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What is the maximum value of $x^2+y^2$, where $(x,y)$ are solutions to:

$$2x^2+5xy+3y^2=2$$ and


Note: Calculus is not allowed. I tried everything I could but whenever I got for example or $x^2+y^2=f(y)$ or $f(x)$ the function $f$ would always be a concave up parabola, so I could not find a maximum for either variable. And by the way I know that you can solve for $x$ and $y$ using the quadratic formula and get $4$ different solutions but I am looking for less messy way. I've asked this question before, but I didn't get the nice answer I wanted. Thanks.

This question came from a math competition from the Math Honor Society, Mu Alpha Theta.

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marked as duplicate by ShreevatsaR, Davide Giraudo, Mark Bennet, Hagen von Eitzen, Asaf Karagila Jul 6 '13 at 9:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If you are not allowed to use calculus, why doesn't that answer work? What are you allowed to use may be a better question? – Amzoti Jul 6 '13 at 1:33
Calculus won't be of help even if it is allowed. – André Nicolas Jul 6 '13 at 2:18
@AndréNicolas I copied and pasted this question from the last time I asked it, and for some reason I thought that you could get $x^2+y^2=f(x)$ $or$ $f(y)$, so if that was the case ( which know I realize is impossible) then you could have used calculus to find the maximum. – Ovi Jul 6 '13 at 3:15
I don't think it's a good practice to ask the same question again. – ShreevatsaR Jul 6 '13 at 9:15
@ShreevatsaR I thought there was a way to find the maximum of $x^2+y^2$ without actually having to find values for $x$ and $y$ and computing $x^2+y^2$ – Ovi Jul 6 '13 at 10:10
up vote 2 down vote accepted

a) $2x^2+5xy+3y^2=2$

b) $6x^2+8xy+4y^2=3$

I would provide an easier way to solve this quadratic

Divide the first eqn with $2$ and the second with $3$ and equate them. Now we can see all the terms have a two degree eqn.

$x^2 + 5/2xy+ 3/2y^2 = 3x^2+ 8/3xy+ 4/3y^2$ divide it by $xy$ Know let $\frac xy$ be $a$ and $\frac yx$ be $\frac 1a$

You would get a simple quadratic equation.

you will get the values of a then equate it to $\frac xy$. Then find the solutions.


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Division is divisive, but multiplication can be kind of fun. So I would multiply the first by $3$, and the second by $2$. – André Nicolas Jul 6 '13 at 2:00
Nice solution, but I was looking for a way to maximize $x^2+y^2$ without solving for four values of $x, y$ and plugging them in. I thought there might be a way to manipulate these equations directly into $x^2+y^2$ and find the maximum from there. – Ovi Jul 6 '13 at 3:02
And I'm confused as to what happens after solving the quadratic. You get $x/y=c$ and $x/y=k$ . If you treat these solutions as a system you are implying that $c=k$ , which would be useless if it was true. – Ovi Jul 6 '13 at 3:13
@Ovi: Let's calculate, not speculate. We get immediately $6x^2+xy-y^2=0$, that is, $(3x-y)(2x+y)=0$. Substitute $3x$ for $y$ in the first equation, solve, two points. Substute $-2x$ for $y$ in the first equation, solve, two more. – André Nicolas Jul 6 '13 at 3:25
@AndréNicolas Oh ok thanks for pointing that out,that was very stupid of me – Ovi Jul 6 '13 at 3:31

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