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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?

Guest Nov 2, 2014

Best Answer 

 #2
avatar+92805 
+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
 #1
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0

please help me!!!!

Guest Nov 3, 2014
 #2
avatar+92805 
+10
Best Answer

 

 

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

 

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
 #4
avatar+87303 
0

Very nice, Melody........no Calculus required !!!!

 

 

CPhill  Nov 3, 2014
 #5
avatar+92805 
0

Thanks Chris  :))

Melody  Nov 3, 2014

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