 Continuing with our discussion of applications at the derivative in this video, I want to focus on how the derivative is useful in the field of business, particularly in the discipline of economics. So let's talk about things like cost, revenue and profit and what that has to do with derivatives. So with the first example, let's suppose that C of X is the total cost a company incurs in producing X units of a certain commodity. Alright, so the function C is this is the so called cost function. How much money does it cost the company to produce X mini graphing calculators or protractors or whatever, whatever it is they're making here. So what is this language of like delta C delta X mean in the situation? Well, if you have one level of production C or X one, so it's like, oh, we made 10,000 protractors, and we change it to some next level of production. Like so we went from 10,000 protractors to 15,000 protractors and be very concerned about the additional cost it takes to compute that are to come to the additional cost to produce a higher level of production. So delta C, which would be C of X two minus C of X one, this would be the cost of going from level X one up to level X two because we're going to subtract the cost of X one we're already spending that much, how much additional would be to go up. Alright, so this delta C can very naturally be a quantity for a business to consider right what's the additional cost to increase our production. Relate of course would be the average rate of change or the average rate of change of the cost function delta X over delta excuse me delta C over delta X. This would be the additional cost on top and then you divide this by the production change. You raise from 10,000 to 20,000 cost divided by 10,000. This would give you the average cost per unit you're producing each additional unit you're producing. If we take the derivative here that is a limit as the production, the change of production goes to zero. So delta X goes to zero, we get the so called marginal cost. Now this is one thing I want to emphasize in this video and we'll come up later on in this lecture series. Whenever you see the adjective marginal used to describe a function in economics that pretty much means you're taking the derivative. Marginal cost is going to be the derivative of cost. Marginal revenue is the derivative of revenue. Marginal profit would be the derivative of the profit function. That's the pattern we see here marginal represents the derivative and why marginal got like a piece of paper and you talk about the margins of that paper. These margins are meant to be a very small region. So a marginal cost will be a very small change to the cost function. That's exactly what the derivative is measuring this infinitesimal change to the to the cost function. So marginal cost would be the limit as delta X approach to zero of the average cost, the average rate of change of cost you should say delta C over delta X. So we have to note this DC over DX. So this is the derivative of cost with respect to production. Now note, if your delta X is equal to one, so you increase production by one unit. If you think about that for a second, if you take delta C over delta X and delta X is just one, you're going to get this delta C over one. This is just delta C. If you compute the average rate of with just a one one commodity additional production level, that's just the same thing as additional cost. But what we also know about the average cost is that average cost when dealt with the when the change of production is small when the change of X is small, the average cost will be approximately the same thing as the derivative. So what this tells economists is that additional costs can be approximated by using the derivative and this is something that's commonly used in practice here. And summarizing that we see that that that the derivative of cost, the marginal cost at production level N is approximately the same thing as the cost of producing in plus one minus the cost of N. So that is to say this additional cost of producing the N plus first commodity thing as the derivative at N, not exactly the same thing, but it's an approximation. And we'll see that in general, computing the marginal cost is actually easier than computing the additional cost. And so that can help when we run these algorithms to make these decisions based upon approximations and such. So let's take a quadratic cost function. This is actually quite reasonable. You get things such like a fixed cost that hey to run the factory every day, there's gonna it's gonna cost $10,000 no matter what doesn't matter how many, how many objects we produce, you know, we have insurance for the day power has to be paid for employees things like that. So let's say that a particular company spends $10,000 to run. So that's a fixed cost, but then there's gonna be some variable costs as well, some of which could be linear some of which could be quadratic or our various other things we won't go into the economics of it but it's like hey, if you're producing, you know, protractors you're going to need some plastic to make the protractors are out of the more protractors you make the more it might cost right. So let's let's just suppose this is our cost function a quadratic cost function to be quite reasonable here. Well the derivative of the cost function by the power rule is going to be fairly straightforward. The derivative C prime of X will just be five plus point zero two X fairly. In fact, many, many business students who learn about this stuff they learn how to take the derivative of a quadratic function without really knowing what derivative is all about it's they're told basically like okay, if you're if your cost function looks like this parabola, you ignore this you take away that you drop this and times everything by two right so they do the power rule without actually knowing it. So they that they've learned about derivatives without necessarily knowing all the details behind it. The derivative rules we learn make calculate marginal costs super easy. And so then once we have the marginal cost we can then ask ourselves