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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?

 Feb 14, 2024
 #1
avatar+128406 
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x^2 -14x  + y^2 + 16y + 26

 

Take the derivative with respect to x and  set to 0

2x - 14   = 0

2x = 14

x = 7

 

Take the derivative with respect to y and  set  to 0

2y + 16 = 0

2y = -16

y = -8

 

The min is    (7)^2  - 14(7) + (-8)^2 + 16(-8) + 26  =  -87

 

cool cool cool

 Feb 14, 2024

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