+0

The temperature of a point in the plane is given by the expression . What is the temperature of the coldest point in the pla

#2
+10

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

Now anything squared has to be  greater or equal to zero so T has to be greater to or equal to -5 degrees. This will occur when x=2 and y=-1

Nov 3, 2014

#2
+10

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

Now anything squared has to be  greater or equal to zero so T has to be greater to or equal to -5 degrees. This will occur when x=2 and y=-1

Melody Nov 3, 2014
#3
+5

http://web2.0calc.com/questions/the-temperature-of-a-point-nbsp-nbsp-in-the-plane-is-given-by-the-expression-nbsp-what-is-the-temperature-of-the-coldest-point-in-the-pla#r3

There is always more than one way to skin a cat Nov 3, 2014
#4
0

Very nice, Melody........no Calculus required !!!!   Nov 3, 2014
#5
0

Thanks Chris  :))

Nov 3, 2014