Maclaurin 2015 Q2
2. Two real numbers \(x\) and \(y\) satisfy the equation \(x^2 + y^2 + 3xy = 2015\).
What is the maximum possible value of \(xy\) ?
Maclaurin Olympiad 2015, Question 2. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: 403
Subtracting \(x^2\) and \(y^2\) from each side of the given equation, we obtain
\(3xy = 2015 - x^2 - y^2.\)
Now adding \(2xy\) to each side, we get
\(\begin{aligned}5xy &= 2015 - x^2 + 2xy - y^2\\[6pt] &= 2015 - (x - y)^2,\end{aligned}\)
which has a maximum value of 2015 when \(x = y\).
Therefore \(xy\) has a maximum value of 403 and this occurs when \(x = y = \sqrt{403}\).
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2015 solutions.