WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to continue our discussion of optimization problems, max/min problems, by doing some more of them.
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Let us jump right on in.
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Our first problem says, a piece of wire is 15 m long, it is cut into 2 pieces.
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One piece is bent into a square and the other into a circle.
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Where should the wire be cut so that the total enclosed area between the square and the circle is maximized?
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Let us draw this out to see what we are looking at.
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We have a piece of wire right here.
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This is pretty much what we are looking for, some distance x.
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I’m going to call this x and that is automatically going to make this 15 – x, because the total length was 15.
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We are going to bend it into a square.
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We are going to bend it into a circle.
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Let us go ahead and find out what the areas are.
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For the square, the perimeter of the square is just going to be x all the way around, if we take.
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We are going to choose one for the square.
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It does not really matter which one, I’m just going to go ahead and choose this one, this one piece for the square.
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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.
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Therefore, the area of the square is actually going to be x/ 4².
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We get x²/ 16, that takes care of the area of the square, as for as a formula is concerned.
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Let us go ahead and do the circle.
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Our circle, the perimeter of our circle is going to be the rest of this, the 15 – x.
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The perimeter = 15 - x which I also know is equal to 2 π r.
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The area of the circle is equal to π r².
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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.
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I get r is equal to 15 - x/ 2 π.
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Take this r, the 15 - x/ 2 π and I put in there.
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What I get is area = π × 15 – x/ 2 π².
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I will go ahead and simplify that.
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We have got area is equal to π/ 4 π² × 225 - 30x + x².
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I hope you are checking my arithmetic.
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Again, I’m notorious for arithmetic mistakes.
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We get finally area = 1/ 4 π ×, I’m going to go ahead and just switch this around.
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X² - 30x + 225, that takes care of the area of the circle.
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Here we have the area of the square.
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Total area, I just add them up.
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Again, I did this because I want some function of one single variable x,
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which is why I used this relation to turn the area into a function of x.
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I have got x²/ 16 + x²/ 4 π - 30x/ 4 π + 225/ 4 π.
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This is this + this, this gives me our total area as a function of x.
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Let me put a couple of things together here.
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I’m going to go ahead and call the area, area sub T.
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I’m going to go ahead and just put some numbers together.
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We have 1/16 + 1/ 4 π x² - 30/ 4 π x + 225/ 4 π.
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I just simplified the equation that I just wrote.
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This is our equation, now we want to take the derivative, set it equal to 0 to find out where the extreme points are.
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at’, when we take the derivative of that, we are going to end up with,
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I’m going to get 1/8 + 1/ 2 π × x - 30/ 4 π, that I set equal to 0.
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Let us just go ahead and solve this.
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That is fine, this simplifies this.
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I get π + 4/ 8 π × x = 30/ 4 π.
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I get 4 π + 16 x is equal to 240.
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That gives me x is equal to 240 divided by 4 π + 16, which comes out to around 8.40.
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When x = 8.40 that gives me an extreme point, a maximum or a minimum.
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We have a little bit of problem.
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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.
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We have a bit of a problem.
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We have a problem, the x = 8.40, it does not maximize the area.
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It actually minimizes the area between the circle of the square.
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It does not maximize the area, it minimizes it.
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Let us see what is going on here.
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It minimizes it.
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When we take the derivative and set it equal to 0, we might have a max, we might have a min.
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We do not know which one, we have to check and see which one by putting all these values into the original function.
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All of the x values that work which are all of the critical points and any endpoints that you have,
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they have to go into the original function to see which one gives you the highest value.
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That is what we have to check.
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In this particular case, let us see what is going on here.
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Let us look at our function.
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Our function is, our total area, we said was equal to 1/16 + 1/ 4 π × x² - 30/ 4 π x + 225/ 4 π.
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Our function is quadratic, but more than that, its leading coefficient is positive.
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What that means, the graph looks like this, not like this.
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Our x = 8.40 actually minimizes the function.
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This total area function is quadratic, it hits a minimum at a certain point between 0 and 15.
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This 8.40 is actually a minimum, it minimizes it.
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But we want to maximized the area.
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The question is how then do we find the max?
