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# General Inequalities

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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
+1

I think the answer is 1/2.

Oct 27, 2019
#2
+1

Nah sorry. i checked and it was wrong. apparently the answer is 1. thank you though

Guest Oct 27, 2019
#3
+105668
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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
+28202
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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
+105668
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Thanks Allan,

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

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