Max and min of a three variable function over a set

Question

I need help for the following exercise, because I am not sure of myself. Consider the function

$$f(x, y, z) = -(xy + zy)^2 + x - z$$

and the set $$A = \{ (x, y, z) \in \mathbb{R}^3; xy + zy + x - z = 0 \}$$

• Does $g(t) = f(e^t, 1, e^t)$ have absolute max/min as $t$ changes in $\mathbb{R}$ ?
• Does $f(x, y, z)$ have absolute max/min over $A$? If yes, find it/them.

My Attempts

So, for what concerns the first request, I just calculated

$$g(t) = -4e^{2t}$$ And this is a monotone decreasing function over $\mathbb{R}$ hence it has no max/min (absolute or relative).

For what concerns the second request, I'm a bit perplexed.

First of all, I don't know how to show if $A$ is bounded and or closed (not asked, but I always want to do this part). So I have no clue if I could use Weierstrass theorem, for not knowing if $A$ is compact cannot make me to proceed.

But alas, in spite of that, I went on.

I wrote the set $A$ in function of $z$ as $z = \frac{x(1+y)}{1-y}$ and I thence wrote $f$ restricted to that

$$f\bigg|_z = -\frac{2xy}{1-y}\left(1 + \frac{2xy}{1-y}\right)$$

At this point I calculated the gradient (tedious):

$$\nabla f = \left(-\frac{4 x y^2}{(1-y)^2}-\frac{2 y \left(\frac{2 x y}{1-y}+1\right)}{1-y},-\frac{2 x \left(\frac{2 x y}{1-y}+1\right)}{1-y}-\frac{2 x y \left(\frac{2 x y}{1-y}+1\right)}{(1-y)^2}-\frac{2 x y \left(\frac{2 x y}{(1-y)^2}+\frac{2 x}{1-y}\right)}{1-y}\right)$$

Now, I found only two solutions: $x = y = 0$ and $y = \frac{1}{1-4x}$ (with $x \neq 0$).

At this point I don't know if I should proceed with the Hessian study, or not. Maybe there is a more conveniente way to attack this problem. I'm just puzzled about the set $A$, I cannot figure it out in terms of what it does represent.

Thank you so much!

Answer 1

The first part of the exercise is correct - $g(t)$ has no minimum nor maximum over $\mathbb{R}$.

For the second part:

1. it can be shown that $A$ is closed and unbounded

   $h(x,y,z) = xy+zy+x−z $ is a continuous function and as ${0}$ is a closed set then $h^{-1}(0)=A$ is a closed set.

   Take $(x,y,z)\in A$ such that $y=1$, $x=0$ and $z\in\mathbb{R}$. Then $f(x,y,z) = -z(z+1)$ diverges to $-\infty$ as $z\rightarrow\infty$. So $f$ is not bounded in $A$.

2. As shown in the example before, $f$ has no minimum in $A$ as we can find points of $A$ arbitrarily small. But it has a maximum

   if $(x,y,z)\in A$ then $(x+z)y=-(x-z)$ and so for every point in $A$, the function $f$ becomes

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

Note that then $f(x,y,z)=-a^2+a=-a(a-1)$ with $a=x-z\in\mathbb{R}$. And this function has a maximum at $a=x-z=\frac{1}{2}$.

Please let me know if something is not clear!