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

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

### Optimization Problems II

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 0:00
- Example I: Optimization Problem 0:13
- Example II: Optimization Problem 17:34
- Example III: Optimization Problem 35:06
- Example IV: Revenue, Cost, and Profit 43:22

### AP Calculus AB Online Prep Course

### Transcription: Optimization Problems II

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

*Today, we are going to continue our discussion of optimization problems, max/min problems, by doing some more of them.*0004

*Let us jump right on in.*0010

*Our first problem says, a piece of wire is 15 m long, it is cut into 2 pieces.*0015

*One piece is bent into a square and the other into a circle.*0020

*Where should the wire be cut so that the total enclosed area between the square and the circle is maximized?*0026

*Let us draw this out to see what we are looking at.*0034

*We have a piece of wire right here.*0038

*This is pretty much what we are looking for, some distance x.*0042

*I’m going to call this x and that is automatically going to make this 15 – x, because the total length was 15.*0048

*We are going to bend it into a square.*0056

*We are going to bend it into a circle.*0058

*Let us go ahead and find out what the areas are.*0060

*For the square, the perimeter of the square is just going to be x all the way around, if we take.*0064

*We are going to choose one for the square.*0075

*It does not really matter which one, I’m just going to go ahead and choose this one, this one piece for the square.*0076

*If that is the case, if the square, if the total perimeter is x that means that one side is going to be x/4.*0082

*Therefore, the area of the square is actually going to be x/ 4².*0091

*We get x²/ 16, that takes care of the area of the square, as for as a formula is concerned.*0097

*Let us go ahead and do the circle.*0105

*Our circle, the perimeter of our circle is going to be the rest of this, the 15 – x.*0108

*The perimeter = 15 - x which I also know is equal to 2 π r.*0115

*The area of the circle is equal to π r².*0123

*I take this one over here, if I have 15 - x is equal to 2 π r, I'm trying to find r in terms of x.*0130

*I get r is equal to 15 - x/ 2 π.*0145

*Take this r, the 15 - x/ 2 π and I put in there.*0153

*What I get is area = π × 15 – x/ 2 π².*0159

*I will go ahead and simplify that.*0171

*We have got area is equal to π/ 4 π² × 225 - 30x + x².*0174

*I hope you are checking my arithmetic.*0191

*Again, I’m notorious for arithmetic mistakes.*0193

*We get finally area = 1/ 4 π ×, I’m going to go ahead and just switch this around.*0197

*X² - 30x + 225, that takes care of the area of the circle.*0205

*Here we have the area of the square.*0213

*Total area, I just add them up.*0216

*Again, I did this because I want some function of one single variable x, *0224

*which is why I used this relation to turn the area into a function of x.*0229

*I have got x²/ 16 + x²/ 4 π - 30x/ 4 π + 225/ 4 π.*0235

*This is this + this, this gives me our total area as a function of x.*0253

*Let me put a couple of things together here.*0261

*I’m going to go ahead and call the area, area sub T.*0267

*I’m going to go ahead and just put some numbers together.*0272

*We have 1/16 + 1/ 4 π x² - 30/ 4 π x + 225/ 4 π.*0274

*I just simplified the equation that I just wrote.*0293

*This is our equation, now we want to take the derivative, set it equal to 0 to find out where the extreme points are.*0295

*at’, when we take the derivative of that, we are going to end up with,*0307

*I’m going to get 1/8 + 1/ 2 π × x - 30/ 4 π, that I set equal to 0.*0319

*Let us just go ahead and solve this.*0336

*That is fine, this simplifies this.*0350

*I get π + 4/ 8 π × x = 30/ 4 π.*0352

*I get 4 π + 16 x is equal to 240.*0363

*That gives me x is equal to 240 divided by 4 π + 16, which comes out to around 8.40.*0375

*When x = 8.40 that gives me an extreme point, a maximum or a minimum.*0393

*We have a little bit of problem.*0401

*This is telling me that my wire which is 15 units long, if I cut it at x = 8.40, I’m going to maximize the area.*0404

