Last time we talked about how to solve related
rates problems.
Today, we’re going to be talking about another
very common application of derivatives, optimization.
This will be the last application we cover
before moving on to integrals where we'll
answer the second fundamental question of
calculus, how to find the area underneath
a curve.
Optimization is all about finding the extremes
of a function.
Basically you're looking for the largest and
smallest values that function attains in a
given range.
These kinds of points are called the function's
extrema, but you'll also hear specific points
referred to as local or global maxima and
local or global minima.
We've already mentioned extrema, and maxima
and minima, which might not be terms we're
familiar with.
Let's visualize what we mean for a second
by looking at the graph of this function.
Let's pretend we were asked to find the extrema
of this function on the range x=-2 to x=2.
On this range we can see that the function
is at it's highest at x=2, and at its lowest
at x=-2.
Because the function is higher at x=2 than
anywhere else, you can say that x=2 is its
global maximum on this range.
Similarly, because the function is lower at
x=-2 than anywhere else, you can say that
x=-2 is its global minimum.
Keep in mind that the highest and lowest points
on a range are not the only points we're interested
in.
We also care about points that are the highest
or lowest points in the neighborhood of points
around them.
For example, this point here is a local minimum
because it is the lowest point in the immediate
vicinity.
In the same way, this point is a local maximum
because it is the highest point in the immediate
vicinity.
Now that we have a visual idea of what we're
looking for, here's a summary of the steps
you'll follow to figure out the largest and
smallest values a graph attains in a given
range.
Let's go through these in more detail by trying
them out on the function we looked at earlier.
First we'll take the derivative of the original
function, then set the derivative equal to
0 and solve for x.
The solutions are the function's critical
points.
Critical points are potential points of extrema.
We can't say yet for sure whether the critical
points represent local or global maximums,
local or global minimums, or none of the above.
We have to test them to find out.
To test them we have to plot them on a number
line.
Then we'll pick one value on the number line
to the left of the left-most critical point,
one value to the right of the right-most critical
point, and one value between every critical
point.
Each of these in-between points will represent
the behavior of the graph on that range.
In other words, -2 will tell us the behavior
of the graph from negative infinity to -1.
0 will represent what the graph does between
-1 and 1/3, and 1 will tell us what the graph
is doing between 1/3 and positive infinity.
Now we'll plug the test values into the derivative
we found earlier.
When we get a positive answer we know that
the function increases on that test point's
range.
If we get a negative answer then we know that
the function decreases on the range that the
test point represents.
As we use each of the test values to test
each range, we'll indicate the result on the
number line with a direction arrow that gives
us a visual representation of the results.
The direction arrows indicate the direction
of the graph, so they represent a very rough
picture of what the graph looks like, which
means we can see from the direction arrows
that this point is a maximum and this one
is a minimum.
Now we know that we have a maximum and a minimum,
but we still don't know whether these points
are local extrema or global extrema.
To find out, we have to plug each of the critical
points and the endpoints of the range into
the original function.
Doing so will tell us the function's actual
value at these points, and we'll be able to
compare all of them to see which are the highest
and lowest.
When we plug the critical points and the endpoints
into the original function, we get the actual
values of the function at those points, and
we can see that the function is highest at
x=2, which makes it the global maximum.
The function is lowest at x=-2, which makes
it the global minimum.
That means that we're left with x=-1 as the
local maximum and x=1/3 as the local minimum.
Now that we've covered the basics of optimization,
let's turn to applied optimization to see
how this tool could be used in the real world.
We're going to review a commonly studied real-world
example, but first, let's talk about the steps
you'll follow to solve an applied optimization
problem.
Generally, the steps you'll follow to solve
an applied optimization problem are:
1.
Write down everything you've been given, and
exactly what you need to find.
2.
Set up your equations, one for the constraint,
and the other to optimize.
3.
Solve the constraint equation for one of the
variables so that you can plug it into the
optimization equation.
4.
After plugging into the optimization equation
from the constraint equation, take the derivative
of the optimization equation to find your
critical point.
Keep in mind that there may be a second critical
point that you have to eliminate.
5.
Finally, make sure to answer the question
you were really asked!
You can use the first or second derivative
test to double check yourself and ensure that
you minimized or maximized appropriately.
Let's work through a common example.
Here's the problem: You need to build an open-topped
rectangular box that has a volume of 972 cubic
inches.
The bottom of the box must be twice as long
as it is wide, and your asked to find the
dimensions of the box that will minimize its
surface area.
The first thing to do is to draw a picture
of our open-topped box.
It's always best to draw a picture of what
we're dealing with.
We know that the length is twice the width,
so we can call the length 2x and the width
x.
We also know that the volume is 972 cubic
inches.
Since we've been asked to find the dimensions
that minimize surface area, we know that the
equation we're going to optimize will be an
equation for the surface area of the box.
The constraint equation will be an equation
involving the volume of the box and it's dimensions,
since those are the constraints within which
we've been asked to work.
If we call the unknown height h, then we can
write the constraint equation as 972=(2x)(x)(h),
because we know that the equation for the
volume of a box is length times width times
height.
The optimization equation for the surface
area is A=xh+xh (for the two ends),+2xh+2xh
(for the two long sides)+2x^2 (for the bottom).
We need to get our optimization area equation
down to one variable, which means we have
to plug in for h.
We can manipulate the constraint equation
to find that h is 972/(2x^2), or 486/(x^2).
Now we can take this value for h and plug
it into the optimization equation.
We can start to calculate critical points
by taking the derivative of our optimization
equation, setting that equal to 0, and solving
for x.
We find that our critical points are x=0 and
x=9.
Because we are talking about the dimensions
of a box, we know that x=0 cannot be a solution,
so x=9 is our only valid critical point.
Since x=9, we can calculate the dimensions
as 9 by 18 by 6.
To ensure that we've found the correct dimensions,
we can confirm that x=9 is in fact a local
minimum of our function.
Using the same process we used in the last
example, we can plot x=9 on our number line,
then pick values on either side of it, like
8 and 10.
We'll plug 8 into the first derivative of
our optimization equation to find that it's
negative.
Then we'll plug 10 into the first derivative
of our optimization equation to find that
it's positive.
The negative, then positive pattern indicates
that 9 is a local minimum of the function,
which means that a critical point of x=9 and
dimensions of 9 by 18 by 6 do in fact minimize
the surface area.
Hopefully that gives you a clearer picture
of how to deal with optimization problems
and their applications.
Next time we’ll start talking about integrals,
and we'll answer the second fundamental question
of calculus: how to find the area beneath
a curve.
I’ll see you then.
