Determine extrema of a multivariate function defined on a set

Question

Given a function $$f(x,y)=2+2x^2+y^2$$ and the set $$A:=\{(x,y)\in\mathbb R^2 | x^2+4y^2\leq 1\}$$ and $f:A\to\mathbb R$ how do I determine the global and local extrema on this set? Normally I would not find this hard since it would just be determining $\nabla f=0$ however I am not sure how this works when $A$ is not something simple like $\mathbb R^2$.

Answer 1

The fact that we are restricted to the set $A$ doesn't change the principle at hand. The Extreme Value Theorem states that if the domain is closed and bounded then there exists a maximum (minimum). Your domain is not closed, so we cannot use this theorem, but if we look at the closed ellipse (call it $\bar{A}$) then we can guarantee a max (min).

We then use the technique we know: check where $\nabla f=0$, and all maxima (minima) occur at either these points or on the boundary. In this case we have

$$\nabla f=(4x,2y)=0\implies x=y=0$$

This gives one critical point. We can analytically check, but based on the fact that this function is always increasing from the origin this must be a minimum. As such, $f$ has a minimum at $(0,0)$.

What about a maximum? This function doesn't have a maximum, because as we approach the border of the ellipse this function keeps getting larger, but never achieves a maximum. If we included the boundary of the ellipse, then we would have a maximum on the boundary, and there are several ways of approaching this problem, such as Lagrange Multipliers or paramaterization. Since this was not in the scope of the question, I'll leave it at that.

Edit If we are to consider the boundary as part of the domain, then the problem can be solved a few ways. I'll detail two of them.

The single variable calculus way of doing it is through parameterizations. We can actually parameterize the boundary of the domain with the following equations

$$x=\cos(t),\;\;y=\frac{1}{2}\sin(t),\;\;0\leq t<2\pi$$

As a single variable problem this then reads

$$g(t)=f(x,y)=2+2x^2+y^2=2+2\cos^2(t)+\frac{1}{4}\sin^2(t)$$

The derivative is

$$0=g'(t)=-4\cos(t)\sin(t)+\frac{1}{2}\sin(t)\cos(t)\implies\sin(t)=0\mbox{ or }\cos(t)=0$$

As such the boundary has critical points whenever $t=0,\pi/2,\pi,3\pi/2$. Plugging in these values for $t$ into $g$ gives a maximum value of $4$, occuring both at $(1,0)$ and $(-1,0)$.

The multivariate approach is to use Lagrange Multipliers. We seek to minimize $f$ subject to the constraint $g(x,y)=x^2+4y^2=1$. This gives the auxiliary equation

$$L(x,y,\lambda)=2+2x^2+y^2-\lambda(x^2+4y^2-1)$$

The derivatives are

$$L_x=4x-2\lambda x=0\implies x=0\mbox{ or }\lambda=2$$ $$L_y=2y-8\lambda y=0\implies y=0\mbox{. or }\lambda=\frac{1}{4}$$ $$L_\lambda=x^2+4y^2-1=0$$

Since $\lambda$ cannot be both $2$ and $1/4$, we have either $x=0$ or $y=0$, which gives points of $(\pm1,0)$ and $(0,\pm1/2)$. This gives us the same maximum.

Answer 2

Since the domain is compact, the function must attain both maximum and minimum values. If you work in the interior of the set, your approach is right and you get a minimum at zero and no maximum value.

Then, over the border, you get that $x^2=1-4y^2$, so $f$ becomes $f(y)=2+2(1-4y^2)+y^2$ which is a real valued function that you only need to optimize over $ y \in [-\frac{1}{2},\frac{1}{2}]$ and you will be done.