*We have a bit of a problem.*0423

*We have a problem, the x = 8.40, it does not maximize the area.*0427

*It actually minimizes the area between the circle of the square.*0439

*It does not maximize the area, it minimizes it.*0443

*Let us see what is going on here.*0458

*It minimizes it.*0461

*When we take the derivative and set it equal to 0, we might have a max, we might have a min.*0466

*We do not know which one, we have to check and see which one by putting all these values into the original function.*0470

*All of the x values that work which are all of the critical points and any endpoints that you have,*0477

*they have to go into the original function to see which one gives you the highest value.*0483

*That is what we have to check.*0488

*In this particular case, let us see what is going on here.*0490

*Let us look at our function.*0495

*Our function is, our total area, we said was equal to 1/16 + 1/ 4 π × x² - 30/ 4 π x + 225/ 4 π.*0506

*Our function is quadratic, but more than that, its leading coefficient is positive.*0528

*What that means, the graph looks like this, not like this.*0544

*Our x = 8.40 actually minimizes the function.*0550

*This total area function is quadratic, it hits a minimum at a certain point between 0 and 15.*0555

*This 8.40 is actually a minimum, it minimizes it.*0562

*But we want to maximized the area.*0567

*The question is how then do we find the max?*0569

*How then do we find the maximum?*0573

*The answer is check the endpoints.*0584

*In all of these max/min problem, let me write this down first because we have a hard time doing two things simultaneously.*0589

*In all of these max/min problems, anytime your domain actually is closed.*0596

*in other words, anytime you have endpoints that are included in the domain,*0601

*you not only have checked the internal points, the critical points. *0605

*When you set the derivative equal to 0 which could be maxes or mins,*0610

*those x values, you also have to check the left endpoint and the right endpoint.*0612

*Putting all of these x values into the original function to see which one gives you the highest value or the lowest value.*0618

*Depending on what you are trying to maximize or minimize, respectively.*0624

*In this particular case, quadratic function, leading coefficient positive, our 8.4 does not maximize it, it minimizes it.*0628

*That means that the maximum, the absolute maximum has to happen at one of the endpoints.*0636

*Let us go ahead and check those now.*0640

*Our domain is 0 to 15, in other words, we can use all of it for the circle, all of the wire, or all of the wire for the square.*0645

*That is all that means.*0659

*If x = 0, then the area of the square is equal to 0/ 4² that = 0.*0661

*The area of the circle = π × 15 – 0.*0679

*Go back to the original equation, / 2 π² is equal to 225 π/ 4 π².*0689

*π cancels, you are left with 225/ 4 π which = 17.90.*0703

*This is one possibility, an area of 17.90.*0710

*Now all of the area belongs to the circle but there is no specification that says it has to be split between the square and the circle.*0715

*It just throws out the problem.*0722

*In this case, if all the wire is used for the circle, the total area is going to be 17.90.*0724

*If x is 15, in other words, if we use all of the wire for the square then the area of the square is going to equal 15/ 4².*0732

*If I done my arithmetic right, this is going to be 14.06.*0755

*In this case, 17.90 is greater than 14.06.*0762

*In order to maximize the total area within the constraints of 0 to 15, the total area, use the entire wire for the circle.*0773

*That is it, that is all that is happening here.*0799

*Basically, what we have done is we found this x = 8.40.*0803

*But given the fact that our equation was actually a parabola that opens upward, at 8.40 minimizes it.*0811

*If I take any x value between 0 and 15, my total area is actually going to end up being minimized not maximized.*0819

*I have to check the endpoints.*0828

*The endpoints tells me that, if I choose the circle to use the entire wire, *0830

*that is going to give me the total maximum area, 17.90/ 14.06.*0837

*If you were to put in the values, 8.40, and calculate the area each,*0843

*which I probably should have done but you can do it yourself, use x = 8.40.*0850

*Calculate the area of the square, the area of the circle, and add them up.*0854

*You will find that it is actually minimized.*0858

*That is the 17.90, when x = 0, that is the real answer.*0860

*Let us go ahead and show you what this looks like here, pictorially.*0865

*We said that our area total, our equation area total was 1/16 + 1/ 4 π x² - 30/ 4 π x + 225/ 4 π.*0870

*This was our equation for total area, this is the graph for that.*0896

*It is minimized at this 8.4, this was our 8.40.*0901

*I actually could did do it here.*0911

*The area here, the total area is going to be 7.88.*0914

*At 15, at this point right here, when x = 15, the area is going to end up being 14.06.*0920

