I have to find max and min of the function $f(x, y) = xy (x+y)$ over the set
$$H = \{(x, y) \in [0, 1]^2; x+y \leq 1\}$$
attempts
Forst of all, I showed that $H$ is bounded (since it's a triangle, I can always find a ball $B$ with radius large enough to contain $H$), and it's also closed because $\partial H \in H$. Hence it's compact, whence I can apply Weierstrass theorem which guarantees the existence of max and min (absolute).
Now I have to study both the critical point inside $H$ and on the boundary.
Inside $H$: I computed the gradiend:
$$\nabla f = (y(2x+y), x(x+2y)) = (0, 0)$$
This means that $(x, y) = (0, 0)$ is the only solution.
But this point is not internal, it's a boundary point. Can I then conclude there is no max/min inside $H$, and pass on the evaluation on the boundary?
In this case (without using Lagrange multipliers), I would evaluate $f(x, y)$ in the restriction $y \leq 1-x$ like
$$f(x, 1-x) = x(1-x)(x + 1-x) = x(1-x)$$
Now it's like a one dimensional function, and $f'(x) = 0$ for $x = 1/2$.
Due to the constraint, then $y = 1-x = 1/2$ too, hence I have $(1/2, 1/2)$ and $(0, 0)$ as candidates.
$$f(0, 0) = 0$$ $$f(1/2, 1/2) = \frac{1}{4}$$
Which means $(0, 0)$ is an absolute min and $(1/2, 1/2)$ is absolute max.
Am I corret? Am I wrong? Did I leave something out? THank you!
Take a moment to reason about your answers.
Can $f$ be negative on $H$? Surely not. So if you've found a point where $f=0$, then surely you've found a minimum (in fact, it's both axes).
Now, is $f$ symmetrical in $x$ and $y$? So, look along the axis of symmetry (the line $x=y$). If you're off-axis and travel along the line $x+y=k$ towards the axis, what happens to $xy$ - does it necessarily get bigger until it reaches a maximum on the axis at $x=y=k/2$? So, is your maximum going to be along $x=y$ somewhere?
So when $x=y$, $f(x,y)=2x^3$ and takes its maximum of $1/4$ when $x=y=1/2$.