the what's the marginal cost at the production level of 500 items here was put into the function C prime of 500. And you plug it in, you do the arithmetic use a calculator to help you if you need to, you end up with the value $15 per item so at a production level of 500 the marginal cost will be $15 per item addition of the additional cost. Now let's actually take a look that if we take this quadratic function right here, if we plug in C of 501 and subtracting that C of 500. So this is right here the cost of producing 500 commodities. And this is the cost of producing 501 items right the difference will be the cost of producing that 501 first item. Plug this into the quadratic formulas and simplified you end up with $15 and one penny. So I mean unless you're like. Unless you're, you know, having an auction battle would like to white shrewd or here that one penny is not going to make much of a difference $15 per item versus $15 and one penny for that additional item with this illustrates in fact that the marginal cost is a pretty good approximation of additional cost absolutely. And by all means, the linear function which is marginal cost is much easier than the difference of quadratic functions. I agree with the economists here that marginal cost is a good tool to help us approximate these these type of calculations. Let's look at another example here. This time let's talk about demand. What is demand after all demand is how much are targeted consumers want the object. And this is closely related to the idea of price that we probably know and if we did not a big deal, we probably know that the more expensive something is people are less inclined to buy it. The cheaper something is the more likely we are to buy it. Personally, I like to use a website called camel camel camel using it to try to buy things, you know, from online right and the thing is the current price for the object where type one, it's more than I'm willing to spend it's like I don't want it. I don't want it $20 worth. I would buy it if it was like $18 maybe. And so I set I set up a watch a price watch with camel cubed to see when it will reach the price that my is going to go with. And so as people decide what price they should sell a certain object for it's like well do we want to sell 100 of these things do we want to sell 200 of these things 1000 of these things three millions of these things because after all we want to sell 3 million of certain things. We can do that we can create that type of demand if we pick the right price and function sometimes called the price function. It looks like the following we say P equals D of X is P for and and what this means is if you want to sell X many items set the price to be P. That's that's how our demand function is going to come into play. Now this is related to what we want to talk about here the idea of revenue. So revenue is the amount of money a company brings in from their sales sales of commodities sales of a service. This is the money they bring into the company this is not the profit we'll talk about that in just a second. But this is the money we bring in by selling our our our products by selling our services whatever the now revenue is computed by the number of items we sell which we call that X multiplied by the price of that. So if you're selling protractors for $2 per protractor and you sell you know 100 protractors with right there 100 times $2 right $2 per protractor we see that the revenue from protractor sales will be 200 that's the basic idea there. But because it demand right price is not some arbitrary thing price is related to the things we want to sell right. If we want to sell 100 protractors that we just set the price to be 200 or $2 so that we sell all 100 we don't want this stock or anything like that. So revenue you can think of it as the number of items sold times by the demand function. Okay. And so for a certain product let's just hypothetically this is just a fictitious product right here. Let's say that the demand function is given by the following price will equal 50,000 minus X over 25,000. Okay, so this is a linear function and write this as well 25,000 goes into 50,000 twice to get to minus over here. We're going to end up with X over 25,000 like so. And so notice here the slope this is a linear demand function which is a reasonable model to use at times. They should have negative slope if you're going to have a net if you're going to have a linear demand. Why say negative slope here you should think of it as this is our X and this is our price right here that as the price falls as the price falls the number of sales will go up. So you can model that in such a way so our slope here would be m equals negative one over 25,000 like so. And how that should be interpreted is as every time you go down $1 you're going to increase the sales by 25,000. So that's that's what we anticipate for this product we're selling right here. Okay. So continuing on. We have our demand function so revenue is going to be the number of things we sold times by the price plug it in the demand function the price function again those are sentence right there you're going to get X times that. Distribute the X like here we end up with the function 50,000 X minus X squared over 25,000 Oregon if you want to break up the fractions. This is going to look like 2x minus one over 25,000 X squared. So revenue is it just like the cost function we saw a moment ago revenue is this quadratic thing not I mean that doesn't have to be the case. If you have a linear demand, then you will have quadratic revenue. Now, let's say we take the derivative of revenue, this would give us the so called marginal revenue will as it's a quadratic function the revenue the marginal revenue is easy to compute. The derivative 2x is going to be a two the derivative of x squared can be a 2x which we see right here times that by the the one over 25,000 we have we end up with negative one over 12,500. So this is the marginal revenue function. Let's plug in the value 10,000 so it's the marginal revenue at production level I should say a sales level of 10,000. The marginal revenue at 10,000 