*That is what the y value is, the y value is the total area because this is our area function, the total area.*0929

*Here, 0, 17.90, this is where the absolute maximum occurs on this domain, from 0 to 15.*0935

*Total area is 17.9 at 0, total area is 7.88 at 8.40, and total area is 14.06, when x = 15.*0949

*If you did not catch the quadratic nature of the function and the positive leading coefficient, coefficient is actually not a problem.*0960

*Just check the endpoints and the critical point x = 8.40.*1004

*You put all of those points into the original function.*1024

*The one that gives you the lowest number, in this case because we are trying to minimize, that is the one you pick.*1027

*If we were trying to maximize it, we are trying to maximize the area, *1031

*that is what you pick, depending on what the problem is asking for.*1037

*The lesson actually of this is closed intervals, you always have to check the endpoints.*1042

*That is all that is going on.*1046

*Let us go onto the next problem here and see what that is.*1051

*It seems a little long, do not worry about it, it is actually pretty straight forward.*1055

*We will draw a nice picture and see what is going on.*1059

*A gas company is on the north shore of a river that is a 1.5 km wide.*1062

*It has storage tanks on the south bank of the river, 7 miles east of a point directly across the river from the company.*1067

*They want to run a pipeline from the company to the storage tanks by first heading east from the company over land,*1076

*to a point p on the north shore, then, going under the water to the storage tanks on the other bank.*1081

*It costs $350,000/km, to run a pipe over land.*1088

*$600,000/km, to do so under water.*1092

*Where should the point p be, in order to minimize the cost of the pipe line.*1096

*Let us draw what is going on here.*1100

*I have a north shore of the riverbank, I have a south shore of the riverbank.*1103

*They tell me that the company, the north shore of the riverbank,*1114

*here is my company, at c.*1118

*River that is 1.5 km wide, this is 1.5 km wide.*1122

*Storage tanks on the south bank of the river 7 miles east of the point directly across the river.*1129

*Across the river and 7 miles east.*1135

*The storage tanks are right here, we will call this s or t, whichever you want.*1139

*They tell me that this distance right here is 7.*1146

*They want to run a pipe line from the company to the storage tanks, *1152

*by first heading east from the company over land to a point p.*1155

*They want to go this way.*1159

*They want to head out this way.*1163

*This is our point p, to the point p in the north shore, then going underwater to the storage tanks to the other bank.*1172

*Then, they want to go under the water there.*1179

*This is the cost, minimize that cost.*1185

*The cost function is as follows.*1190

*I’m going to call this distance x and I’m going to call this distance y.*1198

*$350,000/km, the cost is going to be 350,000 x for x km + 600,000 y under the water.*1203

*That is it, we want to maximize this.*1219

*Notice that it is a function of two variables.*1228

*More than likely, we were going to try to find some relation between the two variables, substitute into either one of them.*1229

*Find an equation in one variable for the cost, take the derivative, so on, and so forth.*1236

*Let us go ahead and mark our domain to double check, to see whether we are dealing with open or closed endpoints.*1243

*X can be 0, in other words, I can run it just straight from c all the way under the water, to the storage tanks.*1254

*It can certainly be 0, or I can go all the way to 7 over land, and then, cut straight across under the river.*1262

*Our domain is 0 and 7, closed.*1270

*We have to check those endpoints.*1273

*We will also check the endpoints.*1278

*Let us talk about what it is that we are going to do with this figure and how we are going to make sense of this.*1289

*I’m going to go ahead and draw a little dotted line straight across.*1294

*I’m going to call that c.*1301

*This angle here, I’m going to call this angle θ.*1305

*This is my p, this is my x, this is my y.*1311

*Just in case, I’m going to go ahead and I got a right triangle right there.*1316

*Perfect, now I’m going to go to the next page and redraw this, just this figure, without anything else.*1323

*Let me go ahead and draw the figure over here.*1334

*I have got my company, I got my point p.*1338

*I got the storage tanks over here.*1348

*I think I will call this c, I think I will call this s, does not really matter.*1350

