MHB Minimizing Cost: Order Size for Optimal Price

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SUMMARY

The discussion centers on minimizing the ordering and transportation cost function $C=100\left(\frac{200}{x^2}+\frac{x}{x+30}\right)$, where $C$ is in thousands of dollars and $x$ is the order size in hundreds. The optimal order size that minimizes cost is determined to be approximately 40.45, translating to 4045 units. Participants utilized derivative calculations and numeric root-finding techniques, specifically Newton's method and the TI-Nspire CAS graphing utility, to arrive at this solution.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and finding critical points
  • Familiarity with numeric root-finding techniques, such as Newton's method
  • Experience using graphing utilities, particularly TI-Nspire CAS
  • Knowledge of cost functions in manufacturing and their optimization
NEXT STEPS
  • Explore advanced applications of Newton's method in optimization problems
  • Learn about other numeric root-finding techniques, such as the bisection method
  • Investigate cost function modeling in different manufacturing scenarios
  • Study the implications of order size on supply chain management and logistics
USEFUL FOR

Mathematicians, operations researchers, supply chain analysts, and anyone involved in cost optimization in manufacturing processes will benefit from this discussion.

karush
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Minimum Cost The ordering and transportation cost $C$ of the components used in manufacturing a product is

$C=100\left(\frac{200}{x^2}+\frac{x}{x+30}\right),\ x\ge1$

Where $C$ is measured in thousands of dollars and $x$ is the order size in hundreds.
Find the order size that minimizes the cost. [Hint: use the roots feature of a graphing utility]

I took the derivative of this and set it to zero but did not get the ans of $4045$ units

from the graph it looks like the min would be 40.45 then times 100 would be the ans

thanks ahead
 
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Re: minumum cost

Differentiating and equating to zero results in:

$\displaystyle 3x^3-40x^2-2400x-36000=0$

Using a numeric root-finding technique such as Newton's method yields the real root at:

$\displaystyle x\approx40.45$

Since x represents hundred of units, the answer is then 4045 units.
 
Re: minumum cost

OK on the TI-Nspire Cas it was

zeros$\left(3x^3-40x^2-2400x-36000\right)\approx\ 40.45$
 
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