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## Differential Calculus (2017 edition)

### Course: Differential Calculus (2017 edition)>Unit 11

Lesson 6: Applied rates of change

# Applied rates of change: marginal costs

Sal solves a problem involving the modeling of cost with a function, which means the function's derivative models the marginal cost. Created by Sal Khan.

## Want to join the conversation?

• To find the marginal cost of producing the 101th gallon why do we use C'(100) instead of C'(101)? When I calculate C'(101) I get 12.14 which is an over estimate of the actual cost by .12. The value of C'(100) is an under estimate of the actual cost by .12. Is it a coincidence that the error is .12 in both cases, just a different direction? • The marginal cost graph that Sal drew represents the cost of the next gallon of wood stain if the price doesn't go up. if everything stays as it is when you produced that 100th gallon, then the next gallon will cost 11.90.
However, because of economics and supply and demand, the the actual cost will go up. That means that when you produce the 101st can of woodstain, it will be 12.02. And the next can will go up not the same amount, but an amount slightly higher, unless the marginal cost doesn't go up (if supply and demand weren't real factors).
• Why is the marginal cost useful? Why not always just calculate the exact cost? • The marginal cost is used (among other things) to determine whether you can afford to increase production. If you have a business and are producing at a certain level, and you can bring a certain amount of capital to bear, then use the marginal cost to determine whether your capital is sufficient to increase production to the desired level.

[If this is sufficient for you, then stop reading here.]

However, to see if this is a worthwhile change, you could calculate the marginal profit. The point at which marginal profit begins to decrease is sometimes called the point of diminishing returns, and even if you can afford to push beyond that level of production, you might not want to.
• I was not clear of why the marginal cost is not equal to the actual cost of producing an additional unit. Can someone explain it simply ? • The marginal cost is C'(x), which is the slope of the tangent line to the cost curve at a production of x units, while the actual cost of producing an additional unit more than x units is C(x+1) - C(x).
From the slope formula, the slope of the secant line between the points
(x, C(x)) and (x+1, C(x+1)) is just C(x+1) - C(x), and this is approximated by the derivative C'(x). Thus the marginal cost C'(x) approximates the cost per additional unit C(x+1) - C(x).
By the mean value theorem, there is a k with x < k < x+1 such that C(x+1) - C(x) = C'(k), so if C' is not changing too rapidly on (x , x+1), the difference between
C'(x) and C'(k) is small and the approximation is good.
• The math presented is straight forward, but who comes up with the formula in the first place, and how is it derived? • Presumably the formula would be worked out by cost accounting experts who seek to identify all the significant ways costs will change when we increase the quantity produced (efficiencies and inefficiencies of scale). For example, up to a point, the wage cost associated with a unit of production may go down as workers spend less idle time, but eventually wage cost may increase with the need to hire more workers or pay overtime. If you continue beyond the first calculus course to study differential equations, you may see problems like this from the reverse perspective: we're given an equation for marginal cost and asked to find the cost equation. I suspect that those problems are more likely to be representative of how these math concepts would be used in the real world.
• I still don't understand why Sal used C'(100) to approximate the value instead of C'(101). Can anyone help me out please? • He uses C'(100) because that tells us the change in cost of producing the next unit (the 101st unit) while C'(101) would tell us the change in cost of producing the 102nd unit. The reason for this is that the value of the derivative at C'(100) is the best approximation of the slope of the curve at x=100. Remember that the definition of slope tells us that for the next unit change in x (from the current location on the curve), we have a corresponding change in y. We want to know the change in y that will result from producing the 101st unit, which is why we approximate the slope from x =100. The slope at x = 101 (i.e. the value of C'(101)) would tell us the marginal cost of producing the 102nd unit.
• Wouldn't it make more sense to use C'((100+101)/2) = C'(100.5) as an estimate for the cost of producing the 101st unit, instead of C'(100)? That way you are taking the marginal cost halfway through your production of the 101st unit, and get a closer approximation to what the actual cost will be. It makes more sense to me than taking it at either end (C'(100) or C'(101)). • I am still confused about this marginal cost, this is a new term to me, completely new. If he wanted a more accurate approximation of the cost of the next gallon, why not use secant definition? When you get the the secant slope, you simply multiply this change in y over x by x which is 1 and the x's will cancel out and you will get distance of y. Am seeing this as triangles that need to be solved so my approach might be wrong here. From what I can tell Sal's method is much simpler since he's using a distance formula, he took a greater coordinate and subtracted a smaller one from it, to get the distance. So what am wondering is... the derivation is just an approximation right? Since the slope changes all of the time, we can't really determine the exact slope with simple derivation, we actually need to calculate distance with the distance formula here, it's just so happens that the slope is exactly the height of the y distance. We have to figure out the the increase of slope from 100 to 101 by ourselves without using the derivation of function? Since the slope will never hit coordinate of F(101) and therefore our slope will be incorrect? Can I essentially view true cost as all the slope of marginal cost + all of the differences of original marginal costs up to that next unit of gallon? Will this be 11.9+something+something....=12.02? This would also mean that the actual cost is a sum of marginal cost + all of the differences of marginal costs of increments of gallons up to the additional 1 gallon - original marginal cost of 100th galon? Could I define it in such way or to be more precise - a way of secant value, because it feels to me like this is the same way I would reach a secant value which should also be correct? • I think everything you said was pretty much correct, except that I don't think I understand this sentence: ` "This would also mean that the actual cost is a sum of marginal cost + all of the differences of marginal costs of increments of gallons up to the additional 1 gallon - original marginal cost of 100th galon?" ` But your main point is correct: the true "marginal cost" is defined as the true cost of producing one more gallon, so MC = C(101) - C(100), and this would equal the slope of the secant line, since the change in x is 1, so slope = Δy/Δx = Δy/1 = Δy. Indeed, using C'(100) is, as you say, simply an approximation. However, the only additional point I would make is that in any realistic situation where you are running a paint company, you will never have some polynomial that will give you the exact, true cost of producing x gallons of paint, for any value of x. So the function, C(x), is already an approximation, and the additional error of using the derivative instead of the secant is relatively small. This is more theoretical than practical.
• why should we approximate the value, when we actually have the function that will give us an absolutely precise one?

additionally, i think the question is misleading: it asks what the marginal cost of the 101st is, but wouldn't the value via linear approximation be achieved by taking the point slope form from the C(100) and then substracting C(100) from the linear approximation of C(100)? so we're really just taking the actual value of the original, not the marginal cost function.   