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Math Help - Quad Functions Max/Min Problem

  1. #1
    Raj
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    Quad Functions Max/Min Problem

    A number is 7 less than double a second number. The sum of the squares of these two numbers is a minimum. Determine the two numbers.

    Need help assigning variables and basically writing the quadratic function. I should be able to complete the square myself though and answer the question.

    Any help is appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Raj View Post
    A number is 7 less than double a second number. The sum of the squares of these two numbers is a minimum. Determine the two numbers.

    Need help assigning variables and basically writing the quadratic function. I should be able to complete the square myself though and answer the question.

    Any help is appreciated.
    Let the first number be x and the second be y.

    Then we are told that x=2y-7, then we are told that x^2+y^2 is a minimum, so (2y-7)^2+y^2 is a minimum.

    RonL
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  3. #3
    Raj
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    Alright i dont know what im doing wrong

    x = y^2 + (2y-7)^2

    x = 5y^2 - 28y + 49

    Then i got
    x = 5(y-14/5)^2 + 49/5


    Last edited by Raj; September 28th 2007 at 02:27 PM.
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  4. #4
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    Quote Originally Posted by Raj View Post
    Alright i dont know what im doing wrong

    x = y^2 + (2y-7)^2

    x = 5y^2 - 28y + 49

    Then i got
    x = 5(y-14/5)^2 + 49/5


    x = y^2 +(2y -7)^2 ---------------(1)
    No.
    Remember that x = 2y -7
    And, the sum of x^2 +y^2 is not x.

    The righthand side (RHS) of (1) will be a parabola in y.
    Normally, a parabola in y is a "horizontal" parabola in that the axis of symmetry is horizontal.
    Then, if you are used to minimum as the lowest vertically, you might get lost with a horizontal parabola because its "lowest" is the point that is the leftmost.
    So we change the 2nd number from y to x. The first number is y.
    And the sum is s.

    Thus,
    y = 2x -7

    s = y^2 +x^2
    s = (2x-7)^2 +x^2
    s = 5x^2 -28x +49 ----------------(i).
    Now the parabola is "vertical", that opens upward. So its vertex is the lowest point, which represent a minimum.

    When you do the "completing the square" on (i), you are trying to transform (i) into the form
    s = a(x-b)^2 +c ------(ii)
    which is still a vertical parabola.
    But this time, (ii) shows already the vertex.
    The vertex is (b,c).
    Or, b is the value of x at the vertex.

    c is the value of s at the vertex.

    We are looking for x and y when s is minimum---at the vertex.

    s = 5x^2 -28x +49
    s = 5[x^2 -(28/5)x +(14/5)^2] -5(14/5)^2 +49
    s = 5[(x -(14/5))^2 ] -196/5 +49
    s = 5[(x-(14/5)^2] +(49/5)

    Hence, x = 14/5 at the vertex.
    And minimum s = 49/5.

    Hence, at the vertex also, y = 2x -7 = 28/5 -7 = (28 -35)/5 = -7/5

    Check,
    y^2 +x^2 = miminum
    (-7/5)^2 +(14/5)^2 =? 49/5
    49/25 +196/25 =? 49/5
    245/25 =? 49/5
    (49*5)/(5*5) =? 49/5
    49/5 =? 49/5
    Yes, so, OK.

    Therefore, the two numbers are -7/5 and 14/5. ---------------answer.
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  5. #5
    Raj
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    tyvm ticol for clarifying the variables
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