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Math Help - Maximum value

  1. #1
    Super Member dhiab's Avatar
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    Maximum value

    Find the maximum value for A :
    A = xy + x\sqrt {1 - y^2 } + y\sqrt {1 - x^2 } - \sqrt {\left( {1 - x^2 } \right)\left( {1 - y^2 } \right)} <br />
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  2. #2
    ynj
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    Quote Originally Posted by dhiab View Post
    Find the maximum value for A :
    A = xy + x\sqrt {1 - y^2 } + y\sqrt {1 - x^2 } - \sqrt {\left( {1 - x^2 } \right)\left( {1 - y^2 } \right)} <br />
    Let x=\sin\theta,y=\sin\phi,then A=\sin\theta\sin\phi+\sin\theta\cos\phi+\cos\theta  \sin\phi-\cos\theta\cos\phi= \sin (x+y)+\cos (x+y)=\sqrt{2}\sin (x+y+\frac{\pi}{4})\leq \sqrt{2}
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by ynj View Post
    Let x=\sin\theta,y=\sin\phi,then A=\sin\theta\sin\phi+\sin\theta\cos\phi+\cos\theta  \sin\phi-\cos\theta\cos\phi= \sin (x+y)+\cos (x+y)=\sqrt{2}\sin (x+y+\frac{\pi}{4})\leq \sqrt{2}
    Hello :
    Remark
    \sqrt {1 - \sin ^2 \theta } = \sqrt {\cos ^2 \theta } = \left| {\cos \theta } \right|<br />
    I'think you have another cases
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  4. #4
    ynj
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    Quote Originally Posted by dhiab View Post
    Hello :
    Remark
    \sqrt {1 - \sin ^2 \theta } = \sqrt {\cos ^2 \theta } = \left| {\cos \theta } \right|<br />
    I'think you have another cases
    yeah, but you can assume x,y>0, since A(x,y)\leq A(|x|,|y|)!!!
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