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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?

 Jul 28, 2019
 #1
avatar+106533 
+1

x^2 + y^2 - 4x + 2y

 

x^2 - 4x  +  y^2 + 2y    complete the square on x and y

 

Take 1/2 of 4  = 2   square this = 4   add and subtract it

Take 1/2 of 2  = 1    square this  = 1    add and subtract it

 

(x^2 - 4x + 4 ) + (  y^2 + 2y + 1)   -  4 -  1        factor the trinomials in the parentheses

 

(x - 2)^2 + (y + 1)^2  - 5

 

Note that  (x - 2)^2   is minimized when x  = 2     and ( y + 1)^2  is minimized when y = -1

 

So

 

(2 - 2)^2  + (-1 + 1)^2  - 5  =

 

0^2  + 0^2  - 5   =

 

-5   =   the lowest temp

 

 

cool cool cool

 Jul 28, 2019
edited by CPhill  Jul 28, 2019
 #2
avatar+1195 
+1

Thanks CPhill

 Jul 28, 2019

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