# Thread: dimensions of a box w/ fixed volume

1. ## dimensions of a box w/ fixed volume

Find the dimensions of a rectangular box, open at the top, with a volume of 32 cubic feet, using the least amount of material.
I'm not really sure even how to start this problem. Any help?

2. Are you familiar with Lagrange multipliers?.

We must minimize $\displaystyle S=xy+2xz+2yz$

subject to the constraint $\displaystyle xyz=32$

If we let $\displaystyle f(x,y,z)=xy+2xz+2yz$ and $\displaystyle g(x,y,z)=xyz$

then, $\displaystyle {\nabla}f=(y+2z)i+(x+2z)j+(2x+2y)k$

and

$\displaystyle {\nabla}g=yzi+xzj+xyk$

If follows that $\displaystyle {\nabla}g\neq{0}$ at any point on the surface

xyz=32, since x, y, and z are all non zero on this surface. So, we must have

$\displaystyle {\nabla}f={\lambda}{\nabla}g$

So, we get the 3 equations:

$\displaystyle y+2z={\lambda}yz$

$\displaystyle x+2z={\lambda}xz$

$\displaystyle 2x+2y={\lambda}xy$

These can be rewritten like:

$\displaystyle \frac{1}{z}+\frac{1}{y}={\lambda}$...[1]

$\displaystyle \frac{1}{z}+\frac{2}{z}={\lambda}$....[2]

$\displaystyle \frac{2}{y}+\frac{2}{z}-{\lambda}$....[3]

From [1] and [2] we get $\displaystyle y=x$

From [2] and [3] we get $\displaystyle z=\frac{x}{2}$

Sub these into the constraint and get:

$\displaystyle \frac{x^{3}}{2}=32$

Now, I will leave finding the values up to you. Not much left to do.

You can also do this with partial derivatives.

3. Hello, ikickyouindanuts!

Find the dimensions of an open-top rectangular box with a volume of 32 ft³,
using the least amount of material.
You need to be familiar with partial derivatives . . .
Code:
          *-----------*
/|          /|
/ |         / |
/  |        /  |H
*-----------*   |
|           |   |
|           |   *
H |           |  /
|           | /W
|           |/
*-----------*
L

The volume is 32 ft³: .$\displaystyle LWH \:=\:32\quad\Rightarrow\quad H \:=\:\frac{32}{LW}\;\;{\color{blue}1]}$

The base has area: .$\displaystyle LW$

The front and back has area: .$\displaystyle 2LH$

The sides have area: .$\displaystyle 2WH$

The total area is: .$\displaystyle A \:=\:LW + 2LH + 2WH\;\;{\color{blue}[2]}$

Substitute [1] into [2]: .$\displaystyle A \;=\;LW + 2L\left(\frac{32}{LW}\right) + 2W\left(\frac{32}{LW}\right)$

. . and we have: .$\displaystyle A \;=\;LW + 64W^{-1} + 64L^{-1}$

Take partial dereivatives and equate to zero:

. . $\displaystyle \begin{array}{cccccccccc}\frac{\partial A}{\partial L} &=& W - 64L^{-2} &=& 0 & \Rightarrow & L^2W &=& 64 & {\color{blue}[3]}\\ \frac{\partial A}{\partial W} &=& L - 64W^{-2} &=&0 & \Rightarrow & LW^2 &=& 64 & {\color{blue}[4]}\end{array}$

Equate [3] and [4]: .$\displaystyle L^2W \:=\:LW^2\quad\Rightarrow\quad L \:=\:W$

Substitute into [3]: .$\displaystyle L^3 \:=\:64\quad\Rightarrow\quad L \:=\:4\quad\Rightarrow\quad W \:=\:4$

Substitute into [1]: .$\displaystyle H \:=\:\frac{32}{4\cdot4} \;=\;2$

. . Therefore: .$\displaystyle (L,W,H) \;=\;(4,4,2)$

4. thanks a lot guys, this really helps. We have covered partial derivatives, and I knew it was possible, but I couldn't figure out where to start.
I have never heard of Lagrange multipliers.