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acacia's question at Yahoo! Answers regarding using Lagrange Multipliers to minimize cost of a box

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MarkFL

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Feb 24, 2012
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Here is the question:

Please help!!? lagrange multiplier method?


Find the dimensions of the box which will minimize the TOTAL COST of manufacturing the following open top box of volume 6ft^3?

Solve by the lagrange multiplier method!
Bottom panel costs $3/ft^2
side panel cost $.50/ft^2
Front and back panels cost $1/ft^2
I have posted a link there to this topic so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello acacia,

I would orient the box such that the width is $x$, the height is $y$ and the length is $z$. Hence, the bottom panel has area $xz$, the side panels have a total area of $2yz$ and the front and back panels have a combined area of $2xy$.

Let all linear measures be given in feet.

Thus, our objective function, the function we wish to minimize is the cost function in dollars, which is given by:

\(\displaystyle C(x,y,z)=3xz+yz+2xy\)

Subject to the constraint on the volume:

\(\displaystyle g(x,y,z)=xyz-6=0\)

Using Lagrange multipliers, we obtain:

\(\displaystyle 3z+2y=\lambda(yz)\)

\(\displaystyle z+2x=\lambda(xz)\)

\(\displaystyle 3x+y=\lambda(xy)\)

Solving for $\lambda$, the first two equations imply:

\(\displaystyle \frac{2y+3z}{yz}=\frac{2x+z}{xz}\)

Cross-multiplying, we obtain:

\(\displaystyle 2xyz+3xz^2=2xyz+yz^2\)

\(\displaystyle 3xz^2=yz^2\)

Since the constraint requires \(\displaystyle 0<z\), we may write:

\(\displaystyle 3x=y\)

In like manner the first and third equations above imply:

\(\displaystyle \frac{2y+3z}{yz}=\frac{3x+y}{xy}\)

Cross-multiplying, we obtain:

\(\displaystyle 2xy^2+3xyz=3xyz+y^2z\)

\(\displaystyle 2x=z\)

Substituting for $y$ and $z$ into the constraint, we obtain:

\(\displaystyle x(3x)(2x)=6\)

\(\displaystyle x^3=1\)

\(\displaystyle x=1\implies y=3,\,z=2\)

Observing that:

\(\displaystyle C(1,3,2)=3(1)(2)+(3)(2)+2(1)(3)=18\)

and another constraint value such as $(x,y,z)=(1,2,3)$ yields:

\(\displaystyle C(1,2,3)=3(1)(3)+(2)(3)+2(1)(2)=19\)

We may then conclude:

\(\displaystyle C_{\min}=C(1,3,2)=18\)