Optimizing Cost per Unit with Calculus

In summary: Just change the value to 1 instead. Homework Statement REFER TO IMAGEIn summary, the student is wondering how to find the average cost per unit in a given equation. They are also wondering if there is a relationship between the average cost per unit and the value of x.
  • #1
Elihu5991
33
0

Homework Statement


REFER TO IMAGE



Homework Equations


SEE ABOVE



The Attempt at a Solution


I haven't been able to start this question. I'm wondering how to find the 'average cost per unit'.
 

Attachments

  • Scan 2.jpeg
    Scan 2.jpeg
    7.7 KB · Views: 431
Physics news on Phys.org
  • #2
What is the definition of 'average'?
 
  • #3
I'm aware of what average means, though how would I find it with calculus? That, I'm not familiar with. I've done averages with actual numbers, not variables. Should I first differentiate it or do something else?
 
  • #4
If it costs C to build x units, what is the cost per unit?
 
  • #5
Your asking me what the question is asking in a simplified manner. I'm asking how to find the average. I'm unfamiliar with finding it with function specified.
 
  • #6
I'm just trying to get you to find the answer yourself. Put it in numbers if it helps.

If it costs $100 to build 10 units, what is the cost per unit?
If it costs $50 to build 1000 units, what is the cost per unit?

How do you solve those? Can you use your solution to find how to express generally (in terms of C and x) the average cost?
 
  • #7
Wouldn't there be a relationship, so I would first have to differentiate it? Thus, that function would then be the one for the average cost per unit?

But it's asking for an actual value for x, which is throwing me off. Would I need to follow common procedure like other optimisation questions?
 
  • #8
Elihu5991 said:

Homework Statement


REFER TO IMAGE



Homework Equations


SEE ABOVE



The Attempt at a Solution


I haven't been able to start this question. I'm wondering how to find the 'average cost per unit'.

For the given const function, what would be the total cost of x units? What would be the corresponding cost per unit? (Forget the word 'average'; just look at the cost per unit).
 
  • #9
Will it be x^3 or cubed x?, since the function is "wrapped" to ^3?
 
  • #10
Elihu5991 said:
Will it be x^3 or cubed x?, since the function is "wrapped" to ^3?

Which message/question are you responding to? Please use the correct "response" panel.
 
  • #11
Forget about the equation. We need to take one step back. Please start with answering post #6.
 
  • #12
@Ray

Sorry I'm on the iPad at the moment, so I can't just do that. I forgot to use the alternative.

I was replying to you.
 
  • #13
@DrClaude

I don't now how to find an actual value of x when there's only one real number in the equation: 10?
 
  • #14
Elihu5991 said:
@DrClaude

I don't now how to find an actual value of x when there's only one real number in the equation: 10?

What are you not seeing? If x = 2, what is the total cost? What is the cost per unit? If x = 10 what are the total cost and cost per unit? If x = 7 what are the total cost and the cost per unit? If x = 25 what are the total cost and the cost per unit?

If you can figure out the cost per unit when the value of x is given, just do the same thing when the value of x is not given.
 

1. What is optimisation using calculus?

Optimisation using calculus is a mathematical method used to find the maximum or minimum value of a function. It involves using derivatives and critical points to determine the optimal value of a variable.

2. What is the purpose of optimisation using calculus?

The purpose of optimisation using calculus is to find the optimal solution for a given problem. This could involve maximizing profits, minimizing costs, or finding the most efficient design for a system.

3. What are the steps involved in optimisation using calculus?

The steps involved in optimisation using calculus include: 1) defining the problem and identifying the variables, 2) setting up the objective function, 3) taking the derivative of the function, 4) setting the derivative equal to zero to find the critical points, and 5) evaluating the critical points to determine the optimal solution.

4. What are the key concepts in optimisation using calculus?

The key concepts in optimisation using calculus include critical points, which are points where the derivative of the function is equal to zero, and the first and second derivatives, which are used to determine whether a critical point is a maximum or minimum value.

5. What are some real-world applications of optimisation using calculus?

Optimisation using calculus has many real-world applications, such as in economics to determine the most profitable production levels, in engineering to design the most efficient structures, and in physics to find the optimal trajectory of a projectile. It is also used in data analysis and machine learning to optimize algorithms and predict outcomes.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
977
  • Calculus and Beyond Homework Help
Replies
1
Views
995
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
6K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
3K
Replies
2
Views
936
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
30
Views
3K
Back
Top