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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?

Jan 23, 2019
edited by Guest  Jan 23, 2019
edited by Guest  Jan 24, 2019

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$$T(x,y) = x^2 -4x + y^2 + 2y = \\ (x-2)^2 -4 + (y+1)^2 - 1 = \\ (x-2)^2 + (y+1)^2 - 5$$

$$\text{Clearly the first two terms are non-negative and are zero at }x=2,~y=-1\\ T(2,-1) = -5\\ \text{and this is as cold as it gets}$$

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Jan 24, 2019