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M6.5 Applications

There are many problems in which derivatives are used in business. We discuss a few of these here.

Economic Order Quantity

In Chapter 6, we show the formula for computing the economic order quantity (EOQ), which minimizes cost when certain conditions are met. The total cost formula under these conditions is

\begin{matrix} Total cost = (Total ordering cost) + (Total holding cost) + (Total purchase cost) \\ T C = \frac{D}{Q} C_{0} + \frac{Q}{2} C_{h} + D C \end{matrix}

$\begin{matrix} Total cost = (Total ordering cost) + (Total holding cost) + (Total purchase cost) \\ T C = \frac{D}{Q} C_{0} + \frac{Q}{2} C_{h} + D C \end{matrix}$

where

\begin{array}{l} Q = o r d e r q u a n t i t y \\ D = a n n u a l d e m a n d \\ C_{o} = o r d e r i n g c o s t p e r o r d e r \\ C_{h} = h o l d i n g c o s t p e r u n i t p e r y e a r \\ C = p u r c h a s e (m a t e r i a l) c o s t p e r u n i t \end{array}

$\begin{array}{l} Q = o r d e r q u a n t i t y \\ D = a n n u a l d e m a n d \\ C_{o} = o r d e r i n g c o s t p e r o r d e r \\ C_{h} = h o l d i n g c o s t p e r u n i t p e r y e a r \\ C = p u r c h a s e (m a t e r i a l) c o s t p e r u n i t \end{array}$

The variable is Q, and all of the others are known constants. The derivative of the total cost with respect to Q is

\frac{d T C}{d Q} = \frac{- D C_{o}}{Q^{2}} + \frac{C_{h}}{2}

$\frac{d T C}{d Q} = \frac{- D C_{o}}{Q^{2}} + \frac{C_{h}}{2}$

Setting this equal to zero and solving results in

Q = \pm \sqrt{\frac{2 D C_{o}}{C_{h}}}

$Q = \pm \sqrt{\frac{2 D C_{o}}{C_{h}}}$

We cannot have a negative quantity, so the positive value is the minimum cost. The second derivative is

\frac{d^{2} T C}{d Q^{2}} = \frac{D C_{o}}{Q^{3}}

$\frac{d^{2} T C}{d Q^{2}} = \frac{D C_{o}}{Q^{3}}$

If all the costs are positive, this derivative will be positive for any value of $Q > 0$ $Q > 0$ . Thus, this point must be a minimum.

The formula for the EOQ model demonstrated here is the same formula used in Chapter 6. This method of using derivatives to derive a minimum cost quantity can be used with many total cost functions, even if the EOQ assumptions are not met.

Total Revenue

In analyzing inventory situations, it is often assumed that whatever quantity is produced can be sold at a fixed price. However, we know from economics that demand is a function of the price. When the price is raised, the demand declines. The function that relates the demand to the price is called a demand function.

Suppose that historical sales data indicate that the demand function for a particular product is

Q = 6, 000 - 500 P

$Q = 6, 000 - 500 P$

where

\begin{array}{l} Q & = & q u a n t i t y d e m a n d e d (o r s o l d) \\ P & = & p r i c e i n d o l l a r s \end{array}

$\begin{array}{l} Q & = & q u a n t i t y d e m a n d e d (o r s o l d) \\ P & = & p r i c e i n d o l l a r s \end{array}$

The total revenue function is

\begin{array}{l} Total revenue & = & Price \times Quantity \\ T R & = & P Q \end{array}

$\begin{array}{l} Total revenue & = & Price \times Quantity \\ T R & = & P Q \end{array}$

Substituting for Q (using the preceding demand function) in this equation gives us

\begin{array}{l} T R & = & P (6, 000 - 500 P) \\ T R & = & 6, 000 P - 500 P^{2} \end{array}

$\begin{array}{l} T R & = & P (6, 000 - 500 P) \\ T R & = & 6, 000 P - 500 P^{2} \end{array}$

A graph illustrating this total revenue function is provided in Figure M6.6. To find the price that will maximize the total revenue, we find the derivative of total revenue:

A graph illustrates a total revenue function using a parabolic curve. — Figure M6.6 Total Revenue Function

Figure M6.6 Full Alternative Text

T R' = 6, 000 - 1, 000 P

$T R' = 6, 000 - 1, 000 P$

Setting this equal to zero and solving, we have

P = 6

$P = 6$

Thus, to maximize total revenue, we set the price at $6. The quantity that will be sold at this price is

Q = 6, 000 - 500 P = 6, 000 - 500 (6) = 3, 000 units

$Q = 6, 000 - 500 P = 6, 000 - 500 (6) = 3, 000 units$

The total revenue is

T R = P Q = 6 (3, 000) = $ 18, 000

$T R = P Q = 6 (3, 000) = $ 18, 000$

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Table of Contents for M6.5 Applications

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Table of Contents for
M6.5 Applications