AP Calculus AB Optimization Problems I

Hello, welcome back to www.educator.com, and welcome back to AP Calculus.0000

Today, we are going to start talking about optimization and optimization problems,0004

otherwise referred to as maxima and minima with practical application.0010

We have talked about maxima and minima in terms of just functions themselves.0015

Now we are going to apply them to real life situations.0019

There is going to be some quantity that we are going to want to maximize or minimize.0023

In other words, optimize, how to make it the best for our particular situation.0029

The calculus of these problems is actually very simple.0037

Essentially, what you are doing is just taking the first derivative.0040

You are setting it equal to 0 and you are solving.0042

The difficulty with these problems is putting all of this information into an equation.0045

It is the normal problems that people had with word problems, ever since we are introduced to word problems.0054

In any case, let us just jump right on in.0063

What I’m going to do is the first problem, I’m just going to launch right into it so that you get a sense of what it is.0066

I’m going to quickly discuss what is necessary for these problems, and then we are just going to do more.0071

The only way to make sense of them is to do as many problems as possible.0076

This is going to be the first of those lessons.0080

This problem says, if 1400 m² is available to make a box with a square base and no top,0086

find the dimensions of the box that gives the greatest volume.0094

I think I’m going to do this in blue again.0102

Probably, the most important thing to do with all of these optimization problems is draw a picture.0106

Always draw a picture.0112

99% of the time you really need to just draw a picture.0117

Let us see what this is asking.0120

I have got myself a box, let me go ahead and draw a little box here.0122

It is telling me that this box has no top.0133

I want to find the dimensions of the box that gives the greatest volume and also tells me that it has a square base.0137

Therefore, I’m going to call this x, I’m going to call this x.0143

It says nothing about the height, I’m just going to call this h.0146

Find the dimensions of the box that gives the greatest volume.0153

The thing that we are trying to maximize is the volume.0156

All of these problems will always be the same.0159

They are going to ask for some quantity that is maximized or minimized.0161

They are going to give you other information that relates to the problem.0166

The first thing you want to do is find just the general equation for what is being maximized.0170

In this case, it is the volume, greatest volume.0175

What we want to do is, volume, I know here is going to be x² × h.0180

We want to maximize that.0185

When you maximize or minimize something, you are finding the places where the derivative is equal to 0.0189

Once you have an equation, you are going to take the derivative of that equation, set it equal to 0, and solve for x.0195

The problem arises, notice that this is a function of two variables.0200

We cannot do that, this is a single variable calculus.0204

We need to find a way to convert this equation into an equation and just one variable, either h or x.0208

That is going to be our task.0215

This is where the problems tend to get more complicated.0217

Let us see what we can, let us write all of this out.0221

We want to maximize v but it is a function of two variables, mainly x and h.0224

Now we use the other information in the problem to establish a relation between these two variables,0254

so that I can solve for one of those variables.0260

Plug into this one and turn it into a function of one variable, that is essentially all of these problems are like that.0263

There is other information in the problem0272

that allows us to establish a relation between x and h, and that is this.0286

They are telling me that I have a total of 1400 m² total, that means the base, the area of the base, and the 4 sides.0311

The base, this side, this side, that side, and that side, let us write that out0323

The area of the base is going to be x².0329

The area of one of these side panels is going to be xh.0334

There are 4 of them, + 4 xh.0337

The sum of those has to be 1400.0342

That is it, we have a second equation, it relates x and h.0346

Let us solve for either x or h and put it in, not a problem at all.0349

What I'm going to do is I'm going to go ahead and solve for h.0355

4 xh = 1400 - x².0358

Therefore, h = 1400 - x²/ 4x.0364

We put this, we put this h into there, and we turn it into a function of one variable x which we can solve.0381

