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Thread: minimum value help

  1. #1
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    minimum value help

    Hello,
    I am stuck with this question ,:

    find the minimum value of$\displaystyle (x + 1)^2 + y^2$where $\displaystyle x + y$are real numbers with $\displaystyle x+y=1$ justify your answer

    can someone help please?
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  2. #2
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    $\displaystyle (x+1)^2 + y^2$ is the distance squared from the point (-1,0) to (x,y)
    So the question is asking the square of the minimum distance from (-1,0) to any point on the line $\displaystyle x + y = 1$.
    i.e. the square of the distance from the point (-1, 0) to the line $\displaystyle x + y = 1$
    you should get the distance to be $\displaystyle \sqrt{2}$,
    and minimum $\displaystyle (x+1)^2 + y^2 = \sqrt{2}^2 = 2$
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  3. #3
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    Hi thanks for your reply...
    I don't understand where the (-1,0) points appeared from....

    But I done it this way, could someone tell me if this way is correct please?
    $\displaystyle x+y=1,
    y=1-x,
    (x+1)^2 + y^2 ,= (x+1)^2 + (1-x)^2
    =,x^2 +1 +1 -x^2$
    = 2
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  4. #4
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    What is the formula for the distance from (x,y) to (-1,0)?
    sqrt((x+1)^2 + y^2) right? So the square of the distance is your formula.

    You made a mistake in your steps:
    (1 + x)^2 = 1 + 2x + x^2
    (1 - x)^2 = 1 - 2x + x^2
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  5. #5
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    could you show me how you got the distance to be the sqrt2 please
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  6. #6
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    Actually, your approach is easier.

    $\displaystyle
    (x+1)^2 + y^2 = (x+1)^2 + (1-x)^2
    = (x^2 + 2x + 1) + (1 - 2x + x^2)
    = 2x^2 + 2 = 2(x^2 + 1)
    $

    Now clearly $\displaystyle 2(x^2 + 1)$ is minimum when $\displaystyle x=0$, which gives 2.

    If you want to know the distance to line approach (don't worry about it, if you haven't learned this):
    Rearranging, the line equation is $\displaystyle y = -x + 1$
    The slope of the perpendicular line has m = 1
    A perpendicular line with m = 1 through point (-1, 0) has the equation $\displaystyle y = x + 1$
    Intersection of the line and the perpendicular line is at (0,1)
    So the minimum distance is from (-1,0) to (0,1) which is sqrt(2).

    If you are confused on some step, draw a diagram and the graphs.
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  7. #7
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    Thanks
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  8. #8
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    Quote Originally Posted by wolfhound View Post
    Hello,
    I am stuck with this question ,:

    find the minimum value of$\displaystyle (x + 1)^2 + y^2$where $\displaystyle x + y$are real numbers with $\displaystyle x+y=1$ justify your answer

    can someone help please?
    $\displaystyle (x+1)^2+y^2=(x+x+y)^2+y^2=(2x+y)^2+y^2=4x^2+4xy+2y ^2$

    $\displaystyle =2\left[x^2+x^2+2xy+y^2\right]=2\left[x^2+(x+y)^2\right]=2\left[x^2+1\right]$

    minimum when x=0.
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