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.