We put this h into v = x² h to get an equation in one variable.0399

In this case, I chose x.0418

Let us go ahead and do that.0425

We have v is equal to x² × h which is 1400 - x²/ 4x.0428

Cancel that, cancel that, multiply through.0444

We end up with 1400 x – x³/ 4.0448

If you want, you can rewrite it as –x³/ 4 + 350x.0458

It is totally up to you how you want to do it.0467

But now I have my equation, I have my v.0469

Now the volume of this box is expressed as a function of a single variable.0479

We know that a function achieves its absolute max or min, in this case, we are talking about a max,0489

I’m just going to leave it as maximum.0515

It achieves its absolute max either at an endpoint of the domain or somewhere in between where the derivative is 0.0518

In other words, a local max/local min.0527

We know that a function achieves its absolute max either at the endpoints of its domain or where f’ is equal to 0.0529

We differentiate this function now.0556

This is the equation of volume, we want to maximize this equation.0562

In order to maximize it, we are going to take the derivative of it, set it equal to 0,0566

and find the places where it either hits a maximum or a minimum.0570

Vx is a function of x is equal to 350x – x³/ 4.0577

V’(x) = 350 - ¾ x².0591

I’m going to set that equal to 0.0598

I have got ¾ x² is equal to 350.0601

When I solve this, I get x² = 1400/3 which gives me x is equal to + or -21.6.0608

We are talking about a distance.0627

Clearly, the negative is not going to be one of the solutions.0628

It is the +21.6 that is going to be the solutions.0632

Let us go over to the next page.0642

First of all, x is a physical length.0644

The -21.6 is not an option.0657

Second, if you rather not think about it physically and have to decide which value that you are going to take, there is another way of doing it.0668

If you prefer a more systematic or analytical approach0680

to excluding a given root or a given possibility, you can do it this way.0702

We said that v(x) is equal to -3/ 4 x³ + 350x.0716

That was the function that we want.0732

That was our original function, -x³.0740

Let me write this again.0747

We said that we had –x³/ 4 + 350x.0752

I will write it this way.0763

I know that when I graph this, I'm looking at this, and this is a cubic function.0765

This is a cubic function and the coefficient of -1/4, the leading coefficient is negative.0772

A normal cubic function begins up here, has two turns and ends down here.0779

This is negative, negative begins up here and ends down here.0790

I already took the derivative and I found that -21.6 and +21.6 are places0804

where it hits a local max or local min because I set the derivative equal to 0.0808

Therefore, I know that -21.6, there is all local min.0814

+21.6, there is a local max.0820

In this particular case, I also know that when x is equal to 0, the function is equal to 0.0823

I know it crosses here.0829

Therefore, I know for a fact that the thing goes like this.0830

Therefore, the maximum is achieved at +21.6.0836

The minimum of the function is achieved at -21.6.0841

We can also use our physical intuition to say that you cannot have, like we did for the first part,0845

like we did for our first consideration, right here.0850

It is a physical length.0853

This is the part of the graph that I'm concerned with.0856

As x gets bigger, there is a certain value of x which happens to be 21.6 where the function –x³/ 4 + 350x is maximized.0859

They gave us the greatest volume.0871

You want to use all the resources at your disposal, if you are dealing with a function.0873

You know what a cubic function looks like, where the negative over the leading coefficient is negative, it looks like this.0877

This tells you systematically, analytically, that -21.6 is not your solution.0885

Not to mention the fact that it physically makes no sense.0891

There are many things that you want to consider.0894

You do not just want to do the calculus.0896

Whatever you get, you want to stop and think about if the calculus makes sense.0899

