Lagrange Multipliers: Minimum and Maximum Values

[goblan]

Homework Statement

I am trying to find the min and max values of f(x,y)=2x^2 + 3y^2 subject to xy=5.

Homework Equations

f(x,y)=2x^2 + 3y^2 subject to xy=5
\mathbf\nablaf=(4x, 6y)
\mathbf\nablag=(y,x)

The Attempt at a Solution

When I go through the calculations, I end up with two critical points, but both yield the same maximum value. Is this correct that there are two maxs and no min? I'm skeptical. Any suggestions are appreciated.

 

[dynamicsolo]

I withdraw my previous remark: the Lagrange equation has an x² term which equals a y² term , so the critical points are symmetrical about the origin, but there are only two (not four as I earlier proposed). Since the function has an elliptic paraboloid for a surface, z is only ≥ 0 , so there won't be maxima. The constraint has a surface which is a hyperbolic cylinder, so the intersection curves turn out to be "upward-opening" parabolas. Calculate the "D-index" (the thingie with the second partial derivatives) to satisfy yourself that the critical points are minima.

 

[Dick]

I withdraw my previous remark: the Lagrange equation has an x² term which equals a y² term , so the critical points are symmetrical about the origin, but there are only two (not four as I earlier proposed). Since the function has an elliptic paraboloid for a surface, z is only ≥ 0 , so there won't be maxima. The constraint has a surface which is a hyperbolic cylinder, so the intersection curves turn out to be "upward-opening" parabolas. Calculate the "D-index" (the thingie with the second partial derivatives) to satisfy yourself that the critical points are minima.

Oooh. Now I deleted my previous post because I felt kind of doubtful. Notice you can actually solve for y=5/x. Substitute that in for y and find the extrema as a single variable problem to check your Lagrange solution.

 

[goblan]

I withdraw my previous remark: the Lagrange equation has an x² term which equals a y² term , so the critical points are symmetrical about the origin, but there are only two (not four as I earlier proposed). Since the function has an elliptic paraboloid for a surface, z is only ≥ 0 , so there won't be maxima. The constraint has a surface which is a hyperbolic cylinder, so the intersection curves turn out to be "upward-opening" parabolas. Calculate the "D-index" (the thingie with the second partial derivatives) to satisfy yourself that the critical points are minima.

Okay, I see what I did wrong. I just assumed the values I got after plugging the critical points in f(x,y) were maximums. I calculated D and see they are minimums. Thanks a lot. I guess I was expecting both a min and a max for some reason.

 

[goblan]

Oooh. Now I deleted my previous post because I felt kind of doubtful. Notice you can actually solve for y=5/x. Substitute that in for y and find the extrema as a single variable problem to check your Lagrange solution.

Done. It worked, thanks.

 

[dynamicsolo]

Oooh. Now I deleted my previous post because I felt kind of doubtful. Notice you can actually solve for y=5/x. Substitute that in for y and find the extrema as a single variable problem to check your Lagrange solution.

Sorry to have dropped a small spanner into the works. When the equations for \lambda led me to 4x² = 6y² , I immediately thought "four-fold symmetry". But the constraint equation imposes a diagonal symmetry, so the number of critical points is forced to be only two, rather than four...

 

[Dick]

Sorry to have dropped a small spanner into the works. When the equations for \lambda led me to 4x² = 6y² , I immediately thought "four-fold symmetry". But the constraint equation imposes a diagonal symmetry, so the number of critical points is forced to be only two, rather than four...

Sure, no problem. I just had a spasm of self-doubt.