Marginal Analysis Applied Calculus

Marginal Analysis   Applied Calculus

Hi, today in applied calculus we’re going to learn about marginal analysis. I always tease my students that I can teach this in 30 seconds or less and ask them to time me. Are you ready? Marginal means derivative. So basically when you’re doing a business application and you read the word marginal, it’s a keyword for derivative. We’re going to go a little bit more in-depth than that and practice with it, but that’s basically it. So marginal means derivative, and before we actually apply that we need to review some of these basic business definitions. So some of them you may know, some may be new. We want to practice with them a little bit. So when you talk about cost for your business, there’s two categories of cost. There’s fixed costs, which are costs that are the same no matter how much you produce and sell. So no matter how good or bad your company is doing these things have to be paid. Like for the the space that you’re using you have to pay your lease, you have to keep the lights on, that sort of thing. So there’s some things that you have to pay no matter what. Variable costs are costs that vary. And so depending how much you’re producing, then that determines how many man-hours you need to run your equipment. It determines how much gas you’re using, depending what you’re talking about if you’re a shipping business. So there are things that vary with how much you are producing and selling. And so total cost is the sum of those two things. It’s the variable cost plus the fixed cost, and this literally means a variable so this one’s going to have your input variable here, x, in it. And that is going to be just the number that is always a constant no matter how much you’re producing. So then we have the price-demand function. So this little p of x, x is the number of items that can be sold if you charge p dollars per item. So the idea, as you can see there, of like supply and demand. If you charge too much for the item, then less people are going to be willing to pay that price and therefore you won’t be able to sell as many of them. Of course if you price your item too low, then you might sell a whole lot but maybe you’re not making the profit that you’re hoping for. So x is the number of items that can be sold at p dollars per item. We need that to build our revenue function. So our revenue function: the product of the number of items sold and the price paid per item. So if I tell you this item costs 10 dollars per item and I sell seven of them, then my revenue is 70, right? Seven at 10 dollars a piece is 70. So a lot of students overthink that. It’s just how many items you sold times how much you sold them for, that’s it. So it’s x times p, but often as discussed up here, p is actually, little p is actually variable and depends. And so it might just be x times a number or it might be x times a function, and then if it’s multiple terms you might need to distribute. And then our last main function will be profit. So it’s a little bit annoying that they use little p and big p when there are so many letters in the alphabet, but they did so we gotta go with it. So they both, you know, make sense but just make sure you’re clear in your writing. So total profit, capital p, is how much you actually make of course. So the issue is yes, you bring in this revenue by selling however many items at whatever price. However you have to pay your costs before you can go put it in your bank, right? Or in your pocket. So the total profit is the total revenue minus the total cost. So these formulas are some that you will need to know, that revenue is x times little p, where little p is that price per item, or price-demand function, and that profit is revenue minus cost. And so lastly some of the questions might talk about a break-even value. That is when you break even, where your revenue covers your cost. So no profit, no loss. The point at which you’ve made just enough from the sale of your products to cover the cost of making those products. So you could either say that it’s when profit is equal to zero, but remember profit was revenue minus cost. So if there’s a zero over here, and you add the c over, then that also means that revenue equals cost. So either one of those is fine. So let’s look at an example. This actually has no calculus in it whatsoever, this is just getting those basic business functions down. So Red Tide is planning a new line of skis. For the first year, the fixed costs for setting up production are 45,000. So that very clearly said that was our fixed cost. The variable cost for producing each pair of skis is $80, and the selling price will be $25 per pair. So they’re telling us that p, little p, there. This we don’t need until we get to a question that actually asks about it. So at first we’re writing the cost function, and we know that that is the variable cost plus the fixed cost. Okay, well remember the variable cost should have a variable in it. X is number of skis and it’s $80 for each pair of skis. So that’s for like the materials and the labor. And the fixed costs of setting up this whole operation was 45,000. So that is our cost function. Then the revenue, total revenue from the sale of x pairs of skis. Well remember… that it’s how many things you sell times the price you sell them for, and they told us that we are selling them for 255 apiece. And so if you wanted to write that as 255x you could. So that’s our revenue function. Now we’re going to need those two to now write a profit function. So we have profit as revenue minus cost, so we’re going to need these two in order to do that on the next page. So profit… is revenue minus cost, and so… Our revenue was the 255x, and then we need to be careful here to use a parentheses because cost had multiple terms. And what was cost, cost was 80x plus 45,000. So the parentheses are really important because it’s going to mean that both of those actually get a negative. So we’ll simplify this and so it’s actually 255x minus 80x, so that’s 175x, and then minus 45,000. So notice that these