Determine the global extreme values for fx y x2 y2 11 on

Determine the global extreme values for f(x, y) = (x^2 + y^2 + 1)^-1 on -1 lessthanorequalto x lessthanorequalto 1 -2 lessthanorequalto y lessthanorequalto 3.

Solution

3.

the function is

f(x,y)=1/(x^2+y^2+1)

g(x,y)=x^2+y^2+1 never vanishes for any real x,y

When , g takes minimum value f take maximum and vica versa

So we can analyse g(x,y) for points of extremum and hence get points of extremum of f(x,y)

x^2+y^2+1 takes global maximum value when

|x| and |y| are both maximum

Hence g(x,y) has global maximum at :x=+-1,y=3

Hence, f(x,y) has global minimum at :x=+-1 ,y=3 with value equal to

f(+-1,3)=1/(1+3^2+1)=1/11

x^2+y^2+1 takes global minimum value when

|x| and |y| are both minimum

Hence g(x,y) has global minimum at :x=0,y=0

Hence, f(x,y) has global maximum at :x=0 ,y=0 with value equal to

f(0,0)=1/(1+0+0)=1