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. $y = −\frac{4x}{3}$
 A: 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 $(4\sin t +3\cos t)^2+(3\sin t -4\cos t)^2=3^2+4^2$
$\implies (4\sin t +3\cos t)^2\le25\iff-5\le4\sin t +3\cos t\le5$
A: 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, $y=\dfrac{18c}{2\cdot25}=\dfrac{9c}{25}\iff x=?$
A: 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 $x$:
$$
y = \frac{\sqrt{1 - 9\cdot 16/225}}{4} = \frac{9}{4\cdot 15} = \frac{3}{20} \Rightarrow \\
\frac{y}{x} = \frac{3}{20} / \frac{4}{15} = \frac{45}{80} = \frac{9}{16}
$$
A: Because of the signs, the right answer is one of a. or c.
With $y=mx$, we have $$x+y=\frac{1+m}{\sqrt{9+16m^2}}.$$
a.  gives $5/12\approx0.417$;
c. gives $3/\sqrt{337}\approx 0.381$.
A: 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. 
A: The shortest solution is using lagrange multipliers. We must maximise $f(x,y) = x+y$ subject to the constraint $g(x,y) = 9x^2 +16y^2 -1$. Now, for extremum, we know that:
$\nabla f = -\lambda .\nabla g$.
substituting, :
$(1,1) = -\lambda.(18x,32y)$ [here, $(a,b)$ denotes the vector $a \hat i + b\hat j$]
Thus clearly, $-\lambda.18x = -\lambda.32y = 1$
Since $\lambda$ cannot be zero here, $y = \frac{9x}{16}$
In case you want to read up on lagrange multipliers, here is the wikipedia article on it: LAGRANGE MULTIPLIERS 
A: Another approach is to use Lagrange multipliers
Maximise $f=x+y $ subject to $9x^2+16y^2=1$
Set up the Lagrangian and set gradient equal zero
\begin{equation}
L = x+y - \lambda (9x^2+16y^2-1)
\end{equation}
\begin{equation}
\nabla L = 0
\end{equation}
From the partial derivatives
\begin{equation}
\frac{\partial L}{\partial x} = 1-18\lambda x=0
\end{equation}
\begin{equation}
\frac{\partial L}{\partial y} = 1-32\lambda y=0
\end{equation}
So $\lambda = 1/(18x)$ and $y=1/(32\lambda)$
Substituting for $\lambda$ in the equation for $y$ we get
\begin{equation}
y=\frac{9x}{16}
\end{equation}
A: with Holder,s Inequality
$$(9x^2+16y^2)\cdot \left[\frac{1}{3^2}+\frac{1}{4^2}\right]\geq \bigg[\bigg(16y^2 \cdot \frac{1}{9}\bigg)^{\frac{1}{2}}+\bigg(9x^2 \cdot \frac{1}{16}\bigg)^{\frac{1}{2}}\bigg]^{2}$$
so $$(x+y)^2 \leq \frac{25}{144}\Leftrightarrow (x+y)\leq \frac{5}{12}$$
and equality hold when $$\frac{9x^2}{\frac{1}{3^2}} = \frac{16y^2}{\frac{1}{4^2}}\Leftrightarrow (9x^2)^2 = (16y^2)^2$$
