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From: Virgil Stokes on 22 Apr 2010 03:31
A very simple question on imposing conditions/constraints.

I know that a, r, and h must be real and non-negative in the following
system of equations:

eq1=Pi*r^2+2*Pi*r*x==0;
eq2=2*Pi*r*h+(2*Pi*h+4*Pi*r)*x==0;
eq3=2*Pi*r^2+2*Pi*r*h-a==0;
sols=Solve[{eq1,eq2,eq3},{r,h,x}]

How can I impose conditions on this system such that only real solutions
are obtained, and r and h are non-negative?

Thank you,
--V


From: Andrzej Kozlowski on 23 Apr 2010 03:50
Just use Reduce instead of Solve:

sols =
Reduce[eq1 && eq2 && eq3 && r >= 0 && h >= 0, {r, h, x}, Reals,
Backsubstitution -> True]

(a == 0 && r == 0 && h == 0) || (a == 0 && r == 0 &&
h > 0 && x == 0) || (a > 0 &&
r == Sqrt[a]/Sqrt[6*Pi] &&
h == Sqrt[2/(3*Pi)]*Sqrt[a] &&
x == -(Sqrt[a]/(2*Sqrt[6*Pi])))


On 22 Apr 2010, at 16:31, Virgil Stokes wrote:

> A very simple question on imposing conditions/constraints.
>
> I know that a, r, and h must be real and non-negative in the following
> system of equations:
>
> eq1=Pi*r^2+2*Pi*r*x==0;
> eq2=2*Pi*r*h+(2*Pi*h+4*Pi*r)*x==0;
> eq3=2*Pi*r^2+2*Pi*r*h-a==0;
> sols=Solve[{eq1,eq2,eq3},{r,h,x}]
>
> How can I impose conditions on this system such that only real solutions
> are obtained, and r and h are non-negative?
>
> Thank you,
> --V
>
>


From: Bob Hanlon on 23 Apr 2010 03:49

Use Reduce

eq1 = Pi*r^2 + 2*Pi*r*x == 0;
eq2 = 2*Pi*r*h + (2*Pi*h + 4*Pi*r)*x == 0;
eq3 = 2*Pi*r^2 + 2*Pi*r*h - a == 0;

sols = Reduce[{eq1, eq2, eq3, r >= 0, h >= 0},
{r, h, x}, Reals] // ToRules

{r -> 1/Sqrt[6*Pi],
h -> Sqrt[2/(3*Pi)],
x -> -(h/(Sqrt[6*Pi]*
(h + Sqrt[2/(3*Pi)])))}


Bob Hanlon

---- Virgil Stokes <vs(a)it.uu.se> wrote:

=============
A very simple question on imposing conditions/constraints.

I know that a, r, and h must be real and non-negative in the following
system of equations:

eq1=Pi*r^2+2*Pi*r*x==0;
eq2=2*Pi*r*h+(2*Pi*h+4*Pi*r)*x==0;
eq3=2*Pi*r^2+2*Pi*r*h-a==0;
sols=Solve[{eq1,eq2,eq3},{r,h,x}]

How can I impose conditions on this system such that only real solutions
are obtained, and r and h are non-negative?

Thank you,
--V


From: DrMajorBob on 24 Apr 2010 04:00
eq1 = Pi*r^2 + 2*Pi*r*x == 0;
eq2 = 2*Pi*r*h + (2*Pi*h + 4*Pi*r)*x == 0;
eq3 = 2*Pi*r^2 + 2*Pi*r*h - a == 0;
sols = Reduce[{eq1, eq2, eq3, r >= 0, h >= 0}, {r, h, x}, Reals]

(a == 0 && r == 0 && (h == 0 || (h > 0 && x == 0))) || (a > 0 &&
r == Sqrt[a]/Sqrt[6 \[Pi]] && h == 2 r && x == -((h r)/(h + 2 r)))

Bobby

On Thu, 22 Apr 2010 02:31:16 -0500, Virgil Stokes <vs(a)it.uu.se> wrote:

> A very simple question on imposing conditions/constraints.
>
> I know that a, r, and h must be real and non-negative in the following
> system of equations:
>
> eq1=Pi*r^2+2*Pi*r*x==0;
> eq2=2*Pi*r*h+(2*Pi*h+4*Pi*r)*x==0;
> eq3=2*Pi*r^2+2*Pi*r*h-a==0;
> sols=Solve[{eq1,eq2,eq3},{r,h,x}]
>
> How can I impose conditions on this system such that only real solutions
> are obtained, and r and h are non-negative?
>
> Thank you,
> --V
>
>


--
DrMajorBob(a)yahoo.com

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