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# Algebra

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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

#1
+466
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$$2(x^2+y^2)=x+y+8\\ 2(x^2-\frac{1}{2}x+y^2-\frac{1}{2}y)=8\\ x^2-\frac{1}{2}x+y^2-\frac{1}{2}y=4\\ x^2-\frac{1}{2}x+\frac{1}{16}+y^2-\frac{1}{2}x+\frac{1}{16}=4+\frac{1}{8}\\ (x-\frac{1}{4})^2+(y-\frac{1}{4})^2=\frac{33}{8}$$

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:

$$-\sin(\theta)-\cos(\theta)$$

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
#2
+115901
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Hi Textot,

What you have done looks very impressive.

I have to think about it some more.

Nov 27, 2021
#3
+466
+1

thanks melody! this problem was pretty hard

textot  Nov 27, 2021