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Lecture Comments (9)

1 answer

Last reply by: Sarmad Khokhar
Sun Apr 30, 2017 7:35 AM

Post by Sarmad Khokhar on April 30, 2017

In Example we could have also taken the second derivative to check our x value.

1 answer

Last reply by: Professor Hovasapian
Wed Jan 18, 2017 7:55 PM

Post by Rohit Kumar on January 8, 2017

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

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

Transcription: Optimization Problems II

Hello, welcome back to, 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 ^- ½1634

× 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, we will see you next time, bye.3308