# Maximum value

• Sep 5th 2009, 01:39 AM
dhiab
Maximum value
Find the maximum value for A :
$\displaystyle A = xy + x\sqrt {1 - y^2 } + y\sqrt {1 - x^2 } - \sqrt {\left( {1 - x^2 } \right)\left( {1 - y^2 } \right)}$
• Sep 5th 2009, 01:50 AM
ynj
Quote:

Originally Posted by dhiab
Find the maximum value for A :
$\displaystyle A = xy + x\sqrt {1 - y^2 } + y\sqrt {1 - x^2 } - \sqrt {\left( {1 - x^2 } \right)\left( {1 - y^2 } \right)}$

Let $\displaystyle x=\sin\theta,y=\sin\phi$,then $\displaystyle A=\sin\theta\sin\phi+\sin\theta\cos\phi+\cos\theta \sin\phi-\cos\theta\cos\phi=$$\displaystyle \sin (x+y)+\cos (x+y)=\sqrt{2}\sin (x+y+\frac{\pi}{4})\leq \sqrt{2} • Sep 5th 2009, 04:49 AM dhiab Quote: Originally Posted by ynj Let \displaystyle x=\sin\theta,y=\sin\phi,then \displaystyle A=\sin\theta\sin\phi+\sin\theta\cos\phi+\cos\theta \sin\phi-\cos\theta\cos\phi=$$\displaystyle \sin (x+y)+\cos (x+y)=\sqrt{2}\sin (x+y+\frac{\pi}{4})\leq \sqrt{2}$

Hello :
Remark
$\displaystyle \sqrt {1 - \sin ^2 \theta } = \sqrt {\cos ^2 \theta } = \left| {\cos \theta } \right|$
I'think you have another cases(Rofl)
• Sep 5th 2009, 06:03 AM
ynj
Quote:

Originally Posted by dhiab
Hello :
Remark
$\displaystyle \sqrt {1 - \sin ^2 \theta } = \sqrt {\cos ^2 \theta } = \left| {\cos \theta } \right|$
I'think you have another cases(Rofl)

yeah, but you can assume $\displaystyle x,y>0$, since $\displaystyle A(x,y)\leq A(|x|,|y|)$!!!