,y,z,\lambda) = 2(xy+xz+yz) - \lambda(xyz-V)$$

taking partials

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

I'll solve for z

$$z=-x, z=-y \Rightarrow -x=-y \Rightarrow x=y$$

$$x(x)y=V$$ and $$x(y)y=V$$

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

Last edited: Jul 15, 2014
14. Jul 16, 2014

### ehild

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

ehild.

15. Jul 16, 2014

### jonroberts74

I did, thank you.

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

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

Last edited: Jul 16, 2014
16. Jul 16, 2014

### ehild

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

Last edited: Jul 16, 2014
17. Jul 16, 2014

### jonroberts74

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

$$y(2-\lambda z) = -2z \Rightarrow y= \frac{-2z}{2-\lambda z}$$

$$x(2-\lambda z) = -2z \Rightarrow x = \frac{-2z}{2-\lambda z}$$

$$2\frac{-2z}{2-\lambda z}+ 2\frac{-2z}{2-\lambda z} = \lambda \frac{-2z}{2-\lambda z}\frac{-2z}{2-\lambda z}$$

I gotta run to class, will finish this after

Last edited: Jul 16, 2014
18. Jul 16, 2014

### vela

Staff Emeritus
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.

19. Jul 16, 2014

### Ray Vickson

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:
$$X+Y = \lambda/2\\ Y + Z = \lambda/2 \\ Z + X = \lambda/2$$

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.

Last edited: Jul 16, 2014
20. Jul 16, 2014

### 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

21. Jul 16, 2014

### 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

22. Jul 16, 2014

### jonroberts74

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

$$A_{x} = 2y - \frac{2V}{x^2}; A_{y} = 2x - \frac{2V}{y^2}$$

now $$x = \frac{V}{y^2}; y = \frac{V}{x^2}$$

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

Last edited: Jul 16, 2014
23. Jul 16, 2014

### ehild

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

ehild

Last edited: Jul 16, 2014
24. Jul 16, 2014

### jonroberts74

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

25. Jul 16, 2014

### ehild

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