Question 1195370: Suppose that the cost function for a product is given by C(x)=0.003x^3+5x+11,719. Find the production level (that is, the value of x) that will produce the minimum average cost per unit _
C(x).
The production level that produces the minimum average cost per unit is x=
(Round to the nearest whole number as needed.)
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
x = number of units produced
C(x) = cost function
C(x) = 0.003x^3+5x+11719
A(x) = average cost function = cost per unit
A(x) = (cost function)/(number of units produced)
A(x) = C(x)/x
A(x) = (0.003x^3+5x+11719)/x
A(x) = (0.003x^3)/x+(5x)/x+(11719/x)
A(x) = 0.003x^2+5+(11719/x)
Apply the derivative
A ' (x) = d/dx[ 0.003x^2+5+(11719/x) ]
A ' (x) = d/dx[ 0.003x^2 ]+d/dx[ 5 ]+d/dx[ 11719/x ]
A ' (x) = 0.006x + 0 - 11719/(x^2)
A ' (x) = 0.006x - 11719/(x^2)
Note: if y = 1/x, then dy/dx = -1/(x^2)
Set the derivative equal to zero and solve for x.
A ' (x) = 0
0.006x - 11719/(x^2) = 0
0.006x = 11719/(x^2)
0.006x*(x^2) = 11719
0.006x^3 = 11719
x^3 = 11719/0.006
x^3 = 1,953,166.66666667
x = cubeRoot(1,953,166.66666667)
x = (1,953,166.66666667)^(1/3)
x = 125.000888882569
This value is approximate.
Rounding to the nearest whole number gets to x = 125 which is the final answer.
Let's check a few values around x = 125
Plug in x = 123
A(x) = (0.003x^3+5x+11719)/x
A(123) = (0.003(123)^3+5(123)+11719)/(123)
A(123) = 145.663422764228
A(123) = 145.663
Plug in x = 124
A(x) = (0.003x^3+5x+11719)/x
A(124) = (0.003(124)^3+5(124)+11719)/(124)
A(124) = 145.636064516129
A(124) = 145.636
Plug in x = 125
A(x) = (0.003x^3+5x+11719)/x
A(125) = (0.003(125)^3+5(125)+11719)/(125)
A(125) = 145.627
Plug in x = 126
A(x) = (0.003x^3+5x+11719)/x
A(126) = (0.003(126)^3+5(126)+11719)/(126)
A(126) = 145.635936507937
A(126) = 145.636
Plug in x = 127
A(x) = (0.003x^3+5x+11719)/x
A(127) = (0.003(127)^3+5(127)+11719)/(127)
A(127) = 145.662590551181
A(127) = 145.663
Those results are approximate with the exception of 145.627 which is exact.
Here's a table to organize things:
| x | A(x) | | 123 | 145.663 | | 124 | 145.636 | | 125 | 145.627 | | 126 | 145.636 | | 127 | 145.663 |
Based on this small sample, we can see that x = 125 leads to A(x) being the smallest.
I'll let you try out other x values.
This is what the graph looks like. I used GeoGebra to make the graph.
If you aren't too familiar with calculus, then you can use the "minimize" function on your calculator to determine the lowest point. That point is approximately (125.0009, 145.627)
In GeoGebra, the function is called "min".
Another useful graphing tool is Desmos. The nice thing about this app is that you can simply click on the lowest point to have the coordinates show up.
You may need to click on the curve itself before clicking on the lowest point.
Here's a link to the interactive graph page.
https://www.desmos.com/calculator/qme1bkefgv
Answer: x = 125
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