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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?

Respuesta :

[tex] x^{2} + y^{2}-4x+2y [/tex]

can be written as follows:

[tex] x^{2} -4x+ y^{2}+2y[/tex]

[tex](x^{2} -4x+4)-4+ (y^{2}+2y+1)-1[/tex]

[tex] (x-2)^{2}+(y+1)^{2} -5[/tex]

since [tex](x-2)^{2}[/tex] and [tex](y+1)^{2}[/tex] are always positive, except for x=2 and y=-1, when they become 0,

the minimum value of this expression is when x=2, and y=-1, that is -5



Answer: -5

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