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How then do we find the maximum?
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The answer is check the endpoints.
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In all of these max/min problem, let me write this down first because we have a hard time doing two things simultaneously.
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In all of these max/min problems, anytime your domain actually is closed.
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in other words, anytime you have endpoints that are included in the domain,
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you not only have checked the internal points, the critical points.
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When you set the derivative equal to 0 which could be maxes or mins,
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those x values, you also have to check the left endpoint and the right endpoint.
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Putting all of these x values into the original function to see which one gives you the highest value or the lowest value.
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Depending on what you are trying to maximize or minimize, respectively.
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In this particular case, quadratic function, leading coefficient positive, our 8.4 does not maximize it, it minimizes it.
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That means that the maximum, the absolute maximum has to happen at one of the endpoints.
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Let us go ahead and check those now.
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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.
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That is all that means.
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If x = 0, then the area of the square is equal to 0/ 4² that = 0.
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The area of the circle = π × 15 – 0.
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Go back to the original equation, / 2 π² is equal to 225 π/ 4 π².
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π cancels, you are left with 225/ 4 π which = 17.90.
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This is one possibility, an area of 17.90.
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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.
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It just throws out the problem.
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In this case, if all the wire is used for the circle, the total area is going to be 17.90.
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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.
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If I done my arithmetic right, this is going to be 14.06.
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In this case, 17.90 is greater than 14.06.
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In order to maximize the total area within the constraints of 0 to 15, the total area, use the entire wire for the circle.
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That is it, that is all that is happening here.
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Basically, what we have done is we found this x = 8.40.
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But given the fact that our equation was actually a parabola that opens upward, at 8.40 minimizes it.
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If I take any x value between 0 and 15, my total area is actually going to end up being minimized not maximized.
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I have to check the endpoints.
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The endpoints tells me that, if I choose the circle to use the entire wire,
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that is going to give me the total maximum area, 17.90/ 14.06.
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If you were to put in the values, 8.40, and calculate the area each,
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which I probably should have done but you can do it yourself, use x = 8.40.
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Calculate the area of the square, the area of the circle, and add them up.
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You will find that it is actually minimized.
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That is the 17.90, when x = 0, that is the real answer.
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Let us go ahead and show you what this looks like here, pictorially.
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We said that our area total, our equation area total was 1/16 + 1/ 4 π x² - 30/ 4 π x + 225/ 4 π.
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This was our equation for total area, this is the graph for that.
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It is minimized at this 8.4, this was our 8.40.
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I actually could did do it here.
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The area here, the total area is going to be 7.88.
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At 15, at this point right here, when x = 15, the area is going to end up being 14.06.
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That is what the y value is, the y value is the total area because this is our area function, the total area.
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Here, 0, 17.90, this is where the absolute maximum occurs on this domain, from 0 to 15.
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Total area is 17.9 at 0, total area is 7.88 at 8.40, and total area is 14.06, when x = 15.
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If you did not catch the quadratic nature of the function and the positive leading coefficient, coefficient is actually not a problem.
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Just check the endpoints and the critical point x = 8.40.
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You put all of those points into the original function.
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The one that gives you the lowest number, in this case because we are trying to minimize, that is the one you pick.
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If we were trying to maximize it, we are trying to maximize the area,
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that is what you pick, depending on what the problem is asking for.
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The lesson actually of this is closed intervals, you always have to check the endpoints.
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That is all that is going on.
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Let us go onto the next problem here and see what that is.
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It seems a little long, do not worry about it, it is actually pretty straight forward.
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We will draw a nice picture and see what is going on.
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A gas company is on the north shore of a river that is a 1.5 km wide.
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It has storage tanks on the south bank of the river, 7 miles east of a point directly across the river from the company.
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They want to run a pipeline from the company to the storage tanks by first heading east from the company over land,
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to a point p on the north shore, then, going under the water to the storage tanks on the other bank.
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It costs $350,000/km, to run a pipe over land.
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$600,000/km, to do so under water.
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Where should the point p be, in order to minimize the cost of the pipe line.
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Let us draw what is going on here.
