Let *f* be a function of one variable
defined for all *x* in some domain *D*. A *global
maximum* of *f* is a point in *D* such that for all *x* in *D*.

For a constant ,
the *neighborhood* of a point is the set of all
points *x* such that .
A point is a *local maximum* of *f* if there exists
such that for all *x* in
where *f*(*x*) is defined.

**Figure 1.1:** local maxima and minima

Similarly one can define local and global minima. In Figure 1.1,
the function *f* is defined for *x* in domain [ *a* , *e* ]. Points
*b* and *d* are local maxima and *b* is a global maximum,
whereas *a*, *c*, and *e* are local minima and *e* is a global minimum.

**Finding extrema**

Extrema, whether they are local or global, can occur in three places:

1. at the boundary of the domain,

2. at a point without a derivative, or

3. at a point with .

The last case is particularly important. So we discuss it
further.
Let *f* be differentiable in a neighborhood of . If
is a local extremum of *f*, then .
Conversely, if , three possibilities may arise:
is a local maximum, is a local minimum, or neither!!!
To decide between these three possibilities, one may use the
second derivative.
Let *f* be twice differentiable in a neighborhood of .

- If and then is a local minimum.
- If and then is a local maximum.
- If and then may or may not be a local extremum.

Figure 1.2 illustrates these three possibilities.

**Figure 1.2:** , and one of the possibilities
with

The revenue at price *x* is

We compute the
derivative of *g* and set it to 0.

Since *f*(*x*);*SPMgt*;0 for all *x*, setting *g*'(*x*)=0 implies 1-0.2*x*=0.
So *x*=5. This is the only possible local optimum. To determine
whether it is a maximum or a minimum, we compute the second derivative.

Putting in *x*=5 shows *g*''(*x*);*SPMlt*;0, so this is a maximum:
the oil cartel maximizes its revenue by pricing gasoline at
$5 per gallon.

We must first calculate the holding cost (cost of stockpiling).
If each order period begins with *Q* cases and ends with zero
cases, and if usage is more or less constant, then the average
stock level is *Q*/2. This means that the average holding cost
is *hQ*/2.

Since each order of *Q* cases lasts *Q*/*d* weeks, the average ordering
cost *per week* is
Thus the average total cost per week is

To find *Q* that minimizes cost, we set to zero the derivative with
respect to *Q*.

This implies that the optimal order quantity is

This is the classical *economic order quantity (EOQ)* model for
inventory management. (You will learn when the EOQ model is
appropriate and when to use other inventory models in 45-765).

Mon Aug 24 13:43:30 EDT 1998