Calculus Problem: Analyzing Cost of Silver Mine Production

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In summary, the cost of producing silver from a silver mine is c = f(x) dollars. The derivative of f(x) is the change in total cost / change in quantity. The statement f ' (600) = 25 means that plugging in a x value of 600 yields a y value of 25. The cost of producing silver will decrease in the long run as economies of scale kick in.
  • #1
Physics1
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The cost of producing x ounces of silver from a silver mine is c = f(x) dollars.

a) What does the derivative of f(x) mean?

Is the derivative the instantaneous cost?

Is f(x) the average cost to get silver out of the ground or the change in cost over change in ounces?


b) What does the statement f ' (600) = 25 mean?

Does this mean that plugging in a x value 600 yields a y value (amount of instantaneous dollars) of 25?

c) Do you think f ' (x) will increase/decrease in short run? Long run? Explain.

I just would say if we go back to silver standard, it would decrease (economies of scale) in both long run and short run.
 
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  • #2
This isn't generic calculus; it is economics. What have you learned so far in Econ?
 
  • #3
EnumaElish said:
This isn't generic calculus; it is economics. What have you learned so far in Econ?

lol. No it's not. I just have to come up with a function (wiggly stuff, exponential, etc.) and convince my teacher that it's ok. I have to come up with something to do part c.
 
  • #5
EnumaElish said:
That's the problem -- "short run" and "long run" are Econ concepts with precise definitions in terms of costs.

It really said short term but I just put short run. It's basically the same thing. I can't do c without a graph.

What about a and b?
 
  • #8
So, you looked at the Wiki link, and decided it isn't useful?
 
  • #9
EnumaElish said:
So, you looked at the Wiki link, and decided it isn't useful?

Well, c is irrelevant to what a and b are asking. It's a math class (not business calc). She just wants us to come up with a graph to convince her on part c. Most likely she won't be picky.
 
  • #10
I suggested the Wiki article for parts a and b.
 
  • #11
EnumaElish said:
I suggested the Wiki article for parts a and b.

Oh, so the derivative of f(x) is the change in total cost / change in quantity.
 
  • #12
Correct.
 
  • #13
EnumaElish said:
Correct.

Let's say there's a function y = f(x). and a tangent line at x = 3. Is that the instantaneous unit (whatever you're measuring)?
 
  • #14
Physics1 said:
Let's say there's a function y = f(x). and a tangent line at x = 3. Is that the instantaneous unit (whatever you're measuring)?
The slope of the tangent line is the change in y resulting from a very small change in x (from 3 to 3 + epsilon), divided by the change in x.
 
  • #15
EnumaElish said:
The slope of the tangent line is the change in y resulting from a very small change in x (from 3 to 3 + epsilon), divided by the change in x.

What about the point of the tangent line that's touching the function?

It passes through points (-2, 3) and (4, -1)

I got y = -(2/3)x + 13/3 for the tangent line using basic algebra.

It says plug in f ' (3) and f(3)

How would I know the original equation y = f(x)?
 
  • #16
Physics1 said:
The cost of producing x ounces of silver from a silver mine is c = f(x) dollars.

a) What does the derivative of f(x) mean?

Is the derivative the instantaneous cost?

Is f(x) the average cost to get silver out of the ground or the change in cost over change in ounces?


b) What does the statement f ' (600) = 25 mean?

Does this mean that plugging in a x value 600 yields a y value (amount of instantaneous dollars) of 25?

c) Do you think f ' (x) will increase/decrease in short run? Long run? Explain.

I just would say if we go back to silver standard, it would decrease (economies of scale) in both long run and short run.

Do you not understand that the derivative of f is the "instantaneous rate of change" of f? If so then all of these should be easy. Certainly "Does this mean that plugging in a x value 600 (in f '(x)) yields a y value (amount of instantaneous dollars) of 25?" is not true: that's what f(x) gives you. f '(x) is not just another name for f(x)!
 
  • #17
Physics1 said:
What about the point of the tangent line that's touching the function?
Ordinarily, a tangent line "barely touches" the function at the point where f' is evaluated (in your example, x = 3). That's different from intersecting the function, e.g. the way the normal line (perpendicular line) does.
It passes through points (-2, 3) and (4, -1)
I am going to assume that means the tangent line intersects f(x) at points p = (-2, 3) and q = (4, -1). If that's what you mean, it doesn't mean anything special. A tangent line might intersect the function at an arbitrary number of points. What is special about the tangent line is that has the same slope as the function at the point where f' is evaluated. For any other point, the tangent line is an arbitrary graphical object.
I got y = -(2/3)x + 13/3 for the tangent line using basic algebra.

It says plug in f ' (3) and f(3)

How would I know the original equation y = f(x)?
How did you derive the tangent line if you didn't know the function f?
 
  • #18
EnumaElish said:
Ordinarily, a tangent line "barely touches" the function at the point where f' is evaluated (in your example, x = 3). That's different from intersecting the function, e.g. the way the normal line (perpendicular line) does.I am going to assume that means the tangent line intersects f(x) at points p = (-2, 3) and q = (4, -1). If that's what you mean, it doesn't mean anything special. A tangent line might intersect the function at an arbitrary number of points. What is special about the tangent line is that has the same slope as the function at the point where f' is evaluated. For any other point, the tangent line is an arbitrary graphical object.How did you derive the tangent line if you didn't know the function f?

Well, actually I'm getting y = -2/3x + (5/3) for the tangent line now. I got it from those two points. Maybe f(x) is just a straight line and those 2 points can be used to find the whole line (slope).
 

1. What is calculus and how is it used to analyze the cost of silver mine production?

Calculus is a branch of mathematics that deals with the study of rates of change and accumulation. In the context of analyzing the cost of silver mine production, calculus is used to determine the optimal amount of silver to mine and sell in order to maximize profits. It involves using derivatives to find the marginal cost of production and integrals to find the total cost over a given time period.

2. What are the key factors that affect the cost of silver mine production?

The cost of silver mine production is affected by various factors such as labor costs, equipment and machinery costs, energy costs, transportation costs, and the quality and quantity of silver ore. Other factors that may also impact the cost include government regulations, market demand, and the location of the mine.

3. How can calculus help in making production decisions for a silver mine?

Calculus is used to analyze and optimize production decisions for a silver mine by determining the optimal level of production that will result in maximum profits. By using calculus, one can find the rate at which costs are changing and make informed decisions on how much silver to mine and sell in order to minimize costs and maximize profits.

4. What are the limitations of using calculus to analyze the cost of silver mine production?

While calculus is a powerful tool for analyzing the cost of silver mine production, it does have some limitations. One limitation is that it assumes a linear relationship between costs and production, which may not always be the case. Additionally, it does not take into account external factors such as fluctuations in market demand or unexpected changes in production costs.

5. Can calculus be applied to other industries besides silver mining?

Yes, calculus can be applied to various industries to analyze and optimize production decisions. It has applications in fields such as economics, engineering, physics, and even biology. Any situation that involves rates of change and accumulation can benefit from the use of calculus to make informed decisions and achieve optimal outcomes.

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