Does your -21.6, does your +21.6 actually makes sense?0904

It does, based on other things that you need to consider.0909

Let us see, where are we, we are not done yet.0916

Let us go ahead.0922

We know that x = 21.6, that is the dimension of our base.0928

For h, h is equal to 1400 - x²/ 4x which is equal to 1400 - 21.6²/ 4 × 21.6.0934

When we do the calculation, we get xh = 10.8.0957

There you go, our box is 21.6 by 21.6 by 10.8.0963

Our unit happens to be in centimeters.0975

There you go, that is it, nice and simple.0978

Let us go ahead and actually show you the particular graph.0983

This is the graph of the function, volume function.0987

This is volume = 1400x – x³/ 4.0992

21.6 is right about there, that is our maximum point.1006

This was the function that we wanted to maximize.1011

In this particular case , we have a certain restriction on the domain.1014

This right here, that is the particular domain of this function.1019

The smallest that x can be is 0, no length.1026

The biggest that x can be is whatever that happens to be, when you set this equal to 0.1030

It turns out that x is equal to about 37.4, that is the other root of this equation.1037

That is the other 0 of that equation so that give us a natural domain.1044

In other words, if x = 0, there is no box.1048

If x = 37.4, there is no box.1052

Between 0 and 37.4, for a value of x, which is the base of the box, x by x, the volume goes up and comes down.1055

There is some x value that maximizes the volume.1067

That x is the 21.6 that we found, local maximum of this function.1070

Again, you can use the graph to help you out to find your domain, to restrict your domain, whatever it is that you need.1077

Let us talk about this a little bit.1087

All optimization problems are fundamentally the same.1089

There is a quantity that is asked to be maximized or minimized.1116

It might be an area, might be a volume.1143

It might be a distance, it might be an angle, whatever it is.1145

There are some quantity that is maximized or minimized.1150

Two, your task is to find a general equation for that quantity, for this quantity.1153

Number 3, if the equation that you get in part 2, if the equation is a function of more than one variable,1172

you use other information in the problem + any other mathematical manipulation you need1197

to find a relation between or among the variables.1236

I say among because you might end up with a general equation that has 3 or 4 variables.1250

And you have to find the relationship among all 3 or 4, not just between the two.1255

Part 4, you use the relations above among the variables1263

to express the desired quantity as a function of one variable, if possible.1285

Again, there might be situations where, we will do when we come up with them, not a problem.1308

I know the thing that you might want to do, this is a little looser but it is always a good idea to do this, if you need to.1317

A lot of this will come up with more experiences in solving these kind of problems.1323

You want to find the domain of the equation.1328

The reason you want to find the domain is,1335

Remember, what we are find here is absolute maximum of a function.1339

The absolute maximum of a function can happen within the domain, at places where it is a local max or min.1344

That is where you set the function, the derivative of a function equal to 0.1349

But you also have to consider the endpoints.1352

If you know the domain, if a domain is a closed interval, like it was in the first problem, 0 and 37.4,1355

you are still going to check those points to see if the value of the function that you get is going to be greater.1363

Because we want to find the absolute maximum.1370

Let us say there were two points in an interval, in the domain.1374

Let us go back to the first problem.1380

You had, 0 you have a 21.6, and you have a 37.4.1381

The 21.6 is the answer but you still have to technically check the 0 and the 37.4.1386

Put those values of x into the original equation.1392

You are going to get 0 for the value of the function.1395

When you put 21.6 in, you are actually going to get a number that is the biggest one among the three.1399

You remember when we were doing absolute maxes and absolute mins,1406

we have to check the values at the endpoints to see if maybe f of those values was actually bigger than what it is at a local max or min.1409

Again, the problems will help make more sense of this.1420

And then, once you have all of this information, you find the absolute max or min.1426

You find the absolute max or min.1434

If your domain is not a closed interval, that does not matter.1439

All you need to do is look for the local maxes and mins.1443

That is where you are going to pick one of those to maximize or minimize, whichever is it that you are trying to do.1446

Again, if you have a closed interval, you have to check the end points of the domain.1452

Most important, draw a picture always.1458

Always draw a picture.1471

Let us do some more examples here.1475

Demonstrate that of all rectangles with a given perimeter, the one with the largest area is a square.1477

