Maximising $f(x,y)=xy^2$ subject to $x^2+y^2 = 1$ using the Lagrange Multiplier

Question

As the title suggests, I'm looking for the maximum of $f(x,y)=xy^2$ on $$ E:= \{(x,y)\in \mathbb{R}^2| x^2+y^2 \leq 1\} $$

As $E$ is a compact I already know $f$ will have a maximum and a minimum there. Because the exercise suggested I use Lagrange Multipliers I'm guessing they are going to lie on the edge where $x^2+y^2 = 1$. Also, I believe, if I find the maximum and minimum there I need not search for additional ones as they would be unique.

Therefore, I define $g(x):= x^2+y^2 - 1$ such that the roots of the functions are the set $Ng:=\{(x,y)\in \mathbb{R}^2| \quad||(x,y)||_2^2=1\}$ and this set is also compact. Furthermore, both $f$ and $g$ are continuously differentiable.

Now because $$\nabla g(x,y)=(2x,2y)\neq 0$$ for values in $Ng$, there exists $\lambda \in \mathbb{R}$ such that

$$ \nabla f(x,y)=\lambda \nabla g(x,y) $$

But can somebody help me make sense of the equivalent system

$$ y^2 = \lambda 2x $$ $$ 2xy = \lambda 2y $$

How do I find the maximum/ minimum points that I'm looking for from this system of equations?

Answer 1

If $x=0$ or $y=0$ then it is easy to find the point.

Otherwise, dividing the two equations you get $$\frac{y^2}{2xy}=\frac{\lambda 2x}{\lambda 2y} \Rightarrow \frac{y}{2x}=\frac{x}{y} \Rightarrow y^2=2x^2$$

You also know that $x^2+y^2=1$.

P.S. If $x_0 <0$ then $f$ cannot have a maximum at $(x_0, y_0)$. And if $x_0 >0$ then it is easy to check that $$f(x_0,y_0) < f(x_0+\epsilon, y_0+\epsilon)$$

This implies that the maximum is indeed achieved on the circle $x^2+y^2=1$.

Answer 2

Here's another approach, wich leads to simpler computations. Since we are working in the region $x^2+y^2=1$, you can replace $f$ by the function $(x,y)\mapsto x(1-x^2)=x-x^3$. So, applying the method of Lagrange multipliers, you get the system$$\left\{\begin{array}{l}1-3x^2=2\lambda x\\0=2\lambda y\\x^2+y^2=1.\end{array}\right.$$Therefore, $y=0$ or $\lambda=0$. If $y=0$, you have the system$$\left\{\begin{array}{l}1-3x^2=2\lambda x\\x^2=1.\end{array}\right.$$In this case, $x=\pm1$. Otherwise, you have the system$$\left\{\begin{array}{l}1-3x^2=0\\x^2+y^2=1.\end{array}\right.$$In this case, the solutions are $\pm\left(\sqrt{\frac13},\sqrt{\frac23}\right)$ and $\pm\left(\sqrt{\frac13},-\sqrt{\frac23}\right)$.

Answer 3

Did you forget to differentiate with respect to $\lambda$? It just gives you the constraint. With this constraint there are three equations in three unknowns: $$\begin{align}y^2=\lambda 2x\end{align}\tag1$$ $$\begin{align}2xy=\lambda 2y\end{align}\tag2$$ $$\begin{align}x^2+y^2=1\end{align}\tag3$$ Let's make a univariate polynomial equation from these equations. Multiply $(1)$ by $x\neq 0$, $(2)$ by $y\neq 0$, add and use $(3)$ $$\begin{align}2\lambda=xy^2+2xy^2=3xy^2\end{align}$$ Use $xy=\lambda y$ and then $(1)$ again $$\Leftrightarrow$$ $$\begin{align}2\lambda=3xy^2=3\lambda y^2=6\lambda^2x\end{align}$$ Divide both $2$, multiply both sides by $y\neq 0$ and use $xy=\lambda y$ $$\Leftrightarrow$$ $$\begin{align}\lambda y=3\lambda^2xy=3\lambda^3y\end{align}$$ $$\Leftrightarrow$$ $$\begin{align}(3\lambda^3-\lambda)y=0\end{align}$$ Since $y\neq 0$ it must be that $3\lambda^3-\lambda=0$. Then either $\lambda=0$ or $3\lambda=-\sqrt 3$ or $3\lambda=+\sqrt 3$.
In the first case $x=\pm 1$. In the second $x=-\sqrt{\dfrac{1}{3}}$ and the third $x=+\sqrt{\dfrac{1}{3}}$.
Maybe my algebra is inelegant here. Suggestion are welcome.

Answer 4

By AM-GM $$xy^2\leq\sqrt{x^2y^4}=\sqrt{\frac{1}{2}\cdot2x^2\cdot y^2\cdot y^2}\leq\sqrt{\frac{1}{2}\left(\frac{2x^2+y^2+y^2}{3}\right)^3}=\frac{2}{3\sqrt3}.$$ The equality occurs for $x=\frac{1}{\sqrt3}$ and $y=\sqrt{\frac{2}{3}},$ which says that the answer is $\frac{2}{3\sqrt3}.$

Done!