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?
PLEASE HELP!!!!
Please anyone??????
\(T(x,y) = x^2 -4x + y^2 + 2y = \\ (x-2)^2 -4 + (y+1)^2 - 1 = \\ (x-2)^2 + (y+1)^2 - 5\)
\(\text{Clearly the first two terms are non-negative and are zero at }x=2,~y=-1\\ T(2,-1) = -5\\ \text{and this is as cold as it gets}\)
