# Minimize the function: `C=(16pic)/3 r^2 +400cr^-1`

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### 1 Answer

The request of the problem is vague since it does not specifies if the function is a function of one variable r or c, or a function of two variables, c and r.

Considering the given function a function of one variable r, you should be able to find the minimum value of the function C, solving the equation `C'(r) = 0` , such that:

`C'(r) = ((16pi*c*r^2)/3)' + (400c/r)'`

Considering c as a constant yields:

`C'(r) = (32pi*c*r)/3 - 400c/(r^2)`

Solving the equation `C'(r) = 0` yields:

`(32pi*c*r)/3 - 400c/(r^2) = 0 => 32pi*c*r^3 - 1200c = 0`

Factoring out `4c` yields:

`4c(8pi*r^3 - 300) = 0 => {(4c!=0),(8pi*r^3 - 300 = 0):}`

`8pi*r^3 - 300 = 0 => r = root(3)(300/8pi) => r = (1/2)root(3)(300/pi)`

**Hence, the function `C(r)` reaches its minimum at **`r = (1/2)root(3)(300/pi).`

Considering the given function a function of one variable c, you should be able to find the minimum value of the function C, solving the equation C'(c) = 0, such that:

`C'(c) = ((16pi*c*r^2)/3)' + (400c/r)'`

`C'(c) = (16pi*r^2)/3 + 400/r != 0` for any value of c (r is considered as a constant)

If you consider the function C is produced by two inputs c and r, then the condition to evaluate the minimum of two variable function `C(r,c)` is: ` (delC(r,c))/(del c) = 0 and (delC(r,c))/(del r) = 0` .