Question

The temperature of a point $(x,y)$ in the plane is given by the expression $x^2 + y^2 - 4x + 2y$. what is the temperature of the coldest point in the plane?

Answer 1

`x^(2) + y^(2)-4x+2y`

can be written as follows:


`x^(2) -4x+ y^(2)+2y`


`(x^(2) -4x+4)-4+ (y^(2)+2y+1)-1`


`(x-2)^(2)+(y+1)^(2) -5`

since
`(x-2)^(2)` and
`(y+1)^(2)` are always positive, except for x=2 and y=-1, when they become 0,

the minimum value of this expression is when x=2, and y=-1, that is -5



Answer: -5