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?
x^2 -14x + y^2 + 16y + 26
Take the derivative with respect to x and set to 0
2x - 14 = 0
2x = 14
x = 7
Take the derivative with respect to y and set to 0
2y + 16 = 0
2y = -16
y = -8
The min is (7)^2 - 14(7) + (-8)^2 + 16(-8) + 26 = -87
+400 
