Kline Calculus Problems - Simple Derivatives and Marginal Cost

In summary: I don't have a background in economics, but I've taken a few courses on it.)In summary, the conversation discusses the concept of marginal cost in economics, which is the rate of change of total cost with respect to the number of units produced. The derivative of the cost function represents the marginal cost, and a negative derivative indicates that the cost decreases as more units are produced. The marginal cost at a specific value of x can be found by taking the derivative of the cost function and plugging in the value of x. The cost of the 16th unit is not necessarily equal to the marginal cost at x = 15. Furthermore, the marginal cost does not always increase with x in realistic situations
  • #1
ghostskwid
8
0
I had questions on 2 Problems in the Text:

1. The total cost C of producing x units of some item is a function of x. Economists use the term marginal cost for the rate of change of C with respect to x. Suppose that:

C = 5x^2 + 15x + 200

What is the marginal cost when x = 15? Would this marginal cost be the cost of the 16th unit?

(((I understand how to take the derivative and find dC/dx. However, I am unsure as to why this represents to cost of the 16th unit. Also, can someone explain to me in simple terms what the derivative of C represents?)))

2. Using the definition of marginal cost in the preceding exercise, suppose that the cost C of producing x units of a toy is C = 3x^2 - 4x + 5. What is the marginal cost at any value of x? Would the marginal cost necessarily increase with x in any realistic situation?

(((Why doesn't the marginal cost always increase with x in any realistic situation?)))


Thanks!
 
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  • #2
ghostskwid said:
I had questions on 2 Problems in the Text:

1. The total cost C of producing x units of some item is a function of x. Economists use the term marginal cost for the rate of change of C with respect to x. Suppose that:

C = 5x^2 + 15x + 200

What is the marginal cost when x = 15? Would this marginal cost be the cost of the 16th unit?

(((I understand how to take the derivative and find dC/dx. However, I am unsure as to why this represents to cost of the 16th unit. Also, can someone explain to me in simple terms what the derivative of C represents?)))

2. Using the definition of marginal cost in the preceding exercise, suppose that the cost C of producing x units of a toy is C = 3x^2 - 4x + 5. What is the marginal cost at any value of x? Would the marginal cost necessarily increase with x in any realistic situation?

(((Why doesn't the marginal cost always increase with x in any realistic situation?)))


Thanks!

Hey ghostskwid and welcome to the forums.

The derivative means the instantaneous rate of change of something with respect to another. In this case it represents how C changes with x at that point. Think of looking at the slope of the function C between a point x and x + dx and what happens is that dx gets smaller and smaller to goes to zero but isn't zero! It's a weird thing to understand but that's the best way to describe it.

Basically in this context if the slope is increasing then the cost is increasing for every x which means there will be a relative increase in cost to produce more stuff and if it decreases then it will cost relatively less.

The thing that businesses want to do is create more things at the cheapest possible rate which means that if we have any dC/dx where it is negative then this means that the businesses can create or produce more things without having to spend as much for each new piece of stuff (in other words it's less per unit to produce more stuff than it is to produce the existing stuff).

Think about a factory creates say a lot of cars or something on an assembly line. They have to pay for running the assembly line, wages, and all that stuff as well as for the materials but once they produce enough to cover things like wages and operating the factory, then it won't cost them as much to produce anything more and this is what businesses with factories want because they will sell their stuff at the same price usually which means they make a lot more profit when they make more stuff if the dC/dx is negative.

By finding the turning point where dC/dx is zero at a minimum, this says for the business what's the best amount to produce to maximize profit in one sense.

This should help you think about the second question.
 
  • #3
Thanks that helps.

Why is the marginal cost at 15 the actual cost of unit 16?

Also in the second problem could the marginal cost ever be negative?
 
  • #4
ghostskwid said:
I had questions on 2 Problems in the Text:

1. The total cost C of producing x units of some item is a function of x. Economists use the term marginal cost for the rate of change of C with respect to x. Suppose that:

C = 5x^2 + 15x + 200

What is the marginal cost when x = 15? Would this marginal cost be the cost of the 16th unit?

(((I understand how to take the derivative and find dC/dx. However, I am unsure as to why this represents to cost of the 16th unit. Also, can someone explain to me in simple terms what the derivative of C represents?)))

