# minimize function subject to constraint

• Oct 14th 2012, 03:23 AM
matmman
minimize function subject to constraint
could you help to minimize function F = x^2+y^2 subject to constraint:

1-x<0
2-0.5x-y<=0
x+y-4<0

and is there difference in solving the problem if we say insted of < say <=

• Oct 15th 2012, 03:01 AM
matmman
Re: minimize function subject to constraint
please any body can help me to solve this proplem any information
• Oct 15th 2012, 04:07 AM
BobP
Re: minimize function subject to constraint
Sketch the area associated with the constraints.

1. \$\displaystyle 1-x<0\$ (or \$\displaystyle x>1\$), gives you an area to the right of the vertical \$\displaystyle x=1.\$

Do the same fo the other two constraints and you should come up with a triangular region.

\$\displaystyle x^{2}+y^{2}\$ is the square of the distance of a point from the origin. It's minimum value will therefore be the square of the distance from the origin to the point in the region which is closest to the origin.
• Oct 15th 2012, 08:59 PM
matmman
Re: minimize function subject to constraint

i draw the reigon as you mension i get trainangle as shown in the attachment and if we draw line from origin to the point which is the nearset to the reigeion we get the function will be minimize at x=1,y=1.5...but
note that is true if x>=1 .However , here in the question x> 1 not x>=1 so that x will never equal 1

is there differeance in the solution if x>1 or x>=1 or there are same
• Oct 15th 2012, 09:25 PM
MaxJasper
Re: minimize function subject to constraint
x>1 as well as x>=1 both result in converging min(x^2+y^2)->3.25 which is a circle through {x,y}={1, 1.5}...so the shape of the region is useless here because its closest POINT to the origin is picked to minimize the function.
• Oct 15th 2012, 09:32 PM
matmman
Re: minimize function subject to constraint
Thanks for all

this mean the point (1,1.5) is the right answer