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

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!

Combine like terms to go from
x%5E2+%2B+y%5E2+-+4x+%2B+2y+%2B+3x%5E2+%2B+4y%5E2+-+12x+%2B+15y+%2B+24
to
4x%5E2+%2B+5y%5E2+-+16x+%2B+17y+%2B+24

Let's complete the square for the x terms.
4x%5E2+%2B+5y%5E2+-+16x+%2B+17y+%2B+24
= %284x%5E2-16x%29+%2B+5y%5E2+%2B+17y+%2B+24
= 4%28x%5E2-4x%29+%2B+5y%5E2+%2B+17y+%2B+24
= 4%28x%5E2-4x%2Bred%280%29%29+%2B+5y%5E2+%2B+17y+%2B+24
= 4%28x%5E2-4x%2Bred%284-4%29%29+%2B+5y%5E2+%2B+17y+%2B+24
= 4%28%28x%5E2-4x%2B4%29-4%29+%2B+5y%5E2+%2B+17y+%2B+24
= 4%28%28x-2%29%5E2-4%29+%2B+5y%5E2+%2B+17y+%2B+24
= 4%28x-2%29%5E2+-+16+%2B+5y%5E2+%2B+17y+%2B+24
= 4%28x-2%29%5E2+%2B+5y%5E2+%2B+17y+%2B+8

Now complete the square for the y terms.
4%28x-2%29%5E2+%2B+5y%5E2+%2B+17y+%2B+8
= 4%28x-2%29%5E2+%2B+%285y%5E2+%2B+17y%29+%2B+8
= 4%28x-2%29%5E2+%2B+5%28y%5E2+%2B+3.4y%29+%2B+8
= 4%28x-2%29%5E2+%2B+5%28y%5E2+%2B+3.4y%2Bred%280%29%29+%2B+8
= 4%28x-2%29%5E2+%2B+5%28y%5E2+%2B+3.4y%2Bred%282.89-2.89%29%29+%2B+8
= 4%28x-2%29%5E2+%2B+5%28%28y%5E2+%2B+3.4y%2B2.89%29+-+2.89%29+%2B+8
= 4%28x-2%29%5E2+%2B+5%28%28y+%2B+1.7%29%5E2+-+2.89%29+%2B+8
= 4%28x-2%29%5E2+%2B+5%28y+%2B+1.7%29%5E2+%2B5%28-2.89%29+%2B+8
= 4%28x-2%29%5E2+%2B+5%28y+%2B+1.7%29%5E2+-14.45+%2B+8
= 4%28x-2%29%5E2+%2B+5%28y+%2B+1.7%29%5E2+-+6.45
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
4x%5E2+%2B+5y%5E2+-+16x+%2B+17y+%2B+24
to
4%28x-2%29%5E2+%2B+5%28y+%2B+1.7%29%5E2+-+6.45
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 %28x-2%29%5E2 can get is 0 and the same goes for the portion %28y%2B1.7%29%5E2
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 %28x-2%29%5E2=0 and %28y%2B1.7%29%5E2+=+0 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.