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I have a north shore of the riverbank, I have a south shore of the riverbank.
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They tell me that the company, the north shore of the riverbank,
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here is my company, at c.
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River that is 1.5 km wide, this is 1.5 km wide.
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Storage tanks on the south bank of the river 7 miles east of the point directly across the river.
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Across the river and 7 miles east.
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The storage tanks are right here, we will call this s or t, whichever you want.
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They tell me that this distance right here is 7.
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They want to run a pipe line from the company to the storage tanks,
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by first heading east from the company over land to a point p.
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They want to go this way.
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They want to head out this way.
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This is our point p, to the point p in the north shore, then going underwater to the storage tanks to the other bank.
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Then, they want to go under the water there.
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This is the cost, minimize that cost.
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The cost function is as follows.
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I’m going to call this distance x and I’m going to call this distance y.
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$350,000/km, the cost is going to be 350,000 x for x km + 600,000 y under the water.
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That is it, we want to maximize this.
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Notice that it is a function of two variables.
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More than likely, we were going to try to find some relation between the two variables, substitute into either one of them.
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Find an equation in one variable for the cost, take the derivative, so on, and so forth.
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Let us go ahead and mark our domain to double check, to see whether we are dealing with open or closed endpoints.
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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.
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It can certainly be 0, or I can go all the way to 7 over land, and then, cut straight across under the river.
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Our domain is 0 and 7, closed.
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We have to check those endpoints.
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We will also check the endpoints.
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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.
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I’m going to go ahead and draw a little dotted line straight across.
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I’m going to call that c.
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This angle here, I’m going to call this angle θ.
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This is my p, this is my x, this is my y.
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Just in case, I’m going to go ahead and I got a right triangle right there.
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Perfect, now I’m going to go to the next page and redraw this, just this figure, without anything else.
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Let me go ahead and draw the figure over here.
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I have got my company, I got my point p.
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I got the storage tanks over here.
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I think I will call this c, I think I will call this s, does not really matter.
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This was c, this was x, this was y, this was our angle θ.
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I’m going to go ahead draw this regular triangle here, something like that.
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This was 1.5, that was the width of the river, and this was 7.
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Let me see if I can come up with some relationship between x and y.
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I’m going to go ahead and take c.
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C is easy to find, that is just Pythagorean theorem, this is a right triangle.
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C is nothing more than 1.5² + 7², under the radical.
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It is going to be 51.25 which is going to be 7.16 km, that is c.
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Now θ, this θ over here, that is actually going to be fixed.
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X is going to change, that is what I’m trying to find.
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How far do I have to go this way?
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But from here to here is a fixed line.
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If I move along this, it is a fixed line, θ stays fixed.
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This θ is the same as that, that angle is the same.
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If I want to find θ, θ is just the inv tan(1.5) divided by 7, which ends up being 12.1°.
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We can go ahead and use the law of cosines.
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Using the law of cosines, we get the following.
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We get y² is equal to x² + 7.16² - 2 × 7.16 × x × cos(12.1).
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That is the law of cosines, I have established a relationship between y, x, c, and θ.
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It is right there.
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Ultimately, what I have got is a relationship between y and x,
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which is what I wanted so that I can put it back into my cost equation.
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I end up with y² = x² + 51.25 - 14x.
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Therefore, y is equal to this x² + 51.25 - 14x, all under the radical.
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We now have a relation between x and y, cost function.
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Therefore, the cost function which is now going to be a function of x,
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is going to equal 350,000x + we said 600,000 × y.
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Y is this, x² +, let me go ahead and write it, -14x + 51.25.
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That is it, that is my cost function.
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Let me go ahead, let us see, should I do it here, should I do it there?
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Let us go ahead and stick with what I have got.
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Now we take c’(x), let me go to the next page.
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I have got, let me rewrite c(x).
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C(x) = 350,000x + 600,000 × x² - 14x + 51.25.
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Our c’(x) is going to equal 350,000 + 600,000 × ½ x² - 14x + 51.25 ^- ½
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× the derivative of what is inside, which is going to be 2x – 14.
