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# This question I tried quite a lot of times

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Let \$x\$ and \$y\$ be real numbers such that \$2(x^2 + y^2) = x + y.\$ Find the maximum value of \$x - y.\$

my go at it:

First I expanded so I got 2x^2+2y^2=x+y, then I subtracted x and y on both sides resulting 2x^2-x+2y^2-y=0

Second I tried to make two factorable parts with x and y, so I made 2x^2-x+1/8+2y^2-y+1/8=1/8+1/8, resulting in

2(x-1/4)^2+2(y-1/4)^2, to make y as small as possible, since this is a square 0 is smallest so y=1/4,

next we find that 2(x-1/4)^2+0=2/8=1/4, square root both sides makes 2(x-1/4)=sqrt(1/4)=1/2

after expanding we get 2x-1/2=1/2, adding 1/2 to both sides we get 2x=1 so x=1/2. Since y=1/4, 1/2-1/4=1/4. However \$1/4\$ is incorrect pls help.

Dec 3, 2020

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Omg Cphill is answering my question again!!!

Dec 3, 2020
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x = 1/2   and  y = 1/2  balances the equation

2*[1/2^2 + 1/2^2] =2 * [ 1/4 + 1/4] = 1

[1/2 + 1/2] = 1

Guest Dec 3, 2020
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But x=1/2 and y=1/2 leads to x-y = 0.

Try x = 1/2 and y = 0.   We still have 2(x^2 + y^2) = x + y, but now x - y = 1/2.

Alan  Dec 3, 2020