SOLUTION: 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?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 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?      Log On


   

Question 1040894: 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 by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
T=x%5E2%2By%5E2-4x%2B2y
T=%28x%5E2-4x%29%2B%28y%5E2%2B2y%29
T=%28x%5E2-4x%2B4%29%2B%28y%5E2%2B2y%2B1%29-4-1
T=%28x-2%29%5E2%2B%28y%2B1%29%5E2-5
Since the two quadratic terms can only be non-negative, the minimum T occurs when x-2=0 and y%2B1=0. The temperature then would be,
T%5Bmin%5D=-5 at (2,-1)