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I get c’(x) is equal to 350,000 + 600,000 divided by 2, I will just leave that alone,
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put this on top, bring this down to the bottom.
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I have got 300,000 × 2x - 14/ x² - 14x + 51.25.
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Of course, the rest of this is just algebra, it is not a problem.
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I will go ahead and go through it all.
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Common denominator, let me get, 350,000 × x² - 14x + 51.25 + 600,000x.
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I have distributed this, -4,200,000/ I knew this is c’(x) = this/ x² - 14x + 51.25,
00:29:04.400 --> 00:29:09.500
all of this is going to equal 0, which means the numerator = 0.
00:29:09.500 --> 00:29:33.500
Therefore, we are going to have 350,000 × √x² - 14x + 51.25 is equal to 4,200,000 - 600,000x.
00:29:33.500 --> 00:29:43.100
Go ahead and divide, I end up with x² - 14x + 51.25, all under the radical.
00:29:43.100 --> 00:29:49.500
It is equal to 12 - 1.7x, square both sides.
00:29:49.500 --> 00:30:05.700
X² - 14x + 51.25 = 144 - 40.8x + 2.89x².
00:30:05.700 --> 00:30:23.100
Rearrange, I end up with 1.89 x² - 26.8x + 92.75.
00:30:23.100 --> 00:30:40.200
This is my c’(x), then, when I solve this which is going to be 0.
00:30:40.200 --> 00:31:09.900
When we solve this quadratic, we get x = 5.929 km and x = 8.068.
00:31:09.900 --> 00:31:15.300
The x = 5.929, that is actually the answer.
00:31:15.300 --> 00:31:19.300
Notice the 8.068 is outside of the domain.
00:31:19.300 --> 00:31:22.000
We will talk a little bit more about that in just a minute.
00:31:22.000 --> 00:31:27.400
We are going to use the 5.929.
00:31:27.400 --> 00:31:42.400
We now put x = 0, x = 5.929, and x = 7.
00:31:42.400 --> 00:31:45.800
0 and 7 are the endpoints, this was the critical point.
00:31:45.800 --> 00:31:57.200
You put these into the cost function.
00:31:57.200 --> 00:32:00.500
We said that the cost function, let me rewrite it, in case we need it.
00:32:00.500 --> 00:32:23.300
It was 350,000x + 600,000 × √x² - 14x + 51.25.
00:32:23.300 --> 00:32:35.400
When we take c(0), we end up with 4.3 × 10 ^$6, $4.3 million.
00:32:35.400 --> 00:32:48.900
When we take c(5.929), we end up with 3.1 × 10 ^$6.
00:32:48.900 --> 00:32:58.200
When we take c(7), we end up with 3.4 × 10 ^$6.
00:32:58.200 --> 00:33:06.900
Clearly, this one minimizes the cost, minimum cost.
00:33:06.900 --> 00:33:18.000
Therefore, we want x to be 5.929 km, that is what is going on.
00:33:18.000 --> 00:33:24.900
Let us go ahead and take a look at what this looks like.
00:33:24.900 --> 00:33:29.200
Our red, this was our original function, this was our cost function.
00:33:29.200 --> 00:33:32.800
It is going to end up minimizing someplace.
00:33:32.800 --> 00:33:34.800
But clearly, our domain is 0 to 7.
00:33:34.800 --> 00:33:37.500
We are only concerned with from here to about here.
00:33:37.500 --> 00:33:40.200
But this is the overall function, if we need it.
00:33:40.200 --> 00:33:48.100
The blue, this was the c’, just to show you where it is.
00:33:48.100 --> 00:33:53.500
That is it, the 5.929, that is this number right here.
00:33:53.500 --> 00:33:57.700
As you can see, that is where it actually minimizes that.
00:33:57.700 --> 00:34:07.000
Also, you should notice that at least here, it only passes through 0 once.
00:34:07.000 --> 00:34:10.000
This c’ that we got, we ended up with a quadratic.
00:34:10.000 --> 00:34:13.600
Whatever it was that we got, it only passes through 0 once.
00:34:13.600 --> 00:34:23.300
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.
00:34:23.300 --> 00:34:27.600
When you do that, you tend to sometimes introduce roots that do not exist.
