You are right. As I said I did change the conditions a bit. I did not use this condition:

h[w]>= 37 when 20<w<2000

Regarding your second point:

I think your solution should be something like a= 0.02 and b= 1.235 or a posible range of values that satisfy the conditions (1), (2) and (3)

I don't think so. As far as I understood the function $h$ would depend on these parameters. That means that the solutions will in general depend on the parameters and $w$, which is exactly what they do. The solution that Mathematica finds indeed does effectively give intervals for the parameters $a$ and $b$, but these intervals depend on $w$, too.

Cheers,

M.

The thing is: it is really easy! I don't think that you'll get it easier than in Mathematica...

But then, every language has some kind of learning curve and requires a bit of dedication to learn.

M.

Hi,

if I haven't made a mistake the result for your particular function and conditions is False, which seems to say that there is no solution. If I modify your conditions a tiny little bit I do get a result:

ClearAll["Global`*"];
h[x_] := a*(x^2) - 10*b + 2;
Reduce[{h[w] <= 42 && h[w] >= 38 && h[w] >= 37 && - 20 <= w <= 2000 && h[w] > 0 && w <= 30007 h[w] >= 1 && w >= 1}, {w, a, b}]

gives:

a \[Element] Reals && 1 <= w <= 2000 && 1/10 (-40 + a w^2) <= b <= 1/10 (-36 + a w^2)

Note, that I had to add $a$ and $b$ to the list of variables to solve for.

Cheers,

M.

Hi Sebastian,

there are some problems with the function that you use.

Reduce[{h[wo)] <= 42, h[wo] >= 38, h[w] >= 37 && 20 <= w <= 2000, h[w] < 0 && w <= 3000, h[w] >= 1 && w >= 1}, {w}]

For example the parenthesis do not match; the argument of the first function is enclosed in a square and on round bracket; in fact the round bracket needs to be deleted.

Reduce[{h[wo] <= 42, h[wo] >= 38, h[w] >= 37 && 20 <= w <= 2000, h[w] < 0 && w <= 3000, h[w] >= 1 && w >= 1}, {w}]

Then the arguments of some $h$ are $wo$ instead of $w$.

Reduce[{h[w] <= 42, h[w] >= 38, h[w] >= 37 && 20 <= w <= 2000, h[w] < 0 && w <= 3000, h[w] >= 1 && w >= 1}, {w}]

Then I concatenate all conditions with a &&.

Reduce[{h[w] <= 42 && h[w] >= 38 && h[w] >= 37 && 20 <= w <= 2000 && h[w] < 0 && w <= 3000&& h[w] >= 1 && w >= 1}, {w}]

Now, the result will depend on the function h. For example I define:

h[x_] := x^2 - 10

With that and your inequalities the set of solutions is empty. So I modified the conditions a little bit like so:

Reduce[{h[w] <= 42 && h[w] >= 38 && h[w] >= 37 && -20 <= w <= 2000 && h[w] > 0 && w <= 30007 h[w] >= 1 && w >= 1}, w]

This results in

4 Sqrt[3] <= w <= 2 Sqrt[13]

Cheers,

M.