*This was c, this was x, this was y, this was our angle θ.*1358

*I’m going to go ahead draw this regular triangle here, something like that.*1364

*This was 1.5, that was the width of the river, and this was 7.*1368

*Let me see if I can come up with some relationship between x and y.*1374

*I’m going to go ahead and take c.*1379

*C is easy to find, that is just Pythagorean theorem, this is a right triangle.*1382

*C is nothing more than 1.5² + 7², under the radical.*1386

*It is going to be 51.25 which is going to be 7.16 km, that is c.*1395

*Now θ, this θ over here, that is actually going to be fixed.*1406

*X is going to change, that is what I’m trying to find.*1413

*How far do I have to go this way?*1417

*But from here to here is a fixed line.*1421

*If I move along this, it is a fixed line, θ stays fixed.*1424

*This θ is the same as that, that angle is the same.*1428

*If I want to find θ, θ is just the inv tan(1.5) divided by 7, which ends up being 12.1°.*1433

*We can go ahead and use the law of cosines.*1447

*Using the law of cosines, we get the following.*1459

*We get y² is equal to x² + 7.16² - 2 × 7.16 × x × cos(12.1).*1466

*That is the law of cosines, I have established a relationship between y, x, c, and θ.*1489

*It is right there.*1496

*Ultimately, what I have got is a relationship between y and x, *1499

*which is what I wanted so that I can put it back into my cost equation.*1503

*I end up with y² = x² + 51.25 - 14x.*1509

*Therefore, y is equal to this x² + 51.25 - 14x, all under the radical.*1520

*We now have a relation between x and y, cost function.*1539

*Therefore, the cost function which is now going to be a function of x,*1553

*is going to equal 350,000x + we said 600,000 × y.*1558

*Y is this, x² +, let me go ahead and write it, -14x + 51.25.*1567

*That is it, that is my cost function.*1582

*Let me go ahead, let us see, should I do it here, should I do it there?*1585

*Let us go ahead and stick with what I have got.*1594

*Now we take c’(x), let me go to the next page.*1596

*I have got, let me rewrite c(x).*1606

*C(x) = 350,000x + 600,000 × x² - 14x + 51.25.*1611

*Our c’(x) is going to equal 350,000 + 600,000 × ½ x² - 14x + 51.25 ^- ½ *1634

*× the derivative of what is inside, which is going to be 2x – 14.*1655

*I get c’(x) is equal to 350,000 + 600,000 divided by 2, I will just leave that alone,*1664

*put this on top, bring this down to the bottom.*1674

*I have got 300,000 × 2x - 14/ x² - 14x + 51.25.*1677

*Of course, the rest of this is just algebra, it is not a problem.*1697

*I will go ahead and go through it all.*1699

*Common denominator, let me get, 350,000 × x² - 14x + 51.25 + 600,000x.*1702

*I have distributed this, -4,200,000/ I knew this is c’(x) = this/ x² - 14x + 51.25,*1721

*all of this is going to equal 0, which means the numerator = 0.*1744

*Therefore, we are going to have 350,000 × √x² - 14x + 51.25 is equal to 4,200,000 - 600,000x.*1749

*Go ahead and divide, I end up with x² - 14x + 51.25, all under the radical.*1773