00:34:27.600 --> 00:34:29.100
You can think about it that way.
00:34:29.100 --> 00:34:34.200
You can think of it as, it is outside the domain so we can ignore that.
00:34:34.200 --> 00:34:36.000
You can think of it as a false root.
00:34:36.000 --> 00:34:41.200
If you go ahead and graph it, you can see that it does not touch 0 again at 8.068.
00:34:41.200 --> 00:34:46.000
Clearly, only the 5.929 is the answer, whatever you need.
00:34:46.000 --> 00:34:49.000
You just have to be vigilant.
00:34:49.000 --> 00:34:53.400
It is not just about putting the numbers in, take the derivative, whatever they are, check the answer.
00:34:53.400 --> 00:34:58.500
You want to still stand back and make sure things make some sort of physical sense.
00:34:58.500 --> 00:35:06.100
You want to take a look at the function, think about the function, use every resource at your disposal.
00:35:06.100 --> 00:35:07.900
Let us go ahead and go to the next problem here.
00:35:07.900 --> 00:35:11.800
A long pipe is being carried down the hallway that is 10 ft wide.
00:35:11.800 --> 00:35:17.200
At the end of the hallway, a right angle turned to the right must be made into the hallway that is 7 ft wide.
00:35:17.200 --> 00:35:20.800
What is the longest pipe that can make that turn.
00:35:20.800 --> 00:35:23.800
Let us draw this out, see what we are looking at here.
00:35:23.800 --> 00:35:36.200
We have got a hallway, it is going to be some hallway this way and it is going to be this way.
00:35:36.200 --> 00:35:43.400
The initial hallway that we are going down is 10 ft wide, that is this one.
00:35:43.400 --> 00:35:50.000
And then, we are going to make a right turn that must near the hallway that is 7 ft wide.
00:35:50.000 --> 00:35:53.000
This is 7, we are carrying this long pipe.
00:35:53.000 --> 00:36:03.500
The only way this pipe is going to make it is like that.
00:36:03.500 --> 00:36:12.800
When we turn it, it has to just barely touch the wall, in order to actually make this turn.
00:36:12.800 --> 00:36:17.000
That is our pipe.
00:36:17.000 --> 00:36:22.100
What is the longest pipe that can make that turn?
00:36:22.100 --> 00:36:24.600
Let us see what we have got.
00:36:24.600 --> 00:36:27.000
I think we are going to break this up.
00:36:27.000 --> 00:36:31.200
You stare at this a little bit, we try different things.
00:36:31.200 --> 00:36:34.200
We try to see what is going to happen.
00:36:34.200 --> 00:36:42.900
Based on this, I decided to call this l1 length 1 and I will call this length 2.
00:36:42.900 --> 00:36:55.800
I ended up calling this θ, this is also θ.
00:36:55.800 --> 00:37:10.500
The total length is equal to l1 + l2.
00:37:10.500 --> 00:37:15.000
Let me do it over here.
00:37:15.000 --> 00:37:23.700
The cos(θ) is equal to 10/l1.
00:37:23.700 --> 00:37:32.400
Therefore, l1 is equal to 10/ cos θ.
00:37:32.400 --> 00:37:40.900
Over here, sin(θ) is equal to 7/l2.
00:37:40.900 --> 00:37:49.000
Therefore, l2 is equal to 7/ sin(θ).
00:37:49.000 --> 00:37:52.300
Therefore, I put these into here.
00:37:52.300 --> 00:38:06.900
Therefore, l is equal to 10/ cos(θ) + 7/ sin(θ).
00:38:06.900 --> 00:38:14.100
Now at least I have a function l, a function of only one variable, θ.
00:38:14.100 --> 00:38:19.500
Our domain in this case, θ is going to be 0 - 90°.
00:38:19.500 --> 00:38:26.100
We do have 0 and we have 90°.
00:38:26.100 --> 00:38:29.700
I’m going to say 0 to π/2.
00:38:29.700 --> 00:38:33.300
Let us stick with radian measure.
00:38:33.300 --> 00:38:34.800
That is that, great.
