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Find the maximum value of \(f(x,y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\) , where \( -1 \le x, y \le 1 \).
 

 Oct 27, 2019
 #1
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+1

I think the answer is 1/2.

 Oct 27, 2019
 #2
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Nah sorry. i checked and it was wrong. apparently the answer is 1. thank you though

Guest Oct 27, 2019
 #3
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This is not the answer you need but

 

if y=0 and x=1 (or vise versa) then the function value is 1

 

So the minimum value is greater or equal to 1.

 Oct 28, 2019
 #4
avatar+28375 
+3

Make use of the symmetry between x and y here.  The maximum (and minimum) of the function will occur at a point where x = y, hence will occur at the maximum of the function \(f(x)=2x\sqrt{1-x^2}\)

 

This occurs when \(x = \frac{\sqrt2}{2}\) and \(f(\sqrt2/2)=1\)

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 Oct 28, 2019
 #5
avatar+107095 
0

Thanks Allan,

I didn't understand but I think I do now  :)

Melody  Oct 28, 2019
edited by Melody  Oct 28, 2019

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