questions were just asking you to write the function, not to plug anything in yet. And I ask you that too in problems to help you build up to where you need to be. So then it says what profit or loss will the company realize if the expected sale, that’s where that 3,000 comes in. So if they sell 3,000, what will happen? And it says profit. So all we are doing, ’cause remember there are numerous functions floating around here, there’s a cost function, there’s a revenue function, and there’s a profit function. It said profit so that’s the one I’m plugging into. So I’m going to plug 3,000 into our newly written and simplified profit function, oops 3,000, and see what we get. So we would multiply and then subtract, and you should get 480,000. ‘Kay, so that’s just taking the function that we worked so hard to build and using it. So if they sold 3,000 pairs of skis, we expect their profit to be $480,000. And then lastly if you’re ever asked about when they will break even, remember that is when profit is zero. You’ve made just enough revenue to cover your costs. So you can either set it up as profit equals zero or revenue equals cost. Doesn’t matter, you should get the same answer. Okay, so our revenue was 255x and our cost was, what was that cost? 80x. 80x plus 45,000. And so I’m trying to find the x that makes this happen. I’m trying to find the number of skis, which is x, that makes this happen. So when I subtract you should see a familiar number… and then we would divide both sides by 175 to find it. Now what you will see is that this is not a whole number, so what you get is 257 point something, maybe one-four? Okay, but the question was how many pairs of skis do you need to sell? I don’t think I can sell 257.14 pairs of skis. That’s like 257 pairs of skis and just a binding or what, right? So there are times where you have done the math right but you have to think about the real world application as to what kind of answer makes sense. So here you may not always want to use standard rounding. And so let’s think about what happens if we were to find our profit if we rounded down to 257 versus if we found our profit if we rounded up. Okay, so I’ve grabbed my calculator and figured out what would happen if we plugged in 257 versus 258. ‘Cause we’ve decided that we need a whole number of skis. And so if you plug 257 into the profit, you actually are still at a loss of $25. Whereas if you plug 258 into profit, you are positive, and so you actually have had a profit. So if I’m trying to report this to my sales manager, and I need a whole number of skis, I’m probably going to err on the side of profit. So err on the side of profit even though… that did not follow standard rounding procedures. So I would say 258 pairs of skis. So if the point was at which point are we good and our revenue has covered our costs, and I had to pick one of these two, I wouldn’t want to be not quite there yet. I’d want to be more than there. I’d want to have had made a little bit of profit and then go from there. Oops, pairs of skis. So remember there was no calculus here whatsoever, it was just practice with the basic business functions. Now we’re getting into the calculus. So marginal, we’ve got three functions floating around. We have cost, revenue, and profit. And for all of them, marginal means the derivative. So marginal cost is the derivative of cost, marginal revenue is the derivative of revenue, and marginal profit is the derivative of profit. So since we know rules about taking derivatives term by terms, it is also the same as if we had done the derivative of revenue minus the derivative of cost. So if this relations ship is true, and we can take derivatives term by term, then that is also true. Okay, so marginal cost is that instantaneous rate of change, that’s another word for derivative, of cost or revenue or profit, whichever one you’re talking about, relative to production, production is that x, how many things you’re producing at a given production level. So let’s give it a shot. So we are finding the marginal profit function if cost and revenue are given here. So the first thing we need to do is have a profit function period, right? So profit… p of x is revenue minus cost, and so that is 2x minus .08x squared minus parentheses, because cost has multiple terms, 189 plus 0.7x. Alright so we would need to distribute and we would have negative .08x squared plus 1.3x minus 189. So that is our profit function, and then marginal profit means derivative, so p prime. So it is the derivative of profit. So p prime of x, I’m just going to apply that new power rule that we learned. So even if I have crazy decimals it’s still works the same way. I bring down the power, multiply it by what was already there, and subtract one from the power. And here if it’s a constant times x it just ends up being that constant ’cause it’s the slope of the line 1.3x. And then remember the derivative of the constant negative 189 is zero. So that would be my marginal profit function. That’s all. So we had to do revenue minus cost and then take a derivative. So we’ve got different keywords and marginal now is another one for derivative. Okay so this is one more formula that you will need to know for this section. And this is one that’s actually directly to calculus, where the others were the foundation of the business material. So this is showing, remember that the derivative talks about the slope of the tangent line. So if this is our curve, the cost function, and this is the point of tangency, that tangent line shares that point and rides alongside it. Well when we’re very, very close to that point of tangency, it’s very hard to even tell the line, the tangent line, from the curve itself. So like in this little area here, they almost run together, you can barely even tell them apart. And so we can use the tangent line as a predictor of the curve. So it’s a nice line versus being some crazy polynomial function, and we can use it to make predictions. Well obviously the further away from the point of tangency you get, the worse they’re going to get, right? So they’re really