MHB What is the Optimal Production Level for Minimizing Average Cost per Unit?

Click For Summary
To minimize the average cost per unit, the average cost function is derived from the total cost function, given as C = 0.5x^2 + 15x + 5000. The average cost per unit is expressed as AC = C/x, which simplifies to AC = 0.5x + 15 + 5000/x. By taking the derivative of the average cost function and setting it to zero, the critical point is found at x = 100 units. This indicates that producing 100 units minimizes the average cost per unit.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
Average Cost. A manufacturer has determined that the total cost $C$ of operating a factory is
$C=0.5x^2+15x+5000$ where $x$ is the number of units produced.
At what level of production will the average cost per unit be minimized?
(The average cost per unit is $\frac{C}{x}$)

taking C' and setting to zero I got $-15$ but how is this going to give the ans which is
$x=100$ units.

thanks for help
 
Last edited:
Physics news on Phys.org
You want to first compute:

$\displaystyle \frac{C}{x}=0.5x+15+5000x^{-1}$ where $\displaystyle 0<x$

Now, equate the derivative of this to zero.
 
MarkFL said:
You want to first compute:

$\displaystyle \frac{C}{x}=0.5x+15+5000x^{-1}$ where $\displaystyle 0<x$

Now, equate the derivative of this to zero.

$(\frac{C}{x})'=\frac{1}{2}-\frac{5000}{x^2}$

$(\frac{C'}{x})'=0$ then x=100$ or $-100$ so $100$ is ans

r
 
Thread 'Problem with calculating projections of curl using rotation of contour'
Hello! I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem. Given: ##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0## ##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1## ##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0## I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...

Similar threads

Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
7
Views
4K
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K