what answer do they want

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homeschool girl

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The temperature of a point [MATH](x,y)[/MATH] in the plane is given by the expression [MATH]x^2 + y^2 - 4x + 2y[/MATH]. What is the temperature of the coldest point in the plane?

I got a circle with the radios of [MATH]\sqrt5[/MATH] with a Centerpoint at [MATH](2,-1)[/MATH] but I don't know how they want me to give my answer, or how temperature comes in and what would be the "coldest". could someone please help, thanks.
 
homeschool girl said:
The temperature of a point [MATH](x,y)[/MATH] in the plane is given by the expression [MATH]x^2 + y^2 - 4x + 2y[/MATH]. What is the temperature of the coldest point in the plane?

I got a circle with the radios of [MATH]\sqrt5[/MATH] with a Centerpoint at [MATH](2,-1)[/MATH] but I don't know how they want me to give my answer, or how temperature comes in and what would be the "coldest". could someone please help, thanks.
Click to expand...
Where does the circle come from? The problem doesn't mention it. "Coldest" means you need to find the minimum of the given function.
 
[MATH]x^2 + y^2 - 4x + 2y[/MATH]
[MATH]x^2 - 4x+y^2 + 2y[/MATH]
i set it equal to zero
[MATH]x^2 - 4x+y^2 + 2y=0[/MATH]
[MATH]x^2 - 4x+4+y^2 + 2y+1=5[/MATH]
[MATH](x-2)^2+(y+1)^2=5[/MATH]
radios=sqrt5

centrer=(2,-1)
 
homeschool girl said:
[MATH]x^2 + y^2 - 4x + 2y[/MATH]
[MATH]x^2 - 4x+y^2 + 2y[/MATH]
i set it equal to zero
[MATH]x^2 - 4x+y^2 + 2y=0[/MATH]
[MATH]x^2 - 4x+4+y^2 + 2y+1=5[/MATH]
[MATH](x-2)^2+(y+1)^2=5[/MATH]
radios=sqrt5

centrer=(2,-1)
Click to expand...
Looks like you found the intersection with the xy plane. What you need is the point x,y where the function is at minimum.
 
I believe what you are being asked to do is to consider the temperature function:

[MATH]T(x,y)=x^2+y^2-4x+2y=(x^2-4x)+(y^2+2y)=(x^2-4x+4)+(y^2+2y+1)-5=(x-2)^2+(y+1)^2-5[/MATH]
Now, what do we know regarding the minimum of the sum of two squares?
 
i remember going over that in class but I don't know how to do it for this one, and it's unclear whether it's asking for maximum x, maximum y, or something in between.
 
The sum of two squares is never negative! So the least possible value of \(\displaystyle (x- 2)^2+ (y+ 1)^2\) is 0. What is the least possible value of \(\displaystyle (x- 2)^2+ (y+ 1)^2- 5\)?
 
homeschool girl said:
i remember going over that in class but I don't know how to do it for this one, and it's unclear whether it's asking for maximum x, maximum y, or something in between.
Click to expand...
You definitely need to review this material.
 
homeschool girl said:
i remember going over that in class but I don't know how to do it for this one, and it's unclear whether it's asking for maximum x, maximum y, or something in between.
Click to expand...

You are being asked for the minimum value of the given expression. Writing the expression as the sum of 2 squares and some constant allows us to observe that the minimum value of the sum of the two squares is zero, and so the minimum value of the expression is the constant.
 
Lets look at this carefully. T(x,y) = x^2+y^2−4x+2y= (x-2)^2 + (y+1)^2 -5.
The coldest temperature will be the least value that T(x,y) can be. That -5 at the end will always be there regardless of which x and y you pick. So we really just want the minimum of (x-2)^2 + (y+1)^2. But that happens when x=2 and y=-1. That gives us 0.

So T(2,-1) = 0^2 + 0^2 - 5 =-5. That is the coldest temperature.
 
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