Uploaded by TheIntegralCALC on 05.06.2012

Transcript:

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.

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.