# Lagrange Multipliers: Minimum and Maximum Values

## 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
$\mathbf\nabla$f=(4x, 6y)
$\mathbf\nabla$g=(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.

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dynamicsolo
Homework Helper
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.

Last edited:
Dick
Homework Helper
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.

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.

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
Homework Helper
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...