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. $y = −\frac{4x}{3}$

HINT:
WLOG $3x=\cos t,4y=\sin t$
$$x+y=\dfrac{\cos t}3+\dfrac{\sin t}4=\dfrac{4\sin t +3\cos t}{12}$$
Now $(4\sin t +3\cos t)^2+(3\sin t -4\cos t)^2=3^2+4^2$
$\implies (4\sin t +3\cos t)^2\le25\iff-5\le4\sin t +3\cos t\le5$