Pick a random rectangle.1487

I’m going to call this x, I’m going to call this y.1491

In short, demonstrate that the one with the largest area is a square.1496

In short, we must show that y is equal to x, that it is a square.1502

Of all rectangles with a given perimeter.1520

The perimeter, that equals 2x + 2y, and they say of a given perimeter, some constant 5, 10, 20, 30, 86.6, whatever.1523

I'm just going to say c, c stands for a constant.1536

One of the largest area is a square.1540

The general equation for area is xy.1544

The largest area, that is the one they want us to maximize right here.1548

Largest area means maximize this, maximize this.1554

It is a function of two variables.1570

It is a function of two variables, I need a relationship between those two variables x and y,1573

in order for me to turn this into a function of one variable.1577

I have a relationship, that is my relationship right there.1582

I’m going to solve for y and plug it into this equation right over here.1585

I’m going to write 2y = c - 2x.1589

I have y = c - 2x/ 2.1595

I’m going to put this into here.1603

I get the area = c/2 – x.1608

I get the area = cx/2 – x².1622

This is my function, that is the function.1628

Now it is a function, I’m trying to maximize it.1634

I now have the area expressed as a function of one variable, x and x².1636

It is taken into account the perimeter.1641

The c, that is where that comes in.1643

Now I have to differentiate.1646

A’ is equal to c/2 - 2x, I set that equal to 0.1648

When I solve for this, I get 2x = c/2 which implies that x = c/4.1656

I found what x has to be.1673

Let us find y.1678

We said that y is equal to c/2 – x, that is equal to c/2 – c/4.1686

C/2 – c/4, it is equal to c/4.1697

Y does equal x which equals c/4.1705

I have demonstrated that, in order to maximize an area of a given rectangle.1710

I have maximized it by finding the derivative of the function of the area.1716

Found the value of x, it turns out that it has to be a square.1721

For a fixed perimeter, the sides have to be the perimeter divided by 4.1725

That is it, square.1730

I have demonstrated what is it that I set out to demonstrate.1733

Let us see here.1740

Notice that I did not explicitly specify a domain.1745

Let us tighten this up a little bit and talk about the domain.1773

Let us tight this up and discuss domain.1781

For a rectangle with a given perimeter c, the domain 0 to c/2.1790

The domain is what the x value can be.1824

If x = 0, if I take the endpoint x = 0.1826

Then, 2 × 0 + 2y is equal to c.1834

2y is equal to c, y = c/2.1847

The area is equal to x × y, that is equal to 0 × c/2, the area is 0.1856

That is this endpoint.1872

If x = c/2, then the perimeter 2 × c/2 + 2y which is equal to c, we get c + 2y = c.1879

We get 2y = 0, we get y = 0.1900

The area equals xy which equals c/2 × 0.1904

Again, the area = 0, our domain is this.1909

X cannot go past c/2 because we already set that the perimeter has to be c.1917

If you have a rectangle where this is c/2 and this is c/2, basically what you have is just a line because there is no y.1926

X, our domain, has to be between 0 and c/2.1943

When we check the endpoints, we got a value of 0.1947

C/4 is in the domain and it happens to be the local max.1952

When you put c/4 into this, you are actually going to get an area that is a number.1957

We know that that number is the maximum, precisely because of how we did it.1973

We took the derivative, we set it equal to 0, and that is what happened.1978

Let us go ahead and actually take a look at this.1984

In both cases, let me actually draw it out.1987

In both cases that we just did for the endpoints, the area was equal to 0.1993

Between 0 and c/2, there is a number such that a is maximized.2005

That number was c/4.2024

What did we say our function was, our a’, our a?2029

We said that our area function of x was equal to -x² + cx/2.2034

This is a quadratic function where the leading coefficient is negative.2041

I know that the graph goes like this.2047

I know that there are some point where I’m going to hit a maximum, that is what is going on here.2049