close together here. As you get further and further away, they get further and further away as well. And so the tangent line will no longer be a good approximation for the curve. But when you’re pretty close, it’s a decent approximation, and can sometimes be easier to work with. So the idea here is that the derivative evaluated here at x, that would be the slope of that tangent line. It tells me if I’m sitting at this point, slope is rise over run, it tells me how to move to get to the next point. It tells me move up this many, over this many, to get to the next point. And so it would predict that we would be at that point. Versus if we actually use the curve, and so use the original function and subtract, we’ll get something slightly different, but they’re pretty close to each other. And that’s why we have this squiggly, that it’s an approximation. So if we wanna know the cost of making just the x plus first item. So what do they mean by this? Let’s put some numbers in. Let’s pretend x is 70, and so the x plus first item is the 71st item, okay, just as an example. So it says c of x is the total cost of producing 70 items, then the marginal cost function approximates the exact cost of producing the 71st. So remember, this is a total cost function. If I just plug into regular c cost, it tells me how much it costs to make numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, all the way up to 70. Okay, or 71, whichever one I’m doing at the time. So if I wanna know just how much it costs to make the 71st item, the total cost is not going to help me because it did one through 71. So what would we do? Well we would find the cost of one through 71, take off the cost of one through 70, and what’s left must be how much it costs to make just that one more, okay? So if we compute it that way and do c of x plus one minus c of x we would land here, and see how much it costs to make that one more item. If we use the tangent line instead, it’s not going to be exact, but it’s an approximation, and again it might be nicer numbers that are easier to work with. And especially because if I know information about x, then I can go ahead and draw a tangent line. But this relies on me already knowing the information about the 71st. Maybe I’m trying to predict it and I don’t know it yet, right? So that’s why we need this is because we can use knowledge we have right now alone, we only need to know about 70 for instance, to predict 71 using this method. Whereas over here, to get the exact cost I have to already have both data points. I have to have already observed both. So this is the only way that we can use it to make a prediction about the 71st. The other would be that it already happened and now we’re computing it after the fact. But it’s very valuable to make predictions, projections, right, in business to be able to plan ahead. So they’re close to each other, not exact, that’s why it is the squiggly, approximately equal there. And the most common mistake with this is students plugging the wrong number in here. So if 70 is still our example, notice that even though the question is asking about the exact cost of making the 71st, it is 70 that I am plugging into the derivative to make my prediction. Why? Because I only know about 70, and I’m using that rise over run, that slope of the tangent line, based on how steep it is right now to predict about the 71st. So make sure you put regular x, or the 70, over there. So let’s try one and see what we’re talking about. Okay, so we’re going to try it using those formulas. We have a total cost function, and actually I just randomly said 70 but it ended up being the right, so 71st food processor, okay? The exact cost is if you take one through 71 and subtract off one through 70, then you would know exactly how much it had costed to make just the 71st. So that’s using the original function, but it requires us to already know both data points. So we are just going to plug 71… into the entire profit function, and then subtract from that carefully with parentheses what we get when we plug 70 into the entire profit function. Okay, so if we calculate that and figure out what those values are, the c of 71 is 5,565 and 40 cents, and the c of 70 is 5,560. And so the difference in between there is that $5.40. So that is exactly how much it costs to make the 71st item. If we are using the marginal to approximate, remember the idea is that we know… about the 70th and we are trying to predict about the 71st. So it’s nice and easy, we just need the derivative. So for this part we’re using, so this was the original for part A, here we are using the derivative. I’m not evil, I’ll always ask them in the same order, so that’ll help you remember. Okay, so this would be a constant, so the derivative of a constant is zero, and then power rule 90 minus 1.2x. And all I’m doing is plugging, be careful, 70, even though this says 71 here, to make a prediction based on slope, based on rise over run, of where I would be at 71. Is it exact? No. That’s why it says approximate. But it’s close, okay? And so we have 90 minus 1.2 times 70, and that is 90 minus 84, and so we get $6. And over here we got $5.40, so they are close, right? So they’re not going to be the exact same, don’t feel like you missed it when they’re different, they’re supposed to be different, but if you get like 20 here, and you got 5.40 there? That’s a problem, so you would want to check your work. But they should be relatively close. So if you were sitting here trying to predict the cost of making the 71st item, and you estimated $6, then once it actually happened and you went back and calculated it, and it was $5.40, well then yeah, that was a pretty good estimate. Okay, so this series of formulas is something that you need to know, that exact cost is the original function of the one 71 minus 70 plugged in. And the marginal cost to approximate means using the derivative, and if it says 71, you’re actually back at 70 making a prediction about the 71st. You can definitely expect a question like that. And then the last business concept that we have is this