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Homework Help: Factorisation with 2 variables

  1. Aug 23, 2006 #1
    Is there a way to find the lowest possible value of the following expression:
    [tex] x^2 + 4xy + 5y^2 - 4x - 6y +7 [/tex] without using calculus?

    I've just got myself a different textbook and am working from the beginning, trying to use only the tools given to me thus far in the book (within reason).

    One way to solve it is to complete the squares for x and y to get
    [itex] (x-2)^2+5(y- \frac{3}{5})^2 +4xy + \frac {6}{5} [/itex] and then find the value of 4xy which keeps the sum of both squares and itself as small as possible..ie differentiate

    Problem is the book has not not even touched upon calculus yet and so I suspect there is a more elementary approach that I have either missed or am not even aware of and this is bugging me :grumpy: ...The furthest I got was the factorisation above and figuring out that 4xy should be negative and its absolute value being as high as possible so that y < 0 whilst x > 2 ... but I still cannot find a basic method of finding the precise values of x and y. Anyone have any suggestions?
    Last edited: Aug 23, 2006
  2. jcsd
  3. Aug 23, 2006 #2


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    Try getting rid of the xy term first... by completing the square!

    (Hint: instead of looking at (x-a)², try looking at (x-ay)²)
  4. Aug 23, 2006 #3
    Thanks for the reply Hurkyl

    By completing the square to eliminate xy I get [tex] (x+2y)^2 +y^2-6y-4x+7 [/tex]...I can go further and wrap up y to get [tex] (x+2y)^2 + (y-3)^2 -4x - 2 [/tex] but I cannot see any useful clues...(apart from what I discovered from the other factorisation), nor can I see a way to force exact values upon x or y :frown:
  5. Aug 24, 2006 #4


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    You can still complete the square to get rid of the x term. (Doing so will introduce another y term, but you already know how to get rid of that)
  6. Aug 24, 2006 #5
    Aha! :biggrin:

    [tex] x^2 + 4xy + 5y^2 -4x -6y + 7 = [/tex]
    [tex](x +2y -2)^2 + y^2 +2y +3 = [/tex]
    [tex] (x +2y -2)^2 +(y+1)^2 +2 [/tex]

    Thanks for steering me in the right direction Hurkyl!
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