Negative minimum of the average variable cost function

Click For Summary

Discussion Overview

The discussion revolves around the question of whether the minimum of the average variable cost function can be negative, using a specific cost function as an example. Participants explore the implications of a negative minimum in the context of variable costs and pricing strategies in economics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents a cost function and derives the average variable cost, concluding that the minimum is negative.
  • Another participant agrees with the interpretation that the minimum of the average variable cost represents the smallest price needed to cover variable costs.
  • Some participants suggest that a negative minimum indicates that variable costs can be covered even if units are sold at a loss or for free, potentially reducing fixed costs.
  • There is a suggestion that discounts on raw materials could lead to scenarios where selling additional units, even at a loss, might be beneficial.

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of the minimum of the average variable cost function, but there is some uncertainty regarding the implications of a negative minimum and whether it guarantees coverage of variable costs in all scenarios.

Contextual Notes

The discussion does not resolve the implications of a negative minimum in all contexts, and assumptions about market conditions and pricing strategies remain unexamined.

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

Can the minimum of the average variable cost function be negative? (Wondering)

Suppose we have the cost function $K(x)=x^3-9x^2+11x+100$.
The variable cost function is then $K_v(x)=x^3-9x^2+11x$. The average variable cost is $k_v(x)=\frac{K_v(x)}{x}=\frac{x^3-9x^2+11x}{x}=x^2-9x+11$.
The first derivative is $k_v'(x)= 2x-9$. The root of $k_v'(x)$ is $x\frac{9}{2}$.
The second derivative is $k_v''(x)= 2$. We have that $k_v''\left (\frac{9}{2}\right )=2>0$.
The minimum of $k_v(x)$ is therefore at $x=\frac{9}{2}$ and the minimum is equal to $k_v\left (\frac{9}{2}\right )=-\frac{37}{4}$.

Is everything correct? (Wondering)

$\min k_v(x)$ is the smallest price that a company has to earn so that the variable costs are covered, right? In this case where the minimum is negative, what does it mean? (Wondering)
 
Physics news on Phys.org
mathmari said:
Hey! :o

Can the minimum of the average variable cost function be negative? (Wondering)

Suppose we have the cost function $K(x)=x^3-9x^2+11x+100$.
The variable cost function is then $K_v(x)=x^3-9x^2+11x$. The average variable cost is $k_v(x)=\frac{K_v(x)}{x}=\frac{x^3-9x^2+11x}{x}=x^2-9x+11$.
The first derivative is $k_v'(x)= 2x-9$. The root of $k_v'(x)$ is $x\frac{9}{2}$.
The second derivative is $k_v''(x)= 2$. We have that $k_v''\left (\frac{9}{2}\right )=2>0$.
The minimum of $k_v(x)$ is therefore at $x=\frac{9}{2}$ and the minimum is equal to $k_v\left (\frac{9}{2}\right )=-\frac{37}{4}$.

Is everything correct? (Wondering)

$\min k_v(x)$ is the smallest price that a company has to earn so that the variable costs are covered, right? In this case where the minimum is negative, what does it mean? (Wondering)

Hey mathmari! (Smile)

The average variable cost is the price it takes to produce 1 additional unit.
As long as this is above the market price, the company can survive in the short run.
So yes, '$\min k_v(x)$ is the smallest price that a company has to earn so that the variable costs are covered'.
And if that is negative, it means we have nothing to worry about in the short run. (Wink)
 
I like Serena said:
The average variable cost is the price it takes to produce 1 additional unit.
As long as this is above the market price, the company can survive in the short run.
So yes, '$\min k_v(x)$ is the smallest price that a company has to earn so that the variable costs are covered'.
And if that is negative, it means we have nothing to worry about in the short run. (Wink)
So, when $\min k_v(x)$ is negative, the variable costs will definitely be covered, right? (Wondering)
 
Last edited by a moderator:
mathmari said:
So, when $\min k_v(x)$ is negative, the variable costs will definitely be covered, right? (Wondering)

I believe so yes. Every additional unit that we sell - even if it means giving them away for free - will reduce the fixed costs.
It's something that could happen if we get a discount when buying more raw materials.
 
I like Serena said:
I believe so yes. Every additional unit that we sell - even if it means giving them away for free - will reduce the fixed costs.
It's something that could happen if we get a discount when buying more raw materials.

Ah ok. Thank you! (Smirk)
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
Replies
4
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K