Find absolute maximum and minimum of the function on the given domain.

Question

Given $f(x,y)=2x^4-xy^2+2y^2,0\le x\le 4, 0\le y\le2$. Find absolute extrema of $f(x,y)$.

I have found $\partial f/\partial x=8x^3-y^2, \partial f/\partial y=-2xy+4y$ and after solving the equation by letting $\partial f/\partial x=0 $ and $\partial f/\partial y=0 $, the critical point are $(0,0), (2,-8) and (2,-8)$. I'm lost how to find the absolute extrema, isn't any alternative way to solve this question?

Answer 1

Recall that we also need to look for the extrema on the boundary, in this case since we have not critical points in the interior of the domain the extrema points are on the boundary, then we need to check

• $f(0,y)$ and $f(4,y)$ for $0\le y\le2$

• $f(x,0)$ and $f(x,2)$ for $0\le x\le4$

Answer 2

Once you've found the extrema, plug them into $f$ and see which one yields the lowest value.

You'll also need to make sure $f$ isn't lower on the boundary of your domain.

Answer 3

You can follow a recipe for these questions.

$\quad(1)$: If the maximum or minimum lies on the interior of the domain, then it must be a critical point (that is, its gradient must vanish).

$\quad\quad(1.1)$: To determine whether a critical point is a local maximum or a local minimum (or saddle), you may directly compute the values and compare them, or else employ a higher order test (e.g. Hessian).

$\quad(2)$: If the maximum or minimum lies on the boundary of the domain, you may use Lagrange multipliers to find them.

$\quad\quad(2.1)$: When the boundary of the domain is $1$-dimensional, you may parametrize and min-max along the parametrization; this reduces the step to a single-variable calculus exercise.

In general, we don't know a priori whether $(1)$ or $(2)$ applies, so we need to check for extrema both in the interior and in the boundary of the domain.

Answer 4

Since your set $E=\{(x,y)\in\mathbb{R}^2:0\leq x\leq4,0\leq y\leq2\}$ is compact and $f$ is continuous on $E$ (being a polynomial), we know that $f$ assumes a global minimum and a global maximum on $E$.

You can find the extrema in the following way.

• First you look for extrema in the interior by calculating the first partial derivatives. Setting both equal to zero and solving yields critical points.
• Then you look at the Hessian Matrix. If it is positive definite (i.e. all eigenvalues are positive), the critical point is a minimum. If it is negative definite (i.e. all eigenvalues are negative), the critical point is a maximum. If there is at least one eigenvalue equal to $0$, the Hessian does not provide any information, further inspection of these points is necessary to classify these as minima/maxima or saddle points. If you have a negative and a positive eigenvalue, you have a saddle point.
• After looking at the interior points, you want to parametrize the boundary. Since $E$ is a square, you have to parametrize $4$ lines. In other cases, the boundary might be a circle or an ellipse, for example. Also then parametrizing works, but in these cases you can also use the Lagrange Multiplier method. If you use a parametrization of the boundary, you have to find the maximum of a one-dimensional function, for which you can just use techniques from one-dimensional calculus. In general, for a function $f:\mathbb{R}^n\to\mathbb{R}$ by parametrizing the boundary you obtain a function $g:\mathbb{R}^{n-1}\to\mathbb{R}$ and you can simplify the problem iteratively until you get a function $\mathbb{R}\to\mathbb{R}$.

This should be enough to solve this exercise.

Answer 5

You can avoid the restrictions making

$$ u = 4\left(\frac{\sin(\phi)+1}{2}\right)\\ v = 2\left(\frac{\sin(\eta)+1}{2}\right) $$

with this change of variables the problem reads

$$ \min\max f(\phi,\eta) = 2u^4-uv^2+2v^2 = 2u^4-(u-2)v^2 $$

The stationary conditions give

$$ \nabla f = (f_{\phi},f_{\eta}) = \left\{ \begin{array}{rcl} 128 (\sin (\phi )+1)^3 \cos (\phi )-2 (\sin (\eta )+1)^4 \cos (\phi )&=&0\\ -8 (\sin (\eta )+1)^3 \cos (\eta ) (\sin (\phi )+1)-4 (\sin (\eta )+1) \cos (\eta )& = & 0 \end{array}\right. $$

or

$$ \left\{ \begin{array}{rcl} \cos(\phi) & = & 0\\ \sin(\eta) & = & 0\\ 64 (\sin (\phi )+1)^3-(\sin (\eta )+1)^4&=&0\\ 2 (\sin (\eta )+1)^2 (\sin (\phi )+1)+1& = & 0 \end{array}\right. $$

Calling

$$ p = \sin(\phi)+1\\ q=\sin(\eta)+1 $$

the system last two equations read

$$ 64p^3-q=0\\ 2p q^2+1=0 $$

and thus the values for $\phi, \eta$ are easily obtained