Help with Lagrange multipliers

70 Views Asked by At

I need to find the absolute minima and maxima of the function $f(x,y) = 12 x^2 + 12 y^2 - x^3 y^3 -5$ in the region bounded by the disk $x^2 + y^2 \le 1$.

I know that $f(x,y)$ has three critical points in its domain, but only one point of the three, namely $(0,0)$, fits in the region bounded by the disk. This point will be a absolute minimum because $f_{xx}(0,0) >0$ and $f_{yy}(0,0) >0$ and $(0,0)$ is the only minimum in the region bounded by the disk.

Therefore I need to find the other "candidate" points for absolute maximums.

The system of equations for Lagrange Multipliers is the following : $$ \begin{split} -3x^2 y^3 + 24x - 2kx &= 0\\ -3y^2 x^3 + 24y - 2ky &= 0\\ x^2 +y^2 -1 &= 0. \end{split} $$ As you can see, this system of equations is more complex that the typical examples on the internet.What ideas do you have to solve this problems?

Thanks.

2

There are 2 best solutions below

3
On BEST ANSWER

Note the function is symmetric in $x,y$ so the answers are expected to be symmetric as well.

Case I. $x = 0$

Then $y = \pm 1$ and you get the points $(0, \pm 1)$.

Case II. $y = 0$

Then $x = \pm 1$ and you get the points $(\pm 1, 0)$.

Case III. $x,y \ne 0$

Then the first equation can be divided by $x$ and the second by $y$. Can you finish this case and individually evaluate the candidates?

UPDATE

Note in case 3, the system becomes $$ \begin{split} -3x y^3 &= 2k - 24\\ -3y x^3 &= 2k - 24\\ x^2 +y^2 &= 1. \end{split} $$ It immediately follows that $xy^3 = yx^3$ and since $x,y\ne 0$ we divide by $xy$ to get $x^2=y^2$. Use the last constraint to solve.

3
On

Doing what gt6989b adviced in Case III, I find that x= y and -x=y (when I divide by x in the first equation, and by y the second equation, then I proceed to equalize the two resultant equations). Therefore I need to solve the equation of the disk with the constraint x= y and -x=y, right?