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plotting - Lagrange Multiplier


We are asked to maximize and minimize f(x,y)=4xy, given the constraint 4x2+y2=8, using the Lagrange Multiplier method. First, I enter the functions f and g.



f = 4 x*y;
g = 4 x^2 + y^2 - 8;

Then I solve using the Lagrange Multiplier method.


sols = Solve[{Grad[f, {x, y}] == λ Grad[g, {x, y}], 
   g == 8}, {x, y, λ}]

Which gives this output.


   {{x -> -1, y -> -2, λ -> 1}, {x -> -1, 
      y -> 2, λ -> -1}, {x -> 1, 

      y -> -2, λ -> -1}, {x -> 1, y -> 2, λ -> 1}}

I can then create the points to investigate.


{x, y} /. sols

Which gives this output.


{{-1, -2}, {-1, 2}, {1, -2}, {1, 2}}

Now I can evaluate f at each point.


f /. sols


Which gives this output.


{8, -8, -8, 8}

So, I have maximum values of 8 at the points (−1,−2) and (1,2), and minimum values of −8 at the points (−2,2) and (1,−2). Now, this can be visualized with the following image.


ContourPlot[f, {x, -2, 2}, {y, -3, 3},
 Contours -> Range[-10, 10],
 Mesh -> {{0}}, MeshFunctions -> Function[{x, y}, g],
 MeshStyle -> Directive[Thick, Yellow],
 Epilog -> {Red, PointSize[Large], Point[{x, y}] /. sols}]


function and extrema


Now, my question. I want to use the Lagrange multiplier technique to maximize f(x,y,z)=xyz subject to the constraint x2+10y2+z2=5.


f = x y z;
g = x^2 + 10 y^2 + z^2;
sols = Solve[{Grad[f, {x, y, z}] == λ Grad[g, {x, y, z}], 
    g == 5}, {x, y, z, λ}];
{x, y, z} /. sols

Gives the following output.



{{0, 0, -Sqrt[5]}, {0, 0, Sqrt[5]}, {0, -(1/Sqrt[2]), 0}, {0, 1/Sqrt[
  2], 0}, {-Sqrt[(5/3)], -(1/Sqrt[6]), -Sqrt[(5/3)]}, {-Sqrt[(5/
   3)], -(1/Sqrt[6]), Sqrt[5/3]}, {-Sqrt[(5/3)], 1/Sqrt[
  6], -Sqrt[(5/3)]}, {-Sqrt[(5/3)], 1/Sqrt[6], Sqrt[5/3]}, {Sqrt[5/
  3], -(1/Sqrt[6]), -Sqrt[(5/3)]}, {Sqrt[5/3], -(1/Sqrt[6]), Sqrt[5/
  3]}, {Sqrt[5/3], 1/Sqrt[6], -Sqrt[(5/3)]}, {Sqrt[5/3], 1/Sqrt[6], 
  Sqrt[5/3]}, {-Sqrt[5], 0, 0}, {Sqrt[5], 0, 0}}

And we can evaluate f at each of these points.


f /. sols


{0, 0, 0, 0, -(5/(3 Sqrt[6])), 5/(3 Sqrt[6]), 5/(
 3 Sqrt[6]), -(5/(3 Sqrt[6])), 5/(
 3 Sqrt[6]), -(5/(3 Sqrt[6])), -(5/(3 Sqrt[6])), 5/(3 Sqrt[6]), 0, 0}

So it appears that I have a maximum value of 53√6 at four separate points and a minimum value of −53√6 at four separate points.


So, it appears I can do this part. The part I need help with is the image. Is it possible to produce contour surfaces of f and show where the constraint g(x,y,z)=5 touches a level surface of f at the minimum and maximum points?


Or, one image that shows touches at the maximum points and a second image that shows touches at the minimum points?



Answer



Here is a visualization using MeshFunctions (as mentioned by Daniel in the comment):



f = x y z;
g = x^2 + 10 y^2 + z^2;

gp = With[{r = 3},
   RegionPlot3D[g < 5, {x, -r, r}, {y, -r, r}, {z, -r, r}, 
    PlotStyle -> Orange, PlotPoints -> 40, Mesh -> None, 
    ViewPoint -> Front, PlotTheme -> "Classic"]
   ];

With[{r = 3}, Manipulate[

  Show[gp,
   ContourPlot3D[x y z == n, {x, -r, r}, {y, -r, r}, {z, -r, r}, 
    PlotPoints -> 40,
    Mesh -> {{5}},
    MeshFunctions -> Function[{x, y, z}, g],
    MeshStyle -> {Thick, Blue},
    ContourStyle -> Opacity[.5], PlotTheme -> "Classic"], PlotLabel -> Row[{"x y z = ", n}]],
  {n, 0.5, 1}]]

manipulate



I did it as a Manipulate so you can show the intersection using time as the additional variable that isolates the solution when the blue circle degenerates to a point. In principle one could also replace this by a Table of surfaces for different values of n in x y z == n, all displayed together with Show. You could also adjust the ViewPoint to your liking, to exhibit the shape of the contour for g from a different perspective.


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