will be two minus one negative one over 12,500 times it by 10,000. I won't bother you with the arithmetic you can see it right here in the end this will simplify to be the number 1.2. How should that be interpreted. This is to say that your marginal revenue will be $1.20 per unit, thus the next item sold at the level of 10,000 will produce an additional revenue of about $1.20 like so. So, again, that's how much that's how much additional revenue we should expect if we were to increase it like so. All right, let's do one last example let's combine the notions of let's combine the notions of let's say that with this example let's keep the same revenue function we had from the previous example. So revenue is equal to 2x minus the one over 25,000 x squared grade. Let's use a different cost function this time let's say that we just have a linear cost function. You have your fixed cost and you have your variable cost and the variable cost is directly proportional to the number of things we sold. So we have a cost function profit is then the difference of revenue and cost so cost is the money exiting the company revenue is the money entering the company. And so that difference would be profit and ideally we want profit to be positive that's we wouldn't want it to be anything else right we want profit right here. And so if we want to compute the profit function for our company we said the difference of these things. So let's take the quadratic revenue that we have from before. Let's take the linear cost which we've introduced for this example. If you subtract these things the polynomials is combining like terms. We see that the profit function is going to be 1.75x minus one or minus x squared over 25,000 minus 2100. Okay, don't worry so much about the formula. This is just this is just we get by doing the mathematics right there. Well then this is a quadratic profit function by the rules we know for derivatives we can very easily easily have a profit so p prime of x that is the so called marginal profit. This will equal 1.75 minus x over 12,500 fairly simple derivative calculation. If you use the rules we've developed here in chapter three. And then if we ask ourselves what's the marginal what's the marginal profit at p prime are at the production level of 15,000 just so we're clear we're using the assumption right here that we're producing 15,000 so our cost function will be 15,000 we're also selling 15,000 work this this model right here is not built into any idea that we didn't sell in we didn't sell everything we were assuming everything is sold and which is built inside of the revenue function. Assuming the demand functions accurate enough we can expect that to be not so outlandish of an assumption so with with a production and sales level of 15,000 the marginal profit will be put into our linear function linear functions are basically the easiest to get. We're going to produce the number 0.55 so what that means what that means for this company is that at the level of 15,000 we're basically making 55 per unit. We can also change that. I should say I should say that we are. At this level that the other instantaneous rate change it's not that we're making 55 units or 55 cents per unit the idea is where our profits increasing at this moment that our profit is that we if we continue to increase, if we continue to produce in production, the next unit will make us an additional that's what we should be thinking of. Think about this one right here if we take 20 21,875 if you put that into your marginal profit function again don't worry about all the details it's just a linear function you end up with a zero. So it's like okay at this level of production we're not we're not gaining nor are we losing profit. Okay. Then if you do 25,000 again for this one if we compute that we're going to end up with a negative 0.25. So what this is suggesting to us at this level of production we're actually losing profit. The reason is that this level of production even though we're selling everything the cost is so much more than the revenue that it's actually going to be cost prohibitive to produce this many 5,000 units if we set the right price you know go Bob Barker on this one the price is right. But if the cost of selling it is less or is more than the revenue we make from selling it, then by all means that's not going to help us get a marginal profit of actually something negative right here. And so notice this one. If you look at the level of 15,000, the marginal profit was telling you like hey, if you makes at this point if you made more than 15,000, you're going to be getting more profit at the 25,000 level. You're going to be getting negative profit. And what's so special about this number 21,875 where did it come up with that number? Well this is the unique number for which the marginal profit is equal to zero right the profit function is neither increasing or decreasing. If we think of our profit function as that parabola we saw we mentioned the quadratic formula but if it's a parabola then that parabola should have this maximum value. There's like this vertex of the parabola which the parabola like again as a vertex as a as this point here it's going to be the maximum value of profit this is the maximum profit. And this maximum value coincides with a horizontal tangent line horizontal tangent line would mean that the derivative profit with respect to x is equal to zero. So finding where the marginal profit is equal to zero will correspond to the maximum profit that's what we want to make this one's too hot this one's too cold. This is the Goldilocks number it's just right. And so, as we study more about derivative applications in our next chapter chapter four we're not quite there yet but we'll be there we'll be there in a few more lectures. We're going to talk a lot more about this optimization problem we're using derivatives as a critical tool to decide what is the right decision. How many protract if whatever this commodity they're selling is the magic number dependent based upon the cost and demand of that product. The best number is 21,875 and that's where we derive from calculus.