Question 1040894
{{{T=x^2+y^2-4x+2y}}}
{{{T=(x^2-4x)+(y^2+2y)}}}
{{{T=(x^2-4x+4)+(y^2+2y+1)-4-1}}}
{{{T=(x-2)^2+(y+1)^2-5}}}
Since the two quadratic terms can only be non-negative, the minimum T occurs when {{{x-2=0}}} and {{{y+1=0}}}. The temperature then would be,
{{{T[min]=-5}}} at ({{{2}}},{{{-1}}})