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Homework Help: Minimization problem: Economics: quantity to order

  1. Sep 16, 2018 at 2:34 PM #1
    1. The problem statement, all variables and given/known data
    Capture1.JPG
    Capture.JPG
    2. Relevant equations
    Minimum/Maximum occurs when the first derivative=0

    3. The attempt at a solution
    $$Q=\sqrt{\frac{2(K+pQ)}{h}}~\rightarrow~Q=\frac{2}{h}(KM+pM)$$
    ##Q'=0~## gives no sense result
     
  2. jcsd
  3. Sep 16, 2018 at 3:50 PM #2

    Ray Vickson

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    You cannot just write ##Q = \sqrt{2(K+pQ)/h}##. You must start the whole modelling procedure over again.

    In the old model, the average cost per week was given as
    $$C(Q) = \frac{KM}{Q} + \frac{1}{2} h Q \; \hspace{2cm} (1) $$
    Solving ##C'(Q)=0## gave the economic order quantity as ##Q = \sqrt{2KM/h}.##

    What is the equation that replaces (1) in the new model?
     
    Last edited: Sep 16, 2018 at 4:20 PM
  4. Sep 18, 2018 at 11:58 PM #3
    In the old model:
    $$A(Q)=\frac{KM}{Q}+cM+\frac{hQ}{2}$$
    Where c is the purchase cost of one item. in the new model:
    $$A(Q)=\frac{K+pQ}{Q/M}+cM+\frac{hQ}{2}=\frac{KM}{Q}+(c+p)M+\frac{hQ}{2}$$
    And differentiating gives the same result
     
  5. Sep 19, 2018 at 8:12 AM #4

    Ray Vickson

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    Correct.

    BTW: why do you call this thread a "min max" problem? It is a "minimization" problem, but there is no "max" involved. In mathematics, operations research, economics and other such fields the term "min-max" has a definite meaning, and it is nothing like how you use it.
     
    Last edited: Sep 19, 2018 at 8:37 AM
  6. Sep 19, 2018 at 11:05 AM #5
    Thank you Ray, you are correct, this is only a minimum problem.
    I just automatically wrote the heading.
    Thanks
     
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