SOLUTION: The temperature of a point (x,y) in the plane is given by the expression x^2 + y^2 - 4x + 2y + 3x^2 + 4y^2 - 12x + 15y + 24. What is the temperature of the coldest point in the pl

Question 1209305: The temperature of a point (x,y) in the plane is given by the expression x^2 + y^2 - 4x + 2y + 3x^2 + 4y^2 - 12x + 15y + 24. What is the temperature of the coldest point in the plane?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Combine like terms to go from

Let's complete the square for the x terms.

=
=
=
=
=
=
=
=

Now complete the square for the y terms.

=
=
=
=
=
=
=
=
=
Each decimal value mentioned is exact and hasn't been rounded.

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After completing the square for both x and y we go from

You can verify this by expanding everything out in that 2nd expression, then simplifying, to arrive back at the 1st expression.
Another way to verify is to use something like WolframAlpha

The smallest that the portion can get is 0 and the same goes for the portion
Think of the parabola y = x^2

Therefore the coldest temperature is

Whether it is Celsius or Fahrenheit, it's not clear.

Side note: The location of this coldest point is (x,y) = (2, -1.7) since this x,y pairing makes and true.
If you plugged x = 2 and y = -1.7 into 4x^2 + 5y^2 - 16x + 17y + 24, or the original expression if you wanted, you should get -6.45 as the result.

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Answer = -6.45
This value is exact. It hasn't been rounded.

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