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?

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Question 1026930: 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?

Found 3 solutions by robertb, richard1234, ikleyn:
Answer by robertb(5830)   (Show Source):
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Let .
Finding the critical points of F:
==> x = 2, and
==> y = -1.
==> critical point is (2,-1).
Also, , , and
Implement the 2nd derivative test for two variables:

==> There is local min at (2,-1). Since it is the only critical point in the domain of the function (which is infinite open), it is also an absolute minimum.
The temperature of the coldest point is thus

Answer by richard1234(7193)   (Show Source):
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Complete the square to get:

It is clear from this form that the minimum temperature occurs when x=2, y=-1, and the minimum temperature is -5.

Answer by ikleyn(52778)   (Show Source):
You can put this solution on YOUR website!
.
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?
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It was just solved before in
https://www.algebra.com/tutors/your-answers.mpl?userid=ikleyn

and coincides with the two newest solutions.