Let x and y be nonnegative real numbers. If x^2 + 3y^2 = 18, then find the maximum value of x + y.

 Feb 14, 2024

We see there are quadratic terms and linear terms, so Cauchy Schwarz is perfect!

Applying Cauchy Schwarz inequatlity, we get,

\(({x}^{2}+3{y}^2)(1+\frac{1}{3})\ge{(x+y)}^{2}\). Substituting our values in,

\(18*\frac{4}{3}\ge{(x+y)}^{2}\), or \(24\ge{(x+y)}^{2}\), and because x and y are nonnegative, or \(x+y\ge0\), we have \(\sqrt{24} \ge x+y\).

But we need to satisfy the cauchy schwarz equality condition, to know for sure this can be achieved. This condition is, satisfied when x = r, \(\frac{\sqrt{3}}{3}r=\sqrt{3}y\), or \(y=\frac{r}{3}\), so \({r}^{2}+\frac{{r}^{2}}{3}=18\), or \({r}^{2}=27/2, r = \frac{3\sqrt{6}}{2}\), this puts x and y as \(x = \frac{3\sqrt{6}}{2}, y=\frac{\sqrt{6}}{2}\). So we know that this solution works, so we know the maximum value is \(\sqrt{24}\).\(\)

 Feb 14, 2024

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