This is my 0, this is my c/2.2055

Let us see what this actually looks like.2058

I have entered the function cx/2 - x².2061

I have taken a particular value of c = 15.2064

I end up with this graph.2068

Notice 0, c/2, 15/2 is 7.5, that 7.5.2069

This right here, this is c/2, this is c/4.2077

That is why I hit my max.2080

This is the function that I’m maximizing.2082

It happens to be the quadratic function where the leading coefficient is negative.2086

Therefore, I know that this is the shape.2090

If I do not know it, let me use a graphical utility to help me out.2092

If I need a graphical utility to help me get the domain, that is fine.2095

I do not necessarily need this, I already know that if my perimeter is c, the most that any one side can be is c/2.2099

Therefore, my domain is 0 to c/2, hope that makes sense.2107

Let us see what we have got here.2115

What is our next one?2119

Find the points on the ellipse 9x² + y² = 9, farthest away from the point 1,0.2120

Let us go ahead and draw this out.2130

I got myself an ellipse.2134

I have got 9x² + y² = 9 x²/ 1² + y²/ 3² is equal to 1.2141

I have got, this is 1, this is 1, this is 1, 2, 3, 1, 2, 3.2158

I have an ellipse that looks like this.2166

Find the points on the ellipse farthest away from the point 1,0.2172

Here is my point 1,0, I need to find the points on the ellipse that are the farthest away from this.2176

Just eyeballing it, I’m guessing it is somewhere around here.2181

We will try to maximize this distance right here.2187

We want to maximize the distance from the point 1,0 to some random point xy on the ellipse, that satisfies this equation.2194

We know we are going to have two answers.2220

We already know that.2222

This is going to be xy1, xy2.2225

Probably you are going to have the same value of x, different values of y.2228

We want to maximize the distance, the distance formula.2233

The distance formula = x2 - x1² + y2 - y1², all under the radical sign.2239

Let me put it in.2252

I have x - 1² - y - 0².2253

This is going to give me x - 1² + y², all under the radical.2270

Let us move on to the next one.2282

I have got d is equal to, I expand the x - 1².2285

I get x² - 2x + 1 + y², all under the radical.2290

I know that 9x² + y² = 9.2300

Therefore, y² = 9 - x².2305

I put that into here, I find my d is equal to x² - 2x + 1 + 9 - 9x².2310

Therefore, I get d = -8x² - 2x + 10.2326

This is my distance function expressed as a single variable x.2337

This is what I want to maximize.2341

Maximize this, we maximize it, we take the derivative and set it equal to 0.2346

D’(x) that is going to equal ½ of -8x² - 2x + 10⁻¹/2 × the derivative of what is inside which is -16x -2.2354

D’(x), when I rearrange this, I get -8x – 1/ √-8x² – 2x + 10.2376

I set that equal to 0.2393

What I get is -8x - 1 = 0.2396

When I solve this, I get 8x = -1, x = -1/8.2400

I have found my x, my x value is -1/8.2410

Now I need to find my y so that I can find what the two points are.2414

I know that the function was 9x² + y² = 9.2424

I’m going to go 9 × -1/8² + y² = 9.2431

I get 9/64 + y² = 9, that gives me y² = 9 - 9/64.2441

I get y² = 567/64.2457

Then, I get y = + or -567/64 that equals + or -2.97.2465

Therefore, I have -1/8 - 2.97, that is one point, I have -1/8 and 0.97.2486

These two points are the points that are on the ellipse, farthest away from the point 1,0.2501

Once again, I have an ellipse, this is 1,0.2510

The points are here and they are here.2516

Those are the points that are farthest away from that.2518

This is the function that we have to maximize.2526

This graph, this is not the ellipse.2528

This is the function we have to maximize.2531

This is the -8x² - 2x + 10, under the radical.2535

This is the function that we maximized.2542

It happens to hit a maximum at -1/8.2545

Be careful, this is not the ellipse, this is the function that you end up deriving, that you needed to maximize.2554