concept of average cost. And so if, and we can do average cost, average revenue, average profit. So if I tell you that it cost me $100 to make 10 things, then you would automatically say well they were about $10 apiece. But it’s not always the case that it actually cost me $10 to make the first one, and the second, and the third. One good example is if you’ve ever ordered T-shirts. So there’s a certain cost for them to make the screen to print for your special design, but then once they have that in place it’s not really that hard for them to make more. So if you only ordered five shirts, it’s going to seem like it’s a lot more per shirt. Whereas if you then do 500, they still only had to set up that print one time, but then they got to print 500. So the per shirt price would go way down. So it is not always true that it costs the same to make each item along the way. But on average, what was it? That’s all. So be careful, there’s no adding things up and dividing by two here, okay? This is saying if it costs that much to make 10 things, then on average, so I had said 100 and 10, then on average they each cost $10, okay? So average cost, average revenue, average profit are just those functions divided by x. The notation for it is a bar over the c or the r or the p. And we can also have marginal average cost, marginal average revenue, and marginal average profit. We need to be careful though, because when we’re doing marginal average cost we need to find average cost first and simplify it before we take the derivative. So I have that warning here. You must first find the average cost or revenue, by dividing by x and simplifying it, then calculate the derivative. The reason is because you cannot just take the derivative of a numerator and the derivative of a denominator. You need something called the quotient rule, which we’re going to get to eventually. So right now the only way around this is for us to actually simplify it and then use the power rule. So we have to simplify it first or we will accidentally be doing bad math because we haven’t learned the quotient rule yet. So you simplify first, and my directions would reflect that to help you with that. Write it and simplify it, now please take the derivative of it. So let’s look at our last example. The total cost in dollars of manufacturing x auto body frames is 20,000 plus 300x. So the first one is find the average cost per unit if 80 frames are produced. So I’m going to break this down even a little bit more, and first just say what is the average cost? Then please plug 80 into it so that you can see each piece. Okay, average cost is just c of x divided by x. That’s 20,000 plus 300x all divided by x. I need to simplify this because eventually I’m going to be doing marginal and I can’t take the derivative as written. So that’s why I’m going to go ahead and simplify it now. And what that means is I need to divide each piece by x. Okay, but that simplifies. So eventually when I go to take the derivative I will want to write it in power rule form. But for now I’ll leave it this way. So 20,000 over x plus 300. And so if I want to know the, and I could use this notation too, they mean the same thing. And it just makes it easier if I now want to know about 80. So I would plug 80 in for my x’s, or my one x over here. And it turns out that that gives us $550. So when 80 of these auto body frames are produced, the average cost of producing each one is $550. So again, that doesn’t mean that number one cost 550 and so did number 80, but if you average out how much it cost then it would be 550. ‘Kay, so then in B, we are finding the marginal average cost. So now for B we are finding marginal average cost, okay? So we have to first know our average cost, but we did that. So our average cost function was 2, or was it 20,000? 20,000 over x plus 300 but I’m about to take a derivative. So to take a derivative I should rewrite this in its power rule form first, remember. So I need to bring that x up, and now it’ll have a negative one power plus 300. And now I’m ready to take its derivative. So the marginal derivative average cost is just the derivative of average cost, but I had to simplify the average cost first. But I lead you through that in the order of the directions. So we have negative 20,000x to the negative 2 and the derivative of the constant 300 is zero. So I could write it as negative 20,000 over x squared, okay. And I think the question said now what about at 80? So if we evaluate this when 80 auto body frames are being produced, then we are just plugging… 80 into that function and we end up with negative 3.125. Okay, so that negative is going to tell us something. If our rate of change, derivative, right, is negative, then we know something is decreasing. And so I think the last part was to kind of write a sentence about what this means. So this is… about $3.13 that it was going negative. And so if we summarize what we found: when 80 frames are being produced and sold… we have the average cost per frame was 550… and it is decreasing because of this negative… at a rate of… $3.13 per frame. So this should remind you of some of the interpreting of the derivative we’ve done before, and you had to write a sentence interpreting and you had phrases like increasing at a rate of or decreasing at a rate of, and we will have that again. So what have we found? We found that when 80 frames were produced and sold, the average cost per frame was 550, but that cost per frame is decreasing, average cost per frame, is decreasing at a rate of $3.13 per frame. So that means if I made the 81st one, it would have lowered my average cost down from 550 to $3.13 less than that, okay? So that’s putting all of that together. But notice we first found the average cost then did some rewriting and took the derivative when we were asked for a marginal average cost. And so it tells us on average how much it is going down now that we’re producing more and more of these items. So that is it for marginal analysis, and we will be starting a new chapter next.

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