<s>[INST] Finding maximum of $x+y$ Let x and y be real numbers satisfying $9x^{2} + 16y^{2} = 1$. Then $x + y$ is maximum when 
a. $y = \frac{9x}{16}$ 
b. $y = −\frac{9x}{16}$
c. $y = \frac{4x}{3}$
d. $y = −\frac{4x}{3}$
 [/INST] By C-S $$\frac{25}{144}=\frac{1}{9}+\frac{1}{16}=\left(\frac{1}{9}+\frac{1}{16}\right)(9x^2+16y^2)\geq(x+y)^2,$$
which gives $x+y\leq\frac{5}{12}$.
The equality occurs, when $\left(\frac{1}{3},\frac{1}{4}\right)||(3x,4y)$, id est, for a. 
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