Let x and y be real numbers such that 2(x^2 + y^2) = x + y + 8. Find the maximum value of x - y.

 Nov 26, 2021

\(2(x^2+y^2)=x+y+8\\ 2(x^2-\frac{1}{2}x+y^2-\frac{1}{2}y)=8\\ x^2-\frac{1}{2}x+y^2-\frac{1}{2}y=4\\ x^2-\frac{1}{2}x+\frac{1}{16}+y^2-\frac{1}{2}x+\frac{1}{16}=4+\frac{1}{8}\\ (x-\frac{1}{4})^2+(y-\frac{1}{4})^2=\frac{33}{8}\)

Notice that for the equation \(x^2+y^2=1\), we can substitute \(x=\cos(\theta)\) and \(y=\sin(\theta)\).

Similarly, here we can substitute \(x=\frac{\sqrt{33}}{2\sqrt{2}}(\cos(\theta))+\frac{1}{4}, y=\frac{\sqrt{33}}{2\sqrt{2}}(\sin(\theta))+\frac{1}{4}\), but with slight adjustments due to the translation and the dilation of the circle.

So, we want to find the maximum value of:

\((\frac{\sqrt{33}}{2\sqrt{2}}(\cos(\theta))+\frac{1}{4})-(\frac{\sqrt{33}}{2\sqrt{2}}(\sin(\theta))+\frac{1}{4})\\ =\frac{\sqrt{33}}{2\sqrt{2}}(\cos(\theta)-\sin(\theta))\)


I know there is a way to find a simpler form of \(\cos(\theta)-\sin(\theta)\), but it's hard to see and I didn't see it, so what I ended up doing is just taking the derivative with respect to theta, which is:


The extreme points are when:

\(-\sin(\theta)=\cos(\theta)\), which only happens when \(\theta = 135+180n\), where n is an integer. From now, It shouldn't be hard to figure out that the maximum value of \(\cos(\theta)-\sin(\theta)\) is \(\sqrt{2}\).

This means that the maximum value of our original expression is \(\frac{\sqrt{33}}{2\sqrt{2}} \cdot \sqrt{2}=\boxed{\frac{\sqrt{33}}{2}}\)

For completeness, try to find which values of x and y give that maximum value.

Edit: I noticed that my previous answer contained a huge mistake lol

 Nov 26, 2021
edited by textot  Nov 26, 2021

Hi Textot,


What you have done looks very impressive.   laugh


I have to think about it some more.  frown

 Nov 27, 2021

thanks melody! this problem was pretty hard

textot  Nov 27, 2021

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