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