- #1
mathmari
Gold Member
MHB
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Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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)
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)