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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 - 12x + 14y + 26. What is the temperature of the coldest point in the plane?

 Apr 4, 2024
 #1
avatar+128794 
+1

Take the derivative with respect to  and  set to 0

2x -16 = 0

x = 8

 

Take the derivative with respect to y and  set to 0

2y + 16 = 0 

y = -8                            

 

The coldest temp is

 

  (8)^2 + (-8)^2-16(8) + 16(-8) + 26  =  -102

 

 

cool cool cool

 Apr 5, 2024

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