MHB Negative minimum of the average variable cost function

[mathmari]

Hey!

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)

 

[I like Serena]

Hey!

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)

 

[mathmari]

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)

 

[I like Serena]

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.

 

[mathmari]

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)