In the final section of this chapter let’s take a look at
some applications of derivatives in the business world. For the most part these are really
applications that we’ve already looked at, but they are now going to be
approached with an eye towards the business world.
Let’s start things out with a couple of optimization
problems. We’ve already looked at more
than a few of these in previous sections so there really isn’t anything all
that new here except for the fact that they are coming out of the business
world.
Example 1 An
apartment complex has 250 apartments to rent.
If they rent x apartments
then their monthly profit, in dollars, is given by,
How many apartments should they rent in order to maximize
their profit?
Solution
All that we’re really being asked to do here is to
maximize the profit subject to the constraint that x must be in the range .
First, we’ll need the derivative and the critical point(s)
that fall in the range .
Since the
profit function is continuous and we have an interval with finite bounds we
can find the maximum value by simply plugging in the only critical point that
we have (which nicely enough in the range of acceptable answers) and the end
points of the range.
So, it looks
like they will generate the most profit if they only rent out 200 of the
apartments instead of all 250 of them.

Note that with these problems you shouldn’t just assume that
renting all the apartments will generate the most profit. Do not forget that there are all sorts of
maintenance costs and that the more tenants renting apartments the more the
maintenance costs will be. With this
analysis we can see that, for this complex at least, something probably needs
to be done to get the maximum profit more towards full capacity. This kind of analysis can help them determine
just what they need to do to move towards that goal whether it be raising rent
or finding a way to reduce maintenance costs.
Note as well that because most apartment complexes have at
least a few units empty after a tenant moves out and the like that it’s possible
that they would actually like the maximum profit to fall slightly under full
capacity to take this into account.
Again, another reason to not just assume that maximum profit will always
be at the upper limit of the range.
Let’s take a quick look at another problem along these
lines.
Example 2 A
production facility is capable of producing 60,000 widgets in a day and the
total daily cost of producing x
widgets in a day is given by,
How many widgets per day should they produce in order to
minimize production costs?
Solution
Here we need to minimize the cost subject to the
constraint that x must be in the
range . Note that in this case the cost function is
not continuous at the left endpoint and so we won’t be able to just plug
critical points and endpoints into the cost function to find the minimum
value.
Let’s get the first couple of derivatives of the cost
function.
The critical
points of the cost function are,
Now, clearly
the negative value doesn’t make any sense in this setting and so we have a
single critical point in the range of possible solutions : 50,000.
Now, as long as
the second derivative is positive and so, in
the range of possible solutions the function is always concave up and so
producing 50,000 widgets will yield the absolute minimum production cost.

Now, we shouldn’t walk out of the previous two examples with
the idea that the only applications to business are just applications we’ve
already looked at but with a business “twist” to them.
There are some very real applications to calculus that are
in the business world and at some level that is the point of this section. Note that to really learn these applications
and all of their intricacies you’ll need to take a business course or two or
three. In this section we’re just going
to scratch the surface and get a feel for some of the actual applications of
calculus from the business world and some of the main “buzz” words in the
applications.
Let’s start off by looking at the following example.
Example 3 The
production costs per week for producing x
widgets is given by,
Answer each of the following questions.
(a) What
is the cost to produce the 301^{st} widget?
(b) What
is the rate of change of the cost at ?
Solution
(a) We can’t
just compute as that is the cost of producing 301 widgets
while we are looking for the actual cost of producing the 301^{st}
widget. In other words, what we’re
looking for here is,
So, the cost of
producing the 301^{st} widget is $295.91.
(b) In this part all we need to do is
get the derivative and then compute .

Okay, so just what did we learn in this example? The cost to produce an additional item is
called the marginal cost and as we’ve
seen in the above example the marginal cost is approximated by the rate of
change of the cost function, . So, we define the marginal cost function to be the derivative of the cost function
or, . Let’s work a quick example of this.
Example 4 The
production costs per day for some widget is given by,
What is the
marginal cost when ,
and ?
Solution
So, we need the derivative and then we’ll need to compute
some values of the derivative.
So, in order to produce the 201^{st} widget it
will cost approximately $10. To
produce the 301^{st} widget will cost around $38. Finally, to product the 401^{st}
widget it will cost approximately $78.

Note that it is important to note that is the approximate cost of producing the item and NOT the n^{th} item as it may
seem to imply!
Let’s now turn our attention to the average cost function. If is the cost function for some item then the
average cost function is,
Here is the sketch of the average cost function from Example
4 above.
We can see from this that the average cost function has an
absolute minimum. We can also see that
this absolute minimum will occur at a critical point with since it clearly will have a horizontal
tangent there.
Now, we could get the average cost function, differentiate
that and then find the critical point.
However, this average cost function is fairly typical for average cost
functions so let’s instead differentiate the general formula above using the
quotient rule and see what we have.
Now, as we noted above the absolute minimum will occur when and this will in turn occur when,
So, we can see that it looks like for a typical average cost
function we will get the minimum average cost when the marginal cost is equal
to the average cost.
We should note however that not all average cost functions
will look like this and so you shouldn’t assume that this will always be the
case.
Let’s now move onto the revenue and profit functions. First, let’s suppose that the price that some
item can be sold at if there is a demand for x units is given by . This function is typically called either the demand function or the price function.
The revenue function
is then how much money is made by selling x
items and is,
The profit function
is then,
Be careful to not confuse the demand function,  lower case p, and the profit function,  upper case P. Bad notation maybe, but
there it is.
Finally, the marginal
revenue function is and the marginal
profit function is and these represent the revenue and profit
respectively if one more unit is sold.
Let’s take a quick look at an example of using these.
Example 5 The
weekly cost to produce x widgets is
given by
and the demand
function for the widgets is given by,
Determine the marginal cost, marginal revenue and marginal
profit when 2500 widgets are sold and when 7500 widgets are sold. Assume that the company sells exactly what
they produce.
Solution
Okay, the first thing we need to do is get all the various
functions that we’ll need. Here are
the revenue and profit functions.
Now, all the
marginal functions are,
The marginal functions
when 2500 widgets are sold are,
The marginal functions when 7500 are sold are,
So, upon producing and selling the 2501^{st}
widget it will cost the company approximately $25 to produce the widget and
they will see an added $175 in revenue and $150 in profit.
On the other hand when they produce and sell the 7501^{st}
widget it will cost an additional $325 and they will receive an extra $125 in
revenue, but lose $200 in profit.

We’ll close this section out with a brief discussion on
maximizing the profit. If we assume that
the maximum profit will occur at a critical point such that we can then say the following,
We then will know that this will be a maximum we also were
to know that the profit was always concave down or,
So, if we know that then we will maximize the profit if or if the marginal cost equals the marginal
revenue.
In this section we took a brief look at some of the ideas in
the business world that involve calculus.
Again, it needs to be stressed however that there is a lot more going on
here and to really see how these applications are done you should really take
some business courses. The point of this
section was to just give a few ideas on how calculus is used in a field other
than the sciences.