# Minimizing with constraints

I would like to minimize the following function with respect to x:

f[x_] := b/Sqrt[x] + ( 4.5 c x^6 Log[x] )^(1/3)


where b and c are constants. That is, I would like to find x, as a function of b and c, that minimizes f.

I've tried using Minimize[f, x]. I have also tried solving df/dx = 0 to find the turning point (and therefore finding the minimum). However, neither of these approaches work.

From what I can tell, one problem is that f doesn't have a minimum for all values of b and c. However, for the ranges of values of b and c that I'm interested in, f does have a minumum. But I can't work out how to put restictions on b and c.. Specifically, I want:

0< c < 1 && b > 0, && x > 1


How can I minimze f with respect to x, with the above constraints? I've worked through Mathemica's help pages, but can't find a working solution to this.

If this problem above isn't solvable, then a simpler but still useful problem would be to minimize f for a given value of b. For example, take b = 10 then find x as a function of c that minimizes f. Again, the constraints 0 < c <1 and x > 1 hold.

• Start by defining f as f[x_, b_, c_] := ... so the dependence on $a,b,c$ is made evident. You could then try Minimize[{f[x, b, c], 0 < c < 1, b > 0, x > 1}, {x, b, c}]. Can you also elaborate on how you know that the function has a minimum under those conditions? – MarcoB Jul 5 '17 at 18:11
• As a simple example plot would suggest, this function has no global minimum in the case b=10 and c=1/2 because the boundary point 1 is not feasible. – Henrik Schumacher Jul 5 '17 at 21:19
• I was a bit vague saying that I know that f does have a minimum, but here's my thinking: first I should say that c=1/n, and my whole goal in this is to find how large n has to be to make f close to zero. Now fix b>0 and n>0. For x->zero the left hand term, b/sqrt(x), tends to infinity. But for x->infinity the right hand term also tends to infinity. But somewhere in the middle, where sqrt(x)>b and x^6<n, both terms are small. Hence there must be intermediate values of x that minimise f! – Paul K Jul 7 '17 at 15:53

There seems to be no way to find the minimum analyticaly.

A way out is to "FindRoot" of the x where the first derivative is zero for given b an c.

f[x_, b_, c_] = b/Sqrt[x] + (45/10 c x^6 Log[x])^(1/3)

dequ[x_, b_, c_] =
FullSimplify[D[f[x, b, c], x] == 0, {0 < c < 1, b > 0, x > 1}]

(*    6^(2/3) c^(1/3) x^(5/2) (1 + 6 Log[x]) == 3 b Log[x]^(2/3)     *)


Find the x value of the minimum for given b and c, ignoring error messages.

fr21[b_, c_] := (xf = x /. FindRoot[dequ[x, b, c], {x, 2, 1, 100}]) //
Quiet

Plot3D[fr21[b, c], {b, 0, 12}, {c, 10^-3, 1}] Control it, showing the minimum for given b,c

pl[b_, c_] :=
Plot[f[x, b, c], {x, 1, 12},
Epilog -> {PointSize[Large], Red,
Point[{fr21[b, c], f[fr21[b, c], b, c]}]}]

Manipulate[pl[b, c], {{b, 10}, 0, 20}, {{c, 1/2}, 0, 1}] • Thanks for this Akku, this is very helpful. – Paul K Jul 7 '17 at 16:02