2. Using the definition of marginal cost in the preceding exercise, suppose that the cost C of producing x units of a toy is C = 3x^2 - 4x + 5. What is the marginal cost at any value of x? Would the marginal cost necessarily increase with x in any realistic situation?

(((Why doesn't the marginal cost always increase with x in any realistic situation?)))


Thanks!



It seems to be that (1) asks the following: the marginal cost at x = 15 is [itex]C'(15)=10\cdot 15+15=165[/itex] .

On the other hand, the cost of the 16th product seems to be C(16)-C(15) = the cost of making 16 items minus the cost of making 15 items, and this gives 170, so no.

DonAntonio

Disclaimer: the person that wrote the above answer is a pure mathematician and thus his messing with mathematicial economics and/or financial stuff must be taken with due care.
 
  • #5
DonAntonio said:
Disclaimer: the person that wrote the above answer is a pure mathematician and thus his messing with mathematicial economics and/or financial stuff must be taken with due care.

I'm not a pure mathematician: my background is in computer programming and I will graduate this year with a double major in statistics and applied mathematics.
 
  • #6
chiro, I believe Don Antonio was referring to himself.

ghostskwid, yes, the marginal cost of the 16th item is just the cost of that item. Since your function, C(x), gives the total cost of manufacturing x items, the marginal cost of the 16th item is C(16)- C(15).

Since you refer to both "simple derivatives" and "marginal cost" in the title of this thread it might be good to point out that the marginal cost of the "x" item is
[tex]C(x+ 1)- C(x)= \lim_{h\to 1}\frac{C(x+h)- C(x)}{h}[/tex]
while the derivative is
[tex]\lim_{h\to 0}\frac{C(x+h)- C(x)}{h}[/tex]

Of course, the "h" going to 1 in the denominator raises much less theoretical issues than it going to 0!
 
  • #7
HallsofIvy said:
chiro, I believe Don Antonio was referring to himself.

ghostskwid, yes, the marginal cost of the 16th item is just the cost of that item. Since your function, C(x), gives the total cost of manufacturing x items, the marginal cost of the 16th item is C(16)- C(15).

Since you refer to both "simple derivatives" and "marginal cost" in the title of this thread it might be good to point out that the marginal cost of the "x" item is
[tex]C(x+ 1)- C(x)= \lim_{h\to 1}\frac{C(x+h)- C(x)}{h}[/tex]
while the derivative is
[tex]\lim_{h\to 0}\frac{C(x+h)- C(x)}{h}[/tex]

Of course, the "h" going to 1 in the denominator raises much less theoretical issues than it going to 0!

Thankyou HallsOfIvy for that. Hopefully the OP will reply so that everything gets cleared up.
 
  • #8
I get 165 when calculating the marginal cost at $15 and 170 when calculating the cost of the 16th product ((C(16) - C (15))). Kline reports the answer as yes it will be.

Could you explain your tex block. How do they differ? I thought the marginal cost was simply the derivative of the function. Thanks
 

1. What is calculus and why is it important?

Calculus is a branch of mathematics that deals with the study of change and is used to analyze and model real-life situations. It is important because it allows us to solve problems and make predictions in various fields such as physics, engineering, economics, and more.

2. What are derivatives and how are they used in calculus?

Derivatives are a fundamental concept in calculus that measures the rate of change of a function. They are used to find the slope of a curve at a specific point, which can help us determine the instantaneous rate of change, the maximum or minimum points, and the concavity of a function.

3. How do you find the derivative of a function?

To find the derivative of a function, we use the rules of differentiation, such as the power rule, product rule, quotient rule, and chain rule. These rules allow us to find the derivative of a function by manipulating its algebraic expression.

4. What is marginal cost and how is it related to derivatives?

Marginal cost is the change in total cost when one additional unit is produced. It is related to derivatives because it is calculated by finding the derivative of the total cost function with respect to the quantity produced. This allows us to determine the cost of producing each additional unit.

5. How can calculus be applied to real-life problems, specifically in economics?

Calculus is widely used in economics to analyze and optimize production, consumption, and pricing decisions. For example, it can be used to determine the optimal production level that will maximize profits for a company, or to calculate the marginal revenue and marginal cost to make pricing decisions.

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