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. $y = −\frac{4x}{3}$

We can solve for $y$
$$
y = \pm \frac{\sqrt{1 - 9 x^2}}{4}
$$
on the ellipse:
$$
\left(\frac{x}{1/3}\right)^2 + \left(\frac{y}{1/4}\right)^2 = 1
$$
Then
$$
f(x) = x + y(x) \le f_+(x) = x + \frac{\sqrt{1 - 9 x^2}}{4}
$$
and for an extremum:
$$
f_+'(x) = 1 + \frac{1}{8 \sqrt{1-9x^2}}(-18 x) = 0 \iff \\
1 = \frac{9x}{4 \sqrt{1-9x^2}} \iff \\
4 \sqrt{1-9x^2} = 9 x \iff \\
16 (1 - 9x^2) = 81 x^2 \iff \\
16 = 225 x^2 \iff \\
x^2 = \frac{16}{225} \iff \\
x = \pm \frac{4}{15}
$$
and thus picking the positive solution for $x$:
$$
y = \frac{\sqrt{1 - 9\cdot 16/225}}{4} = \frac{9}{4\cdot 15} = \frac{3}{20} \Rightarrow \\
\frac{y}{x} = \frac{3}{20} / \frac{4}{15} = \frac{45}{80} = \frac{9}{16}
$$
