+1207 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?
+129852 x^2 + y^2 - 4x + 2y
x^2 - 4x + y^2 + 2y complete the square on x and y
Take 1/2 of 4 = 2 square this = 4 add and subtract it
Take 1/2 of 2 = 1 square this = 1 add and subtract it
(x^2 - 4x + 4 ) + ( y^2 + 2y + 1) - 4 - 1 factor the trinomials in the parentheses
(x - 2)^2 + (y + 1)^2 - 5
Note that (x - 2)^2 is minimized when x = 2 and ( y + 1)^2 is minimized when y = -1
So
(2 - 2)^2 + (-1 + 1)^2 - 5 =
0^2 + 0^2 - 5 =
-5 = the lowest temp
