Find the maximum of $2xy^2 - 2x^2$

Question

I got the constraint $g(x,y)=x^2+y^2$ (since the domain is $x^2+y^2\ge 1$ and $x\ge 0$) (it is a half circle domain) and the function $f(x,y)=2xy^2-2x^2$. I got the critical point to be $(0,0)$. The Hesse gave me $0$ which is of course no conclusion. I have a problem expressing $x$ and $y$ in terms of $\lambda$ but here is where my problem occurs. I get $\lambda=y^2-2$ and $\lambda=-y^2-2$. What am I doing wrong, is it the start?

Answer 1

$f_x=2y^2-4x,f_y=4xy$ so $2y^2-4x=2\lambda x$ and $4xy=2\lambda y$. The Lagrange condition in 2D is usually handled most easily by dividing one equation by the other to eliminate $\lambda$. In this case you then obtain $\frac{y}{2x}-\frac{1}{y}=\frac{x}{y}$. Clearing denominators you get $y^2/2 - x = x^2$. That in combination with the constraint $x^2+y^2=1$ is enough to find some points of interest, I think.

However this division is not in general legitimate; you needed $4xy \neq 0$ and $2 \lambda y \neq 0$, so you needed none of $x$ or $y$ or $\lambda$ to be zero.

So you need to go back and check those cases. For $\lambda=0$ you get either $x$ or $y$ being zero. In the case $x=0$ you get $y=0$, which violates the constraint. In the case $y=0$ you get $-4x=2\lambda x$ so either $\lambda=-2$ (contradiction) or $x=0$ (violates the constraint). So $\lambda \neq 0$.

If $x=0$ then immediately $2y^2=0$ so $y=0$, again violating the constraint.

If $y=0$ then $-4x=2\lambda x$ so either $x=0$ (violating the constraint) or $\lambda=-2$. In the latter case $x$ can be anything and still satisfy the second equation, so by using the constraint you conclude that $(1,0)$ is another point of interest.

Answer 2

Before we start using Lagrange multipliers, lets look at the objective function.

$f(x,y) = x^2y - 2x^2$

If we set $y$ equal to zero, we can make $x$ large and get a very negative value for $f(x).$

If we set $x$ equal to something moderately small, say 1, and let $y$ become large, we get a very large positive value for $f(x).$ The function is unbounded in both directions within the limits of the constraint.

There is no absolute minimum or maximum.

We can use the method of Lagrange multipliers, to find critical points on the boundary of the constraint.

The constraints $x^2 + y^2 \ge 1$ and $x\ge 0$ give a half-plane, excluding a half-disk.

I am inclined to ignore the constraint $x\ge 0$ for the moment.

$F(x,y,\lambda) = 2xy^2 - 2x^2 + \lambda (x^2 + y^2 - 1)$

$\frac {\partial f}{\partial x} = 2y^2 - 4x + 2\lambda x = 0\\ \frac {\partial f}{\partial y} = 4xy + 2\lambda y = 0\\ \frac {\partial f}{\partial \lambda} = x^2 + y^2 - 1 = 0$

from the second line:

$2\lambda y = -4xy$

Either $y = 0$ or $\lambda = -2x$

$y=0$ gives us $(1,0)$ as a critical point.

Setting $\lambda = -2x$ in the first line:

$2y^2 - 4x - 4x^2 = 0\\ y^2 = 2x^2 + 2x$

Substituting into the equation of the circle:
$3x^2 + 2x - 1 = 0\\ (3x - 1)(x+1)\\ x = \frac 13, -1$

$x = \frac 13$

$(\frac 13 , \pm \frac {\sqrt 8}{3})$ is a second set of critical points to check.