fx y x2 2xy for x y belongsto D Find maxmin of f on D Solu

f(x, y) = x^2 - 2xy for (x, y) belongsto D: Find max/min of f on D.

First we compute maximum and min on boundaries

On x axis ie y=0

f(x,0)=x^2

Taking minimum value =0 and maximum equal to 1

ON y axis ie x=0

f(0,y)=0

On x+y=1

f=x(x-2y)=x(x-2(1-x))=x(3x-2)=3x^2-2x

f\'=6x-2

f\'=0 at x=1/3

f\'\'=6

So, x=1/3 is point of minima

Minimum value on x+y=1 of f(x,y) is

(1-2)/3=-1/3

Now in the interior

f_x=2x-2y=0 gives x=y

f_y=-2x=0 gives x=0

x=0 critical point gives f=0 so we can ignore this point as we have already found ponts giving larger and smaller values than 0

So x=0 is neither point of max or min

Next critical point is x=y

f(x,y)_{x=y}=x^2-2x^2=-x^2

Smallest value is clearly 0 on origin

It decreases as x increases till it intersects x+y=1

Intersection of x+y=1 and y=x is

x=1/2

So minimum is -(1/2)^2=-1/4

-1/3<-1/4

HEnce point of minimum is

(1/3,2/3) which gives minimum f=-1/3

Point of maximum is (1,0) which givs maximum =1