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Thread: 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!

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

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

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

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

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

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

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

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  3. #3
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    Without differentiation, of course

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

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

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

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

    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 $\displaystyle \sqrt {\left( {2x - 1} \right)^2 + 27} = 0$

    But in this way how can I get x = 1/2?
    Since$\displaystyle \forall x \in \mathbb{R}, (2x-1)^2 \geq 0$, we have $\displaystyle \sqrt{\left( {2x - 1} \right)^2 + 27} \geq \sqrt{27}$
    with equality at 2x-1 = 0.
    Thus min is achieved at $\displaystyle x = \frac12$
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  6. #6
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    Thanks.
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