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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?

 Jun 8, 2024
 #1
avatar+129399 
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Take the derivative  with respect to x and  set it = 0

2x -16  = 0

x = 8

 

Take the derivative with respect to y and  set to 0

2y + 16  = 0

y = -8

 

The minimum is

 

(8)^2 + (-8)^2 - 4(8) + 2(-8) - 12(8) + 14(-8) + 26 =   -102

 

cool cool cool

 Jun 8, 2024

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