Find minimum and maximum on range

Question

$f(x,y)=x^{4}-x^{2}+y^{2}$

$B={(x,y)\in \mathbb R, x^{2}+y^{2}\leq 1 }$

I should find minimum and maximum of this function on the range B.

I tried it with Lagrange Multiplier and I got these points $P(-1,0)$ and $P(1,0)$. If I put that in function I will get $0$.

So what is my minimum and maximum in this case? Is this good?

Answer 1

First we look for candidates in the interior of the disk. Setting the partials equal to $0$, we find that the candidates are $x=0,y=0$ and $x=\pm \frac{1}{\sqrt{2}}$, $y=0$.

Next we look for candidates on the boundary of the disk. Lagrange multipliers are fine, or else use the fact that $y^2=1-x^2$ to note that we are maximizing/minimizing $(x^2-1)^2$ on the boundary. So the maximum on the boundary is clearly $1$, and the minimum is $0$.

Now put together the information gathered in the preceding two paragraphs.

Answer 2

Okay, so let's split this up into two problems:

Extrema in $x^2+y^2<1$:

$$\nabla f(x,y)=(4x^3-2x,2y)=0$$

So $x=\pm\frac{\sqrt{2}}{2}$ or $x=0$, and $y=0$

Extrema on $x^2+y^2=1$:

$$x=\cos\theta$$ $$y=\sin\theta$$

$$\frac{d}{d\theta}\left[\cos^4\theta-\cos^2\theta+\sin^2\theta\right]=4\sin\theta\left(\cos\theta-\cos^3\theta\right)$$

So $\theta \in \left\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\right\}$, meaning that the extrema are located at $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$.

After analyzing all of these points, we find that the actual minima are located at $\left(\pm\frac{\sqrt{2}}{2},0\right)$ where the value of $f(x,y)$ is $-\frac{1}{4}$. The maxima are located at $(0,\pm1)$, where the value of $f(x,y)$ is $1$.

Answer 3

I am adding this just to demonstrate another way to use Lagrange multipliers.

Because of the symmetry of the constraint disk $ \ x^2 \ + \ y^2 \ \le \ 1 \ $ , we can also apply the Lagrange-multiplier method through the interior of the disk* by considering the set of concentric circles $ \ x^2 \ + \ y^2 \ = \ c^2 \ $ , with $ \ 0 \ \le \ c^2 \ \le \ 1 \ $ , as a collection of constraint curves. [That's a lot of inadvertent alliteration...]

$ \ ^* $ Since this is a technique for matching normal vectors of constraint "curves" with gradients of function "surfaces", it usually can only be applied to boundaries of regions, as noted in the other answers.

As you likely already found, the "Lagrange equations" are

$$ 4 \ x^3 \ - \ 2 \ x \ = \ \lambda \ \cdot \ 2 \ x \ \ \Rightarrow \ \ 2 \ x \ ( \ 2 \ x^2 \ - \ [\lambda \ + \ 1] \ ) \ = \ 0 \ \ , $$

$$ 2 \ y \ = \ \lambda \ \cdot \ 2y \ \ \Rightarrow \ \ 2 \ y \ ( \ 1 \ - \ \lambda \ ) \ = \ 0 \ \ . $$

The second equation gives us either $ \ y \ = \ 0 \ $ or $ \ \lambda \ = \ 1 $ .

We'll look at the latter case first, for which the first Lagrange equation becomes $$ \ 2 \ x \ ( \ 2 \ x^2 \ - \ 2 \ ) \ = \ 0 \ \ \Rightarrow \ \ x \ = \ 0 \ \ \text{or} \ \ x^2 \ = \ 1 \ \ . $$

Along the $ \ y-$ axis ( $ \ x = 0 \ $ ) , the function is $ \ f(0, \ y) \ = \ y^{2} \ $ , which has the value $ \ c^2 \ $ on the constraint circles. Thus there is a minimum $ \ f(0, \ 0) \ = \ 0 \ $ at the origin $ \ (c = 0 ) \ $ and there are maxima $ \ f(0, \ \pm 1) \ = \ 1 \ $ on the unit circle $ \ \partial B \ \ ( c = 1) \ $ .

The solution pair $ \ x^2 \ = \ 1 \ $ is only applicable on the unit circle (since the constraint circles will require $ \ y^2 \ = \ c^2 \ - 1^2 \ $ ) . So we have $ \ f(\pm 1, \ 0) \ = \ (\pm 1)^4 \ - \ (\pm 1)^2 \ + \ 0^2 \ = \ 0 \ $ .

We now return to the first case, $ \ y \ = \ 0 \ $ : along the $ \ x-$ axis, the points on the constraint curves are given by $ \ x^2 \ + \ 0^2 \ = \ c^2 \ \ \Rightarrow \ \ x \ = \ \pm c \ $ , so the value of the function at these points is $ \ f(\pm c, \ 0) \ = \ (\pm c)^4 \ - \ (\pm c)^2 \ + \ 0^2 \ = \ c^4 \ - \ c^2 \ $ . We can apply the usual method for finding the critical points of a single-variable function to produce

$$ \frac{df}{dc} \ = \ 0 \ = \ 2 \ c \ (2 \ c^2 \ - \ 1 ) \ \ \Rightarrow \ \ c^2 \ = \ 0 \ , \ \frac{1}{2} \ \ . $$

For $ \ c^2 \ = \ 0 \ $ , we again obtain the function value at the origin. On $ \ x^2 \ + \ y^2 \ = \ \frac{1}{2} \ $ , we have

$$ \ f \left(\pm \frac{1}{\sqrt{2}}, \ 0 \right) \ = \ \left(\frac{1}{2} \right)^2 \ - \ \left(\frac{1}{2} \right) \ + \ 0^2 \ = \ - \frac{1}{4} \ \ . $$

These two points then locate the absolute minima of our function (as it turns out, not only on the unit disk, but "everywhere"). The maxima determined earlier, $ \ f(0, \ \pm 1) \ = \ 1 \ $ , are the absolute maxima on the disk.

This is not greatly different from what we do for a "critical-point" analysis; it is intended to show that, in some situations, a single method can be used for both interior and boundary of a region, rather than needing to examine the boundary separately.