Maximize

For those who want some elaboration:

Searching for the maximum value of $\dfrac1{x^2+3xy+5y^2}$ is equivalent to searching for the mininimum value of $x^2 + 3xy + 5y ^2$ . Substituting of $x = \sin\theta, y= \cos\theta$ , we get $1 + 4\cos^2\theta + 3\cos\theta \sin\theta = 1 + 2(\cos2\theta + 1) + \frac32 (2\sin\theta \cos\theta) = 3 + 2\cos2\theta + \frac32 \sin2\theta$ .

Apply the Trigonometric R method : $3 + \sqrt{2^2 + \left(\frac32\right)^2} \sin(2\theta + \alpha)$ . Minimum occurs when the sine function equal to $-1$ , the expresion becomes $3 - 2.5 = 0.5$ .

So the maximum value is $1\div 0.5 = 2$ .

i also solved it in the same way