Question

solving for minimum cost:

I have two formulas that I understand are correct, and well I'm lost as to solving for the variables with the one constant, the volume.

V = xyz
C(xyz) = xy+yz+xz

With the V set to any number I'm at a loss....I'm able to set the V myself and I've tried 34.

Answer 1

I know that I can do the equation 34=xyz and solve for say z and then place that into the other equation but after that I'm at a loss.

Answer 2

\[z=\frac{ 34 }{ xy }\]
\[C(xyz) = 2(xy+y(\frac{ 34 }{ xy })+x(\frac{ 34 }{ xy }))\]
But what to do with this is where I'm lost.

Answer 3

Do I take the derivative or something like that? Set something equal to 0?

Answer 4

This looks like a Lagrange multiplier question... have you learned about that yet?

Answer 5

don't understand it very well....we skimmed that area in class.

Answer 6

Hmm, let's see if there's a simpler way then :)

From what I can tell, you're making boxes of dimensions \(x\times y\times z\), and the cost comes from the material used to produce them. Cost is thus based on the box's surface area.

Are you sure about the cost function? The surface area of a box would be \(2xy+2xz+2yz\), or if it's a box with an open top, \(xy+2xz+2yz\) or some variant. Is it a special kind of box with three faces?

Answer 7

it's a closed rectangular box and the sides are actually labeled b, c, d so the final cost formula would be C(xyz)=2(bxy+cyz+dxz)

Answer 8

I just thought trying to solve for xyz first would be easier then dealing with b c and d

Answer 9

Those \(b,c,d\)s are throwing me off, could you label the box accordingly?

Answer 10

X = width
Y = height
z = length
B = front and back
C = left and right
D = top and bottom

Answer 11

So by labels, you mean actual labels like with a brand name? Are they constant then?

Answer 12

yep along with the volume

Answer 13

Okay, so your substitution would work if we could also get rid of the \(y\). Are one of the faces square by any chance, or do we have any info about the ratio between certain dimensions?

If not, I'm afraid Lagrange is the way to go.

Answer 14

no ratio

Answer 15

Okay, well Lagrange multipliers aren't too difficult to work with, from what I can remember. You want to minimize a certain function subject to a constraint - in this case, cost subject to a constant volume.
\[\begin{cases}C(x,y,z)=2bxy+2cyz+2dxz\\
V(x,y,z)=xyz&\text{where }V=34\end{cases}\]
The method involves solving the following system:
\[\begin{cases}\nabla C(x,y,z)=\lambda\nabla V(x,y,z)\\
V(x,y,z)=34\end{cases}\]
That is,
\[\begin{cases}\langle2by+2dz,2bx+2cz,2cy+2dx\rangle=\lambda\langle yz,xz,xy\rangle\\xyz=34\end{cases}\]
Follow so far?

Answer 16

yeah

Answer 17

Great. The first equation has you matching up the respective components:
\[\begin{cases}2by+2dz=\lambda yz\\
2bx+2cz=\lambda xz\\
2cy+2dx=\lambda xy\\
xyz=34\end{cases}\]
Multiplying the first, second, and third equations by \(x\), \(y\), and \(z\), respectively, gives
\[\begin{cases}2bxy+2dxz=\lambda xyz\\
2bxy+2cyz=\lambda xyz\\
2cyz+2dxz=\lambda xyz\\
xyz=34\end{cases}~~\iff~~
\begin{cases}2bxy+2dxz=34\lambda\\
2bxy+2cyz=34\lambda\\
2cyz+2dxz=34\lambda\end{cases}\]
Now each equation shares a term with another equation, so you should be able to solve by elimination.

Answer 18

Awesome thanks.

Answer 19

No problem. Here's a resource in case you need to see some examples carried out: http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx

Answer 20

Proper elimination would give you
\[\begin{cases}2dxz-2cyz=0\\
2bxy-2dxz=0\\
2bxy-2cyz=0\end{cases}~~\Rightarrow~~
\begin{cases}dx=cy\\
by=dz\\
bx=cz\end{cases}\]
You should be able to find nice substitutions for a single variable now and plug into the volume equation.

For example, solving in terms of \(x\) gives you \(y=\dfrac{d}{c}x\) and \(z=\dfrac{b}{c}x\), so the volume equation is
\[\begin{align*}V&=xyz\\\\
34&=x\left(\frac{d}{c}x\right)\left(\frac{b}{c}x\right)\\\\
34&=\frac{bd}{c^2}x^3\\\\
x^3&=\frac{34c^2}{bd}\\\\
x&=\sqrt[3]{\frac{34c^2}{bd}}\end{align*}\]
This in turn gives
\[d\sqrt[3]{\frac{34c^2}{bd}}=cy~~\iff~~y=\sqrt[3]{\frac{34d^2}{bc}}\]
and
\[b\sqrt[3]{\frac{34c^2}{bd}}=cz~~\iff~~z=\sqrt[3]{\frac{34b^2}{cd}}\]
The answer's kinda ugly but there's some neat symmetry between the dimensions.