+964 Let $x$ and $y$ be real numbers. If $x^2 + 3y^2 = 18$, then find the maximum value of $x + y$.
+129895 x^2 + 3y^2 =18
Take the derivative with respect to x and set to 0
2x = 0
Take the derivative with respect to y and set to 0
6y = 0
This implies that 2x = 6y
y = (1/3)x
So
x^2 + 3 [(1/3) x]^2 = 18
(4/3]x^2 =18
x^2 = 54/4
x = sqrt [54/4] = sqrt [27/2] = 3sqrt (3) /sqrt (2) = (3/2)sqrt 6
y = (1/2)sqrt 6
Max is
x + y = (3/2)sqrt (6) + (1/2)sqrt (6) = 2sqrt 6
