minimization problem

hirnyk — Fri, 12 Mar 2010 22:51:13 Z

The infimum of the function ((w^2)/2)+((a/3)*x^3)+(2by)+((2a/3)*y^3) under the constraint w+x+2y=1 is equal to -infinity. Put w=0, then y=(1-x)/2 and the function a/3*x^3-a/12*x^3 approaches -infinity if x approaches - infinity. The objective function is not convex, therefore Maple has a problem with solving it. Your notation of the objective function makes difficulties and does not make a good impression.

I forgot to include the

sha — Fri, 12 Mar 2010 23:28:42 Z

I forgot to include the constraint that w, x, y and z are all between zero and one. 0=

0=

sha — Fri, 12 Mar 2010 23:32:08 Z

0=< w=< 1 0=< x=< 1 0=< y=< 1 0=< z=< 1

minimization problem

hirnyk — Sat, 13 Mar 2010 02:09:39 Z

Sorry, as yet Maple does not solve such symbolic problems with inequalities. Mathematica 7 at my comp cannot do it too. Try to solve this problem by hand, considering its restrictions on the polygons of the boundary of the set {(x,y,w): w+x+2*y=1,0 <=x<=1, 0<=y<=1,0<=w<=1 }. It is known that the minimum of a continuous function on a compact exists.

partial solution of minimization problem

hirnyk — Mon, 15 Mar 2010 21:50:07 Z

The problem can be reduced to two variables:

minimize (1-2*x-y)^2/2+a/3*x^3+2b*y+2a/3*y^3 under the three consraints 0<=x<=1, 0<=y<=1, 0<=1-2*x-y.

Next, we find the critical points of the objective function by Maple or Mathematica .There are at most four critical

points. We choose the local minimums among them and verify the belonging to the feasible region. Then we

consider the restrictions of the problem on the three closed intervals of the boundary. For example,

minimize (1-y)^2/2+2*b*y+2*a/3*y^3 if x=0, 0<=y<=1,

and find the solutions by Mathematica 7. This is one of a few cases when Mathematica is stronger , that is

life. In the case above, min=1/2 at x=0, y=0 for a>0 and b>0 . In two other cases the answer is too long to be

cited here. At last, we find the minimum of the values of the objective function at all these points. It is a long tale

because the depedance on a and b, but it can be done.

Critical points

hirnyk — Wed, 24 Mar 2010 18:59:06 Z

We find the coordinates of the critical points of a function of two variables by setting f'_x and f'_y

both equal to 0 (for more details see a textbook). In your case

>solve({-2 + 2*x + 2*y + a*x^2 = 0, -1 + 2*x + y + 2*b + 2*a*y^2 =0},{x,y});

MaplePrimes - answers and comments on Question, minimization problem

minimization problem

I forgot to include the

0=

minimization problem

partial solution of minimization problem

Critical points