Quadratic Maximum Value 3
Let $ x $ and $ y $ be real numbers such that
\[3x^2 - 18x + 4y^2 - 32y + 91 = 300.\]Find the maximum value of $ x^2 + y^2 + 2xy - 14x - 14y + 49 $.
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Solution
In the equation
\[3x^2 - 18x + 4y^2 - 32y + 91 = 300,\]we can complete the square in $ x $ and $ y $ to get
\[3(x - 3)^2 + 4(y - 4)^2 = 300.\]We want to find the maximum value of
\[x^2 + y^2 + 2xy - 14x - 14y + 49 = (x + y)^2 - 14(x + y) + 49 = (x + y - 7)^2.\]By Cauchy-Schwarz,
\[\left( \frac{1}{3} + \frac{1}{4} \right) [3(x - 3)^2 + 4(y - 4)^2] \ge ((x - 3) + (y - 4))^2 = (x + y - 7)^2,\]so
\[(x + y - 7)^2 \le \frac{7}{12} \cdot 300 = 175.\]Equality occurs when $ 3(x - 3) = 4(y - 4) $ and $ 3(x - 3)^2 + 4(y - 4)^2 = 300 $. We can solve to get $ x = \frac{1}{7} (21 \pm 20 \sqrt{7}) $ and $ y = \frac{1}{7} (28 \pm 15 \sqrt{7}), $ so the maximum value is $ \boxed{175} $.
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