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Lagrange Multipliers: Minimum and Maximum Values

  • Thread starter goblan
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  • #1
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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
[itex]\mathbf\nabla[/itex]f=(4x, 6y)
[itex]\mathbf\nabla[/itex]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.
 

Answers and Replies

  • #2
dynamicsolo
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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.
 
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  • #3
Dick
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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.
 
  • #4
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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.
 
  • #5
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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.
 
  • #6
dynamicsolo
Homework Helper
1,648
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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 [itex]\lambda[/itex] 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...
 
  • #7
Dick
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Sorry to have dropped a small spanner into the works. When the equations for [itex]\lambda[/itex] 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.
 

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