We needed to maximize the distance.2563

It actually gives me the x value.2567

Once I have the x value, I put it back into the original equation for the ellipse to find out where the y values are for the ellipse.2570

There are a lot to keep track of.2579

My best advice with all of math and science is go slowly, that is all.2582

Let us do one last example here.2590

Find the dimensions of the rectangle of the largest area.2592

We are going to be maximizing area, know that already.2595

That can be inscribed in a circle of a given radius r.2598

Let us draw it out.2602

We have a circle and we are going to try to inscribe some random rectangle in it.2604

Probably, not going be the best drawing in the world, sorry about that.2612

It tells me that the radius is r.2614

We are going to maximize area.2620

I’m going to call this x, and I’m going to call this side y, of the rectangle.2622

Area is equal to x × y.2627

We have our general equation, we want to maximize this.2631

I have two variables, I need to find the function of one variable.2639

I have to find the relationship between x and y.2642

I have a relationship between x and y.2645

If I draw this little triangle here, this side is y divided by 2 and this side is x/2.2648

Therefore, I have by the Pythagorean theorem, x/ 2² + y/ 2² = r².2663

I have got x²/ 4 + y²/ 4 = r² which gives me x² + y² = 4r²,2674

which gives me y² = 4r² - x², which gives me a y equal to √4r² - x².2688

This is what I plug into here, to this.2701

Therefore, I get an area which is equal to x × 4r² - x².2706

Now I have a function of one variable.2716

Take the derivative and set it equal to 0.2719

A’(x) is equal to this × the derivative of that, x × ½ of 4r² – x²⁻¹/2 × the derivative of what is inside.2722

4r² is just a constant at 0.2737

It is only -2x + that × the derivative of that.2740

We get just 4r² - x² × 1.2745

I rearranged this to get, 2 and 2 cancel, -x².2752

I get -x²/ 4r² - x², under the radical, +√4r² - x².2762

Then, I find myself a nice common denominator.2777

I end up with a’(x) is equal to -x² + 4r².2779

I hope the algebra is not giving you guys any grief.2788

I just found the common denominator, over √4r² - x².2792

This is the derivative we said is equal to 0.2797

When we set it equal to 0, I have the top -x² - x².2801

I end up with a’(x) =, this is 0 so the denominator goes away.2806

I’m left with -2x² + 4r² = 0.2812

2x² = 4r², x² = 2r².2820

Therefore, x is equal to r√2.2829

Again, I take the positive because I’m talking about a distance here.2836

X = r√2.2842

We know what y is, we said that y is equal to √4r² - x² which is equal to 4r² - r√2²,2844

all under the radical, which is equal to 4r² - 2r², all under the radical.2858

That equals √2r² which is equal to r√2.2869

Y is also equal to r√2.2878

Again, I will just say y is equal to x.2891

In other words, the rectangle of largest area that you can describe in a circle is a square,2899

where the sides of the square are equal to the radius of the circle × √2.2908

That is what we have found.2914

Let us go ahead and show you what it looks like.2919

I have the function, the area function that I try to maximize.2922

√4r² - x².2926

I picked a particular value of r radius of the circle happens to equal 2.2931

This is the function.2935

Again, x is the physical distance, really, our domain is here and here.2938

I set the function equal to 0 to find the end points.2946

When I check the endpoints, when I put the endpoints into the area function, I'm going to get an area of 0.2949

0 does not work, 0 does not work.2955

However, there is a point someplace here.2957

What we found is r√2.2960

When I put r√2 into the function for area, I end up getting the largest area.2964

This graph confirms it.2974

The maximum of this graph, the maximum of the area function happens at √2.2975

It happens where the derivative of this function = 0.2981

I hope that helped.2987

Do not worry about it, in the next lesson we are going to be continuing to do more optimization problems,2988

more complicated optimization problems.2993

Thank you so much for joining us here at www.educator.com.2995

We will see you next time, bye.2998