Because it is the sum of three square terms, there is no maximum, only the minimum.

If f(x, y) is to be the minimum, the term (x- 1) 2 should be the minimum, that is, when x= 1, (x- 1) 2 = 0.

So,

The minimum value of f (x, y) = (x-1) 2+y 2+(y-2) 2 is the minimum value of y 2+(y-2) 2.

Let g (y) = y 2+(y-2) 2. If you have studied derivatives, you can take the test directly.

G`(y)=2y+2(y-2)=0, and the solution is y= 1.

Therefore, when the minimum value of f (x, y) = (x-1) 2+y 2+(y-2) 2 is x = 1 and y = 1:

f( 1, 1)=( 1- 1)^2+ 1^2+( 1-2)^2=2

If you haven't studied derivatives, you need the sum of squares.

g(y)=y^2+(y-2)^2=y^2+y^2-4y+4

=2y^2-4y+4

=2(y- 1)^2+2

If the minimum value of g(y) is guaranteed, the square term is 0, so the minimum value of g(y) is 2.

So the minimum value of f(x, y) is 2.