For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

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## Table of Contents

## Transcription

### Optimization Problems I

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Example I: Find the Dimensions of the Box that Gives the Greatest Volume
- Fundamentals of Optimization Problems
- Example II: Demonstrate that of All Rectangles with a Given Perimeter, the One with the Largest Area is a Square
- Example III: Find the Points on the Ellipse 9x² + y² = 9 Farthest Away from the Point (1,0)
- Example IV: Find the Dimensions of the Rectangle of Largest Area that can be Inscribed in a Circle of Given Radius R

- Intro 0:00
- Example I: Find the Dimensions of the Box that Gives the Greatest Volume 1:23
- Fundamentals of Optimization Problems 18:08
- Fundamental #1
- Fundamental #2
- Fundamental #3
- Fundamental #4
- Fundamental #5
- Fundamental #6
- Example II: Demonstrate that of All Rectangles with a Given Perimeter, the One with the Largest Area is a Square 24:36
- Example III: Find the Points on the Ellipse 9x² + y² = 9 Farthest Away from the Point (1,0) 35:13
- Example IV: Find the Dimensions of the Rectangle of Largest Area that can be Inscribed in a Circle of Given Radius R 43:10

### AP Calculus AB Online Prep Course

### Transcription: 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 problem*0272

*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 out*0323

*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 approach*0680

*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 places*0804

*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 need*1197

*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 variables*1263

*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

0 answers

Post by Maya Balaji on November 11 at 01:28:34 PM

Hello Professor. For question 1- I'm not sure why you would check the endpoints of the domain (variable at 0, volume at 0) to see if they are plausible absolute maximums, because technically this domain is not a closed interval. The volume can never be 0, and the length can never be 0- so these would not be included in the domain- so it would not be a closed interval, correct?- and you must only check endpoints if it is a part of a closed interval (please correct me if this isn't true!). Thank you.