Respuesta :
Answer:
q = sqrt(a/b)
Step-by-step explanation:
A warehouse selling cement has to decide how often and in what quantities to reorder. It is cheaper, on average, to place large orders, because this reduces the ordering cost per unit. On the other hand, larger orders mean higher storage costs. The warehouse always reorders cement in the same quantity, q. The total weekly cost, C, of ordering and storage is given by C=aq+bq, where a, b are positive constants.
Minimum C?
Why, find where dC/dq is zero.
C = a/q +bq -------a and b are positive constants,
C = a*q^(-1) +b*q
Differentiate both sides with respect to q,
dC/dq = a[-1 *q^(-2)] +b
dC/dq = -a/(q^2) +b
Make dC/dq zero,
0 = -a/(q^2) +b
a/(q^2) = b
a/b = q^2
q = sqrt(a/b) ----the value of q that will give minimum C.
It is minimum because the 2nd derivative of C, [2a/(q^3)], is positive
The total weekly cost function can be used by the warehouse to find the
most economic order quantity.
[tex](a) \ The \ term \ that \ represents \ the \ ordering \ cost \ is \ \underline{ \dfrac{a}{q}}[/tex]
The term that represents the storage cost is b·q.
[tex](b) \ The \ value \ of \ \mathbf{q} \ that \ gives \ the \ minimum \ total \ cost \ is \ \underline{ \mathbf{q} = \sqrt{\dfrac{a}{b} }}[/tex]
Reasons:
(a) The ordering cost is reduced when the quantity ordered, q, is increased
The storage cost increases as the quantity ordered, q, increases
[tex]The \ given \ weekly \ cost \ is \ C = \dfrac{a}{q} + b \cdot q[/tex]
Let O represent the ordering cost, we have;
[tex]O \propto \dfrac{1}{q}[/tex]
Therefore;
[tex]The \ ordering\ cost\ O, = \dfrac{k}{q}[/tex]
Where;
k = Constant
[tex]\dfrac{k}{q} \ is \ similar \ to \ \dfrac{a}{q} \ in \ the \ cost \ function[/tex]
Therefore, the term that represents the ordering cost is [tex]\underline{\dfrac{a}{q}}[/tex],
Let SC represent the storage cost, we have;
SC ∝ q
∴ SC = k·q
Using the the method of obtaining the term for the ordering cost, we have;
The term that represents the storage cost is b·q
b) At the minimum total cost, we have;
[tex]\dfrac{dC}{dq} = \dfrac{d}{dq} \left( \dfrac{a}{q} + b \cdot q \right) = b - \dfrac{a}{q^2} = 0[/tex]
Therefore;
[tex]The \ value \ of \ \mathbf{q} \ that \ gives \ the \ minimum \ total \ cost \ is \ \underline{\mathbf{q} = \sqrt{\dfrac{a}{b} }}[/tex]
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