Finding maximum of $x+y$

Question

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. $y = −\frac{4x}{3}$

Answer 1

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 $(4\sin t +3\cos t)^2+(3\sin t -4\cos t)^2=3^2+4^2$

$\implies (4\sin t +3\cos t)^2\le25\iff-5\le4\sin t +3\cos t\le5$

Answer 2

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, $y=\dfrac{18c}{2\cdot25}=\dfrac{9c}{25}\iff x=?$

Answer 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 $x$: $$ y = \frac{\sqrt{1 - 9\cdot 16/225}}{4} = \frac{9}{4\cdot 15} = \frac{3}{20} \Rightarrow \\ \frac{y}{x} = \frac{3}{20} / \frac{4}{15} = \frac{45}{80} = \frac{9}{16} $$

Answer 4

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

Answer 5

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}

Answer 6

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 $3/\sqrt{337}\approx 0.381$.

Answer 7

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 $$\frac{9x^2}{\frac{1}{3^2}} = \frac{16y^2}{\frac{1}{4^2}}\Leftrightarrow (9x^2)^2 = (16y^2)^2$$

Answer 8

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.