*It is equal to 12 - 1.7x, square both sides.*1783

*X² - 14x + 51.25 = 144 - 40.8x + 2.89x².*1789

*Rearrange, I end up with 1.89 x² - 26.8x + 92.75.*1806

*This is my c’(x), then, when I solve this which is going to be 0.*1823

*When we solve this quadratic, we get x = 5.929 km and x = 8.068.*1840

*The x = 5.929, that is actually the answer.*1870

*Notice the 8.068 is outside of the domain.*1875

*We will talk a little bit more about that in just a minute.*1879

*We are going to use the 5.929.*1882

*We now put x = 0, x = 5.929, and x = 7.*1887

*0 and 7 are the endpoints, this was the critical point.*1902

*You put these into the cost function.*1906

*We said that the cost function, let me rewrite it, in case we need it.*1917

*It was 350,000x + 600,000 × √x² - 14x + 51.25.*1920

*When we take c(0), we end up with 4.3 × 10 ^$6, $4.3 million.*1943

*When we take c(5.929), we end up with 3.1 × 10 ^$6.*1955

*When we take c(7), we end up with 3.4 × 10 ^$6.*1969

*Clearly, this one minimizes the cost, minimum cost.*1978

*Therefore, we want x to be 5.929 km, that is what is going on.*1987

*Let us go ahead and take a look at what this looks like.*1998

*Our red, this was our original function, this was our cost function.*2005

*It is going to end up minimizing someplace.*2009

*But clearly, our domain is 0 to 7.*2013

*We are only concerned with from here to about here.*2015

*But this is the overall function, if we need it.*2017

*The blue, this was the c’, just to show you where it is.*2020

*That is it, the 5.929, that is this number right here.*2028

*As you can see, that is where it actually minimizes that.*2033

*Also, you should notice that at least here, it only passes through 0 once.*2038

*This c’ that we got, we ended up with a quadratic.*2047

*Whatever it was that we got, it only passes through 0 once.*2050

*This other root, this 8.068 that we got, that was actually a false root that showed up because we ended up squaring a radical.*2053

*When you do that, you tend to sometimes introduce roots that do not exist.*2063

*You can think about it that way.*2067

*You can think of it as, it is outside the domain so we can ignore that.*2069

*You can think of it as a false root.*2074

*If you go ahead and graph it, you can see that it does not touch 0 again at 8.068.*2076

*Clearly, only the 5.929 is the answer, whatever you need.*2081

*You just have to be vigilant.*2086

*It is not just about putting the numbers in, take the derivative, whatever they are, check the answer.*2089

*You want to still stand back and make sure things make some sort of physical sense.*2093

*You want to take a look at the function, think about the function, use every resource at your disposal.*2098

*Let us go ahead and go to the next problem here.*2106

*A long pipe is being carried down the hallway that is 10 ft wide.*2108

*At the end of the hallway, a right angle turned to the right must be made into the hallway that is 7 ft wide.*2112

*What is the longest pipe that can make that turn.*2117

*Let us draw this out, see what we are looking at here.*2121

*We have got a hallway, it is going to be some hallway this way and it is going to be this way.*2124

*The initial hallway that we are going down is 10 ft wide, that is this one.*2136

*And then, we are going to make a right turn that must near the hallway that is 7 ft wide.*2143

*This is 7, we are carrying this long pipe.*2150

*The only way this pipe is going to make it is like that.*2153

*When we turn it, it has to just barely touch the wall, in order to actually make this turn.*2163

*That is our pipe.*2173

*What is the longest pipe that can make that turn?*2177

*Let us see what we have got.*2182

*I think we are going to break this up.*2184

*You stare at this a little bit, we try different things.*2187

*We try to see what is going to happen.*2191

*Based on this, I decided to call this l1 length 1 and I will call this length 2.*2194

*I ended up calling this θ, this is also θ.*2203

*The total length is equal to l1 + l2.*2216

*Let me do it over here.*2230

*The cos(θ) is equal to 10/l1.*2235

*Therefore, l1 is equal to 10/ cos θ.*2244

*Over here, sin(θ) is equal to 7/l2.*2252

*Therefore, l2 is equal to 7/ sin(θ).*2261

*Therefore, I put these into here.*2269

*Therefore, l is equal to 10/ cos(θ) + 7/ sin(θ).*2272

*Now at least I have a function l, a function of only one variable, θ.*2287

*Our domain in this case, θ is going to be 0 - 90°.*2294

*We do have 0 and we have 90°.*2299

*I’m going to say 0 to π/2.*2306

*Let us stick with radian measure.*2310

*That is that, great.*2313

*L is just function of θ.*2315

*Now I’m going to take dl dθ or l’.*2317

*What I get is the following.*2322

*Dl dθ is equal to 10 sin θ/ cos² θ.*2326

*This is quotient rule, or if you want to bring this and write this is 10 cos⁻¹, however you want to do it.*2338

*This × the derivative of that - that × the derivative of this/ this².*2347

*That is 10 sin θ cos² θ.*2354

*And then this one, this × the derivative of that 0 - that × the derivative of this/ this².*2356

