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Help how to solve

 

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?

 Dec 19, 2023
 #1
avatar+126665 
+1

Simplify as

 

x^2 -16x +y^2 + 10y  + 26

 

Take the partial derivative with respesct to  x  and set to 0

2x -16  = 0      →   x =  8

Take the partial derivative with respect to y  and set to 0

2y + 10  = 0    →   x =  -5

 

The minimum   is   (8)^2 - 16(8) + (-5)^2 + 10(-5) + 26 =  -63

 

cool cool cool

 Dec 19, 2023

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