00:38:34.800 --> 00:38:37.500
L is just function of θ.
00:38:37.500 --> 00:38:41.800
Now I’m going to take dl dθ or l’.
00:38:41.800 --> 00:38:46.600
What I get is the following.
00:38:46.600 --> 00:38:58.300
Dl dθ is equal to 10 sin θ/ cos² θ.
00:38:58.300 --> 00:39:07.600
This is quotient rule, or if you want to bring this and write this is 10 cos⁻¹, however you want to do it.
00:39:07.600 --> 00:39:14.200
This × the derivative of that - that × the derivative of this/ this².
00:39:14.200 --> 00:39:16.600
That is 10 sin θ cos² θ.
00:39:16.600 --> 00:39:25.000
And then this one, this × the derivative of that 0 - that × the derivative of this/ this².
00:39:25.000 --> 00:39:33.100
You are going to get -7 cos θ/ sin² θ.
00:39:33.100 --> 00:39:39.400
We are going to set this derivative equal to 0.
00:39:39.400 --> 00:39:43.600
Now we have this, now we just need to solve this equation.
00:39:43.600 --> 00:39:49.600
I have got a common denominator there.
00:39:49.600 --> 00:39:54.700
Actually, let me go ahead and rewrite it, it is not a problem.
00:39:54.700 --> 00:40:14.500
I have got dl dθ is equal to 10 sin θ/ cos² θ - 7 cos θ/ sin² θ, we want that equal to 0.
00:40:14.500 --> 00:40:16.000
How do we solve this?
00:40:16.000 --> 00:40:19.900
Let us do a little common denominator here.
00:40:19.900 --> 00:40:46.000
I will do a common denominator, I'm going to get to 10 sin³ θ - 7 cos³ θ/ cos² θ sin² θ that = 0.
00:40:46.000 --> 00:41:03.700
It is the numerator that equal 0, I have got 10 sin³ θ – 7 cos³ θ, that is going to be equal to 0.
00:41:03.700 --> 00:41:15.400
I have got 10 sin³ θ = 7 cos³ θ, switch things around.
00:41:15.400 --> 00:41:23.500
Sin³ θ/ cos³ θ = 7/10.
00:41:23.500 --> 00:41:30.100
This is 10³ θ = 7/10.
00:41:30.100 --> 00:41:42.400
When I take this, I get 10 θ = 0.8879.
00:41:42.400 --> 00:41:52.700
When I take the inverse, I get θ is equal to 41. 6°, that is one of the critical points.
00:41:52.700 --> 00:42:15.500
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.
00:42:15.500 --> 00:42:20.000
That gives me the length of l1.
00:42:20.000 --> 00:42:38.600
L2 was equal to 7/ sin θ which is 7/ sin(41.6), which ends up being 15.06 ft.
00:42:38.600 --> 00:42:47.600
Our total l is going to be 28.13 ft.
00:42:47.600 --> 00:42:49.400
That is it, it is that simple.
00:42:49.400 --> 00:43:02.100
When you put in the 0 and the 90, I should have done it, I apologize.
00:43:02.100 --> 00:43:07.500
You are going to get numbers that are not going to give you an ultimate length.
00:43:07.500 --> 00:43:12.900
The longest that it can be is going to be the 28.13 ft.
00:43:12.900 --> 00:43:21.000
For this particular one, I will let you take care of the 0 and the π/2 in for θ.
00:43:21.000 --> 00:43:26.100
Let us go to our final example here, revenue, cost, and profit.
00:43:26.100 --> 00:43:31.500
Show that the marginal revenue = marginal cost, when profit is maximized.
00:43:31.500 --> 00:43:42.600
B, if the cost function is this function and the price function is this function, what level of production will maximize profit?
00:43:42.600 --> 00:43:51.300
Level of production means how many units sold, how many units should I make, when all of them are sold, in other words, x?
00:43:51.300 --> 00:43:53.400
Let us talk about this a little bit.
00:43:53.400 --> 00:44:00.000
Let us talk about some variables here.
00:44:00.000 --> 00:44:02.100
We will talk generally about what these things are.
