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Q equals square root of fraction numerator 2 d K over denominator h end fraction end root
The formula above is used to estimate the ideal quantity, Q, of items a store manager needs to order given the demand quantity, d; the setup cost per order, K; and the storage cost per item, h. Which of the following correctly expresses the storage cost per item in terms of the other variables?

  1. h equals square root of fraction numerator 2 d K over denominator Q end fraction end root
  2. h equals fraction numerator square root of 2 d K end root over denominator Q end fraction
  3. h equals fraction numerator 2 d K over denominator Q squared end fraction
  4. h equals fraction numerator Q squared over denominator 2 d K end fraction

hintHint:

Hint:
The storage cost per item is given by h. We need to modify the given equation to find an expression for h. The first step is to remove the square root by squaring both sides. Next, we just need to bring h on the left-hand side and the rest of the quantities on the right-hand side. This should give us the required result

The correct answer is: h equals fraction numerator 2 d K over denominator Q squared end fraction


    Given,
    Q equals square root of fraction numerator 2 d K over denominator h end fraction end root
    where
    Q is the quantity,
    K is the setup cost per order,
    d is the demand quantity,
    h is the storage cost per item.
    Now
    Q equals square root of fraction numerator 2 d K over denominator h end fraction end root
    Squaring both sides, we have
    Q squared equals fraction numerator 2 d K over denominator h end fraction
    Multiplying h both sides,

    Q squared h equals 2 d K
    Dividing by Q squared on both sides, we get
    h equals fraction numerator 2 d K over denominator Q squared end fraction
    Thus, the value of h in terms of other variables is given by,
    h equals fraction numerator 2 d K over denominator Q squared end fraction
    The correct option is C)

    Note:
    Whenever there is an equation with a square root involved, first we try to remove that square root to get a clear picture of the equation. It is easier to operate on an equation with no square roots or log or inverse.

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