jkwiens.com: Answer: Optimize Material Cost of an Ice Tray

Answer: Optimize Material Cost of an Ice Tray

Saturday, 24 of November , 2007 at 11:54 am

Answer: The simplest way to calculate the optimal dimensions of the ice tray will be to use Lagrangian Multipliers which states that the extrema of a function , which is subjected to constraints , will satisfy the following conditions:

$\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$
In this example, the function will be the cost function:

In order to construct , it is a little more complicated. Since the volume of each compartment needs to be $12 cm^{3}$ , we can represent this constraints with the following formula: $V = \frac{x}{6}\frac{y}{2}z = 12$ . Therefore would be:
$g(x,y,z) = \frac{x}{6}\frac{y}{2}z - 12 = 0$
Since the horizontal cross-section is a square, we can use the constraint $\frac{x}{6} = \frac{y}{2}$ to simply the equations. The equations will be reduced to:
$f(x,z) = \frac{x^{2}}{3} + \frac{16}{3}xz$
$g(x,z) = x^{2}z - 432 = 0$
Now we use the relation,
$\nabla f(x,z) = \lambda \nabla g(x,z)$
to calculate the optimal dimensions of the ice tray.
$\nabla f(x,z) = \lambda \nabla g(x,z)$
$\nabla (\frac{x^{2}}{3} + \frac{16}{3}xz) = \lambda \nabla (x^{2}z - 432)$
$\begin{pmatrix} \frac{2x}{3} + \frac{16}{3}z = \lambda 2xz \\ \frac{16}{3}x = \lambda x^{2} \end{pmatrix}$
$\lambda = \begin{pmatrix} \frac{1}{3z} + \frac{8}{3x} \\ \frac{16}{3x} \end{pmatrix}$
Therefore,
$\frac{1}{3z} + \frac{8}{3x} = \frac{16}{3x}$
$16 = \frax{x}{z} + 8$

Next, insert this relation in .
$g(x,z) = x^{2}z - 432 = 0$
$g(x,z) = (8z)^{2}z - 432 = 0$
$z = \frac{3}{2} \sqrt[3]{2}$
Therefore, using the relationships we calculated above:
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 12 \\ 4 \\ \frac{3}{2} \end{pmatrix}\sqrt[3]{2}$

Category: Answers, Calculus

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