00:44:02.100 --> 00:44:15.700
X is going to be the number of units sold, or the number of units produced, the number of units.
00:44:15.700 --> 00:44:31.000
Our price function, the price function, we will generally use a p(x), that is this one.
00:44:31.000 --> 00:44:40.300
Our revenue function, revenue function is just your revenue, how much money you are actually bringing in?
00:44:40.300 --> 00:44:46.300
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.
00:44:46.300 --> 00:44:55.900
It is going to be x × p(x), the number of units × the price, that is the revenue function.
00:44:55.900 --> 00:45:04.000
Cost function, your cost function is how much it costs you to make each unit?
00:45:04.000 --> 00:45:12.400
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.
00:45:12.400 --> 00:45:18.500
That is that is difference, there is a price and there is a cost that you actually incur for making this thing.
00:45:18.500 --> 00:45:22.900
The cost function, we will usually just call it c(x), whatever that happens to be.
00:45:22.900 --> 00:45:25.600
In this case, our cost function is this.
00:45:25.600 --> 00:45:33.700
If I make 500 units, I put 500 in for x, that gives me the cost that I pay,
00:45:33.700 --> 00:45:39.100
that I have to incur as the manufacturer, in order to make 500 units of this thing.
00:45:39.100 --> 00:45:43.900
Hopefully, I make that cost back + any more, that is going to be my profit.
00:45:43.900 --> 00:45:54.100
There you go, your profit function P(x), it is equal to your revenue function,
00:45:54.100 --> 00:46:01.900
how much money you bring in total - your cost function, how much you actually spend.
00:46:01.900 --> 00:46:03.400
I hope that make sense.
00:46:03.400 --> 00:46:13.300
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.
00:46:13.300 --> 00:46:16.600
Your profit ends up being $700,000, it is that simple.
00:46:16.600 --> 00:46:21.700
Revenue – cost, profit = revenue – cost.
00:46:21.700 --> 00:46:30.900
When we speak of marginal in the world of money and finance, as far as mathematics is concerned,
00:46:30.900 --> 00:46:42.000
marginal just means take the derivative.
00:46:42.000 --> 00:46:48.300
Marginal cost c’, marginal profit p’, what is p’?
00:46:48.300 --> 00:46:51.600
It is marginal, it is r’ – c’.
00:46:51.600 --> 00:46:55.500
It is marginal revenue - marginal cost.
00:46:55.500 --> 00:47:01.100
Marginal just means it is a derivative, it is a rate of change.
00:47:01.100 --> 00:47:04.900
Let us do part a, very simple actually.
00:47:04.900 --> 00:47:11.200
Part A says, show that the marginal revenue = marginal cost, when profit is maximized.
00:47:11.200 --> 00:47:13.300
The profit function, we already know what that is.
00:47:13.300 --> 00:47:21.700
The profit function is equal to the revenue function, the money that I bring in - the cost function, the money that I spend.
00:47:21.700 --> 00:47:35.500
Profit is maximized when p’ = 0.
00:47:35.500 --> 00:47:46.900
Profit is maximized when p’(x) = 0.
00:47:46.900 --> 00:47:52.900
P’ is nothing more than r’ – c’.
00:47:52.900 --> 00:47:57.200
P’ is just r’ – c’, the derivative is linear.
00:47:57.200 --> 00:48:03.800
R’ – c’ is equal to 0, I just solved this equation.
00:48:03.800 --> 00:48:10.700
R’ = c’, marginal revenue = marginal cost.
00:48:10.700 --> 00:48:16.100
That takes care of part A, very simple.
00:48:16.100 --> 00:48:23.000
Let us go ahead and deal with part B.
00:48:23.000 --> 00:49:05.900
A derivative is a rate of change, r’(x) marginal revenue is the rate at which the revenue changes per unit sold.
00:49:05.900 --> 00:49:14.600
How fast is my revenue increasing or decreasing, if I sell one more unit?
00:49:14.600 --> 00:49:42.600
C’(x) marginal cost is the rate at which cost changes per unit produced.
00:49:42.600 --> 00:49:51.900
In other words, how fast is the cost that I'm incurring changing if I make one more unit?
