SOLUTION: Need to find the graph of below and the set of admissible points.Also how do we get to know whether it is bounded or not? min f(x,y)=(x^2 + y^2) subject to xy>=4 Need to kno

Algebra ->  Linear-equations -> SOLUTION: Need to find the graph of below and the set of admissible points.Also how do we get to know whether it is bounded or not? min f(x,y)=(x^2 + y^2) subject to xy>=4 Need to kno      Log On


   



Question 1102452: Need to find the graph of below and the set of admissible points.Also how do we get to know whether it is bounded or not?
min f(x,y)=(x^2 + y^2) subject to xy>=4
Need to know the process for this.I have looked at graph of xy=4,but don't know anything after that!

Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


I'm guessing by the "graph of below" you mean the graph of xy>=4. But you said you did that.... And once you have that graph, it is easy to find the set of admissible points, and to see whether the graph is bounded or not.

So I'm not sure why you are asking those questions.

Here is a graph of xy=4:

graph%28400%2C400%2C-8%2C8%2C-8%2C8%2Cy=4%2Fx%29

The points that satisfy the constraint xy>=4 (what I assume you are calling the admissible points) are all the points that are either on that graph or above the graph in the first quadrant, or on that graph or below the graph in the third quadrant.

Clearly that set of points is unbounded.

As for minimizing the value of f(x,y)=x^2+y^2 subject to the constraint xy>=4, notice that the graph of x^2+y^2=r^2 is a circle centered at the origin and having radius r.

So in effect we are looking for the smallest circle with center at the origin that touches the graph of xy=4. The symmetry of the graph of xy=4 tells us that the smallest circle with center at the origin that touches the graph of xy=4 will touch it at the points (2,2) and (-2,-2).

And at each of those two points, the value of f(x,y)=x^2+y^2 is 2^2+2^2=4+4=8.

So, with my interpretation of your question, the minimum value of f(x,y)=x^2+y^2, subject to the constraint xy>=4, is 8.

Answer by ikleyn(52890) About Me  (Show Source):
You can put this solution on YOUR website!
.
After reading your post I still have no clear idea what really and exactly do you want !?

It is not the way to write/to present a mathematical request to the forum in such a form.

There is one simple rule which may help you in such deals and in the life:


      --------------------------------------------------------
      ASK ONE AND ONLY ONE QUESTION at a time in your written request.
      --------------------------------------------------------


Now, if you want to find min f(x,y) = (x^2 + y^2) subject to xy >= 4,
then below is very helpful wording visualization/description:

     the function f(x,y) = (x^2 + y^2)  represents a paraboloid in 3D space  R%5E3.

     It is a rotational paraboloid with the axis "Z" as the axis of rotation.


     Your task is to find its minimum over the domain xy >= 4 in the coordinate plane  (x,y)  (coordinate plane  Z = 0).


         In 3D,  the domain xy >= 4  is the cylindrical unbounded (in z-coordinate) 3D volume/body.

         When I say "cylindrical (in z-coordinate) 3D volume/body", I mean that this volume/body has vertical generating line 
         over the curve/the hyperbola  xy = 4  in  QI  and  QIII  quadrants in the plane Z = 0.


     If you imagine it MENTALLY  in your head, it will become clear to you, that the minimum you are looking for 

     is over the points  (2,2)  and  (-2,-2)  of the (x,y) plane.

     Exactly where the circle x^2 + y^2 = 8 or the radius 2%2Asqrt%282%29  centered at the origin of the coordinate plane 

     touches the hyperbola xy = 4 in the plane (x,y).


     And this minimum is equal to (2^2 + 2^2) = 4 + 4 = 8.