The temperature of a point (x,y) in the plane is given by the expression x^2 + y^2 - 4x + 2y - 12x + 14y + 26. What is the temperature of the coldest point in the plane?
Take the derivative with respect to x and set it = 0
2x -16 = 0
x = 8
Take the derivative with respect to y and set to 0
2y + 16 = 0
y = -8
The minimum is
(8)^2 + (-8)^2 - 4(8) + 2(-8) - 12(8) + 14(-8) + 26 = -102
+153 