00:49:51.900 --> 00:49:59.100
I asked my people in my company, if I make one more unit, how much more is it going to cost me?
00:49:59.100 --> 00:50:04.000
That is c’, it is the derivative of the cost function.
00:50:04.000 --> 00:50:15.700
Let us do B, our profit is equal to our revenue - our cost.
00:50:15.700 --> 00:50:18.000
Revenue is how much you bring in.
00:50:18.000 --> 00:50:22.200
It is how many you sell × the price that you are selling it at.
00:50:22.200 --> 00:50:42.200
That = x × the price function pp, P profit, p price, -c(x) = x × 1500.
00:50:42.200 --> 00:50:50.000
1500 - 5x, I think was the price function, - the cost function, - the whole cost function.
00:50:50.000 --> 00:51:10.700
You have to put them in brackets, [14,000 + 400x - 1.3 x² + 0.0035 x³].
00:51:10.700 --> 00:51:41.000
Therefore, our profit function is equal to 1500x - 5 x² - 14,000 - 400x + 1.3 x² - 0.0035 x³.
00:51:41.000 --> 00:52:01.100
Simplify, we end up with -0.0035 x³ - 3.7 x² + 1100x - 14,000.
00:52:01.100 --> 00:52:05.000
I take the derivative, try to maximize profit.
00:52:05.000 --> 00:52:09.200
What value of x, what production level will allow me to maximize my profit?
00:52:09.200 --> 00:52:13.400
My profit function is equal to my revenue – cost, I have that function.
00:52:13.400 --> 00:52:34.100
Simplify that function, now I take the derivative of that function, -0.0105 x² - 7.4 x + 1100.
00:52:34.100 --> 00:52:53.000
I set the derivative equal to 0.
00:52:53.000 --> 00:53:08.300
When I solve this quadratic equation, I end up with x = 126.1, x = -830.9.
00:53:08.300 --> 00:53:16.800
Clearly, I cannot make a negative amount of things.
00:53:16.800 --> 00:53:32.400
Producing 126 units maximizes profit.
00:53:32.400 --> 00:53:38.100
Let us take a look at what this looks like.
00:53:38.100 --> 00:53:42.900
This right here, this is our p(x).
00:53:42.900 --> 00:53:47.400
Remember, our p(x) was a cubic equation.
00:53:47.400 --> 00:53:48.100
This is not a quadratic equation, it is actually cubic.
00:53:48.100 --> 00:53:51.700
I have just taken this piece of it so that you see the piece of it matters.
00:53:51.700 --> 00:53:58.600
Obviously, the minimum I can make is 0 things.
00:53:58.600 --> 00:54:02.200
I'm not going to make 0 things, I'm going to make more than that.
00:54:02.200 --> 00:54:08.200
This, as I make more and more, this was my profit function.
00:54:08.200 --> 00:54:12.700
At some point, I'm going to hit a maximum.
00:54:12.700 --> 00:54:16.100
This right here, that is p’.
00:54:16.100 --> 00:54:19.500
It is p’(x), that is the one that we set to 0.
00:54:19.500 --> 00:54:26.400
Notice it crosses 0 at 126.
00:54:26.400 --> 00:54:36.900
When I make 126 items, my profit is maximized and the maximum is something like that, whatever that number is.
00:54:36.900 --> 00:54:39.600
600,000, somewhere near 600,000.
00:54:39.600 --> 00:54:41.700
That is what is going on here.
00:54:41.700 --> 00:54:45.000
I have zoomed in on this, this is not a quadratic function.
00:54:45.000 --> 00:54:48.000
It is the high point of a cubic function.
00:54:48.000 --> 00:54:53.100
This function actually goes down and comes back up the other end.
00:54:53.100 --> 00:54:55.500
I’m not concerned about this part, the negative part.
00:54:55.500 --> 00:55:00.800
I’m concerned because I have to make at least one unit.
00:55:00.800 --> 00:55:07.700
Clearly, maximizes at x = 126.
00:55:07.700 --> 00:55:10.000
Thank you so much for joining us here at www.educator.com, we will see you next time, bye.