*You are going to get -7 cos θ/ sin² θ.*2365

*We are going to set this derivative equal to 0.*2373

*Now we have this, now we just need to solve this equation.*2379

*I have got a common denominator there.*2383

*Actually, let me go ahead and rewrite it, it is not a problem.*2389

*I have got dl dθ is equal to 10 sin θ/ cos² θ - 7 cos θ/ sin² θ, we want that equal to 0.*2395

*How do we solve this?*2414

*Let us do a little common denominator here.*2416

*I will do a common denominator, I'm going to get to 10 sin³ θ - 7 cos³ θ/ cos² θ sin² θ that = 0.*2420

*It is the numerator that equal 0, I have got 10 sin³ θ – 7 cos³ θ, that is going to be equal to 0.*2446

*I have got 10 sin³ θ = 7 cos³ θ, switch things around.*2464

*Sin³ θ/ cos³ θ = 7/10.*2475

*This is 10³ θ = 7/10.*2483

*When I take this, I get 10 θ = 0.8879.*2490

*When I take the inverse, I get θ is equal to 41. 6°, that is one of the critical points.*2502

*Now I have got l1 which is equal to 10/ cos θ which is equal to 10/ cos(41.6) is equal to 13.37 ft.*2513

*That gives me the length of l1.*2535

*L2 was equal to 7/ sin θ which is 7/ sin(41.6), which ends up being 15.06 ft.*2540

*Our total l is going to be 28.13 ft.*2558

*That is it, it is that simple.*2567

*When you put in the 0 and the 90, I should have done it, I apologize.*2569

*You are going to get numbers that are not going to give you an ultimate length.*2582

*The longest that it can be is going to be the 28.13 ft.*2587

*For this particular one, I will let you take care of the 0 and the π/2 in for θ.*2593

*Let us go to our final example here, revenue, cost, and profit.*2601

*Show that the marginal revenue = marginal cost, when profit is maximized.*2606

*B, if the cost function is this function and the price function is this function, what level of production will maximize profit?*2611

*Level of production means how many units sold, how many units should I make, when all of them are sold, in other words, x?*2622

*Let us talk about this a little bit.*2631

*Let us talk about some variables here.*2633

*We will talk generally about what these things are.*2640

*X is going to be the number of units sold, or the number of units produced, the number of units.*2642

*Our price function, the price function, we will generally use a p(x), that is this one.*2656

*Our revenue function, revenue function is just your revenue, how much money you are actually bringing in?*2671

*The amount of money that you are bringing in is going to be the price of one unit × the number of units that you sell.*2680

*It is going to be x × p(x), the number of units × the price, that is the revenue function.*2686

*Cost function, your cost function is how much it costs you to make each unit?*2696

*In other words, if I'm making toasters and it cost me $10.00 to make one toaster, let us say I end up selling it for $15.00.*2704

*That is that is difference, there is a price and there is a cost that you actually incur for making this thing.*2712

*The cost function, we will usually just call it c(x), whatever that happens to be.*2718

*In this case, our cost function is this.*2723

*If I make 500 units, I put 500 in for x, that gives me the cost that I pay,*2725

*that I have to incur as the manufacturer, in order to make 500 units of this thing.*2734

*Hopefully, I make that cost back + any more, that is going to be my profit.*2739

*There you go, your profit function P(x), it is equal to your revenue function, *2744

*how much money you bring in total - your cost function, how much you actually spend.*2754

*I hope that make sense.*2762

*You bring in $1,000,000 but if you end up spending $300,000 to make those things that you just sold, you lose that money.*2763

*Your profit ends up being $700,000, it is that simple.*2773

*Revenue – cost, profit = revenue – cost.*2776

*When we speak of marginal in the world of money and finance, as far as mathematics is concerned, *2782

*marginal just means take the derivative.*2791

*Marginal cost c’, marginal profit p’, what is p’?*2802

*It is marginal, it is r’ – c’.*2808

*It is marginal revenue - marginal cost.*2811

*Marginal just means it is a derivative, it is a rate of change.*2815

*Let us do part a, very simple actually.*2821

*Part A says, show that the marginal revenue = marginal cost, when profit is maximized.*2825

*The profit function, we already know what that is.*2831

*The profit function is equal to the revenue function, the money that I bring in - the cost function, the money that I spend.*2833

