<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] HINT:
Let $x+y=c\iff x=c-y$
$$1=9x^2+16y^2=9(c-y)^2+16y^2\iff25y^2-18cy+9c^2-1=0$$
As $y$ is real, the discriminant must be $\ge0$
$$\implies(18c)^2\ge4\cdot25(9c^2-1)\iff144c^2\le25\iff-5\le12c\le5$$
For maximum $c=x+y=\dfrac5{12}$
consequently, $y=\dfrac{18c}{2\cdot25}=\dfrac{9c}{25}\iff x=?$
</s>