# Lagrange Multiplier

We are asked to maximize and minimize $f(x,y)=4xy$, given the constraint $4x^2+y^2=8$, 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}]


Now, my question. I want to use the Lagrange multiplier technique to maximize $f(x,y,z)=xyz$ subject to the constraint $x^2+10y^2+z^2=5$.

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 $\dfrac{5}{3\sqrt6}$ at four separate points and a minimum value of $-\dfrac{5}{3\sqrt6}$ 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?

• Maybe use ContourPlot3D with suitable z range spec, and likewise add z to the MeshFunctions setting. – Daniel Lichtblau Mar 27 '15 at 16:38
• Related answer – garej Feb 21 '16 at 10:16

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}]]


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.

• Fantastic answer! This will be quite helpful for our students. – David Mar 27 '15 at 18:24