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Math Help - There is a problem from sine & cosine chapter

  1. #1
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    There is a problem from sine & cosine chapter

    In ∆ABC, AB = (2 x) cm, BC = (x + 1) cm and angle ABC = 120˚:

    a) Show that AC^2 = x^2 x +7.
    b) Find the value of x for which AC has a minimum value.

    Ive done 1st part. But how can I solve 2nd part. Please help me.
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  2. #2
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    Hello, geton!

    \text{In }\Delta ABC,\;AB \,= \,2 - x,\;BC \,= \,x + 1,\;\angle ABC = 120^o

    a) Show that: .  AC^2 \:= \:x^2 - x +7

    b) Find the value of x for which AC has a minimum value.

    You already have: . AC \:=\:(x^2-x+7)^{\frac{1}{2}}

    Minimize the function: . AC' \;=\;\frac{1}{2}(x^2-x+7)^{-\frac{1}{2}}(2x-1)

    We have: . \frac{2x-1}{2\sqrt{x^2-x+7}} \:=\:0\quad\Rightarrow\quad 2x-1 \:=\:0

    Therefore: . x \:=\:\frac{1}{2}

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  3. #3
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    Without differentiation, of course

    2 \cdot \overline {AC}  = \sqrt {4x^2  - 4x + 28}  = \sqrt {\left( {2x - 1} \right)^2  + 27} .

    The rest follows.
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    Without differentiation, of course

    2 \cdot \overline {AC}  = \sqrt {4x^2  - 4x + 28}  = \sqrt {\left( {2x - 1} \right)^2  + 27} .

    The rest follows.
    You mean  \sqrt {\left( {2x - 1} \right)^2  + 27} = 0

    But in this way how can I get x = 1/2?
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  5. #5
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    Quote Originally Posted by geton View Post
    You mean  \sqrt {\left( {2x - 1} \right)^2  + 27} = 0

    But in this way how can I get x = 1/2?
    Since  \forall x \in \mathbb{R}, (2x-1)^2 \geq 0, we have \sqrt{\left( {2x - 1} \right)^2  + 27} \geq \sqrt{27}
    with equality at 2x-1 = 0.
    Thus min is achieved at x = \frac12
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    Thanks.
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