We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
87
2
avatar+1040 

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+103122 
+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+1040 
+1

Thanks CPhill

 Jul 28, 2019

7 Online Users