*Profit is maximized when p’ = 0.*2842

*Profit is maximized when p’(x) = 0.*2855

*P’ is nothing more than r’ – c’.*2867

*P’ is just r’ – c’, the derivative is linear.*2873

*R’ – c’ is equal to 0, I just solved this equation.*2877

*R’ = c’, marginal revenue = marginal cost.*2884

*That takes care of part A, very simple.*2891

*Let us go ahead and deal with part B.*2896

*A derivative is a rate of change, r’(x) marginal revenue is the rate at which the revenue changes per unit sold.*2903

*How fast is my revenue increasing or decreasing, if I sell one more unit?*2946

*C’(x) marginal cost is the rate at which cost changes per unit produced.*2954

*In other words, how fast is the cost that I'm incurring changing if I make one more unit?*2982

*I asked my people in my company, if I make one more unit, how much more is it going to cost me?*2992

*That is c’, it is the derivative of the cost function.*2999

*Let us do B, our profit is equal to our revenue - our cost.*3004

*Revenue is how much you bring in.*3016

*It is how many you sell × the price that you are selling it at.*3018

*That = x × the price function pp, P profit, p price, -c(x) = x × 1500.*3022

*1500 - 5x, I think was the price function, - the cost function, - the whole cost function.*3042

*You have to put them in brackets, [14,000 + 400x - 1.3 x² + 0.0035 x³].*3050

*Therefore, our profit function is equal to 1500x - 5 x² - 14,000 - 400x + 1.3 x² - 0.0035 x³.*3071

*Simplify, we end up with -0.0035 x³ - 3.7 x² + 1100x - 14,000.*3101

*I take the derivative, try to maximize profit.*3121

*What value of x, what production level will allow me to maximize my profit?*3125

*My profit function is equal to my revenue – cost, I have that function.*3129

*Simplify that function, now I take the derivative of that function, -0.0105 x² - 7.4 x + 1100.*3133

*I set the derivative equal to 0.*3154

*When I solve this quadratic equation, I end up with x = 126.1, x = -830.9.*3173

*Clearly, I cannot make a negative amount of things.*3188

*Producing 126 units maximizes profit.*3197

*Let us take a look at what this looks like.*3212

*This right here, this is our p(x).*3218

*Remember, our p(x) was a cubic equation.*3223

*This is not a quadratic equation, it is actually cubic.*3227

*I have just taken this piece of it so that you see the piece of it matters.*3228

*Obviously, the minimum I can make is 0 things.*3232

*I'm not going to make 0 things, I'm going to make more than that.*3238

*This, as I make more and more, this was my profit function.*3242

*At some point, I'm going to hit a maximum.*3248

*This right here, that is p’.*3253

*It is p’(x), that is the one that we set to 0.*3256

*Notice it crosses 0 at 126.*3259

*When I make 126 items, my profit is maximized and the maximum is something like that, whatever that number is.*3266

*600,000, somewhere near 600,000.*3277

*That is what is going on here.*3279

*I have zoomed in on this, this is not a quadratic function.*3282

*It is the high point of a cubic function.*3285

*This function actually goes down and comes back up the other end.*3288

*I’m not concerned about this part, the negative part.*3293

*I’m concerned because I have to make at least one unit.*3295

*Clearly, maximizes at x = 126.*3301

*Thank you so much for joining us here at www.educator.com, we will see you next time, bye.*3308

1 answer

Last reply by: Professor Hovasapian

Wed Jan 18, 2017 7:55 PM

Post by Rohit Kumar on January 8 at 12:10:56 AM

What would the domain be for the last example and how would you find it?

1 answer

Last reply by: Professor Hovasapian

Tue Apr 12, 2016 3:33 PM

Post by Acme Wang on April 12, 2016

In example IV, does the price function denote the price per unit?

0 answers

Post by Gautham Padmakumar on December 5, 2015

Also, you made an error in using the quadratic formula. Its supposed to be

x = 8.01 and x = 6.09

1 answer

Last reply by: Professor Hovasapian

Thu Dec 17, 2015 12:26 AM

Post by Gautham Padmakumar on December 5, 2015

Isn't it a problem to work with different units like that in Example 2? We have both 1.5 km and 7 miles?

Thank you