dimensions of a box w/ fixed volume

**ikickyouindanuts** · May 11th 2008, 03:52 PM

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?

**galactus** · May 11th 2008, 04:38 PM

Are you familiar with Lagrange multipliers?.

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

subject to the constraint $xyz=32$

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

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

and

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

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

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

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

So, we get the 3 equations:

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

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

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

These can be rewritten like:

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

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

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

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

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

Sub these into the constraint and get:

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

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

You can also do this with partial derivatives.

**Soroban** · May 11th 2008, 06:55 PM

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: . $LWH \:=\:32\quad\Rightarrow\quad H \:=\:\frac{32}{LW}\;\;{\color{blue}1]}$

The base has area: . $LW$

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

The sides have area: . $2WH$

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

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

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

Take partial dereivatives and equate to zero:

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

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

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

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

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

**ikickyouindanuts** · May 11th 2008, 09:30 PM

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.

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