The Method of Lagrange Multipliers

Apr. 22, 2025 • Updated Apr. 30, 2025

The method of Lagrange multipliers

The method of Lagrange multipliers in optimization is¹:

In mathematical optimization, the method of Lagrange multipliers is a strategy for finding the local maxima and minima of a function subject to equation constraints (i.e., subject to the condition that one or more equations have to be satisfied exactly by the chosen values of the variables). It is named after the mathematician Joseph-Louis Lagrange.

…

The basic idea is to convert a constrained problem into a form such that the derivative test of an unconstrained problem can still be applied. The relationship between the gradient of the function and gradients of the constraints rather naturally leads to a reformulation of the original problem, known as the Lagrangian function or Lagrangian. In the general case, the Lagrangian is defined as:

\[\mathcal{L}(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{\lambda}})\equiv f(\boldsymbol{\mathrm{x}})+\langle\boldsymbol{\mathrm{\lambda}},\boldsymbol{\mathrm{g}}(\boldsymbol{\mathrm{x}})\rangle\notag\]

for functions $f$, $\boldsymbol{\mathrm{g}}$; the notation $\langle\cdot,\cdot\rangle$ denotes an inner product. The value $\boldsymbol{\mathrm{\lambda}}$ is called the Lagrange multiplier.

In simple cases, where the inner product is defined as the dot product, the Lagrangian is:

\[\mathcal{L}(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{\lambda}})\equiv f(\boldsymbol{\mathrm{x}})+\boldsymbol{\mathrm{\lambda}}\cdot\boldsymbol{\mathrm{g}}(\boldsymbol{\mathrm{x}})\notag\]

The method can be summarized as follows: in order to find the maximum or minimum of a function $\boldsymbol{\mathrm{f}}$ subject to the equality constraint $\boldsymbol{\mathrm{g}}(\boldsymbol{\mathrm{x}})=\boldsymbol{0}$, find the stationary points of $\mathcal{L}$ considered as a function of $\boldsymbol{\mathrm{x}}$ and the Lagrange multiplier $\boldsymbol{\mathrm{\lambda}}$. This means that all partial derivatives should be zero, including the partial derivative with respect to $\boldsymbol{\mathrm{\lambda}}$:

\[\left\{ \begin{split} \dfrac{\partial\mathcal{L}}{\partial\boldsymbol{\mathrm{x}}}=\boldsymbol{\mathrm{0}}\\ \dfrac{\partial\mathcal{L}}{\partial\boldsymbol{\mathrm{\lambda}}}=\boldsymbol{\mathrm{0}}\\ \end{split}\right.\]

or equivalently:

\[\left\{ \begin{split} &\dfrac{\partial f(\boldsymbol{\mathrm{x}})}{\partial \boldsymbol{\mathrm{x}}}+\boldsymbol{\mathrm{\lambda}}\cdot\dfrac{\partial\boldsymbol{\mathrm{g}}(\boldsymbol{\mathrm{x}})}{\partial \boldsymbol{\mathrm{x}}}&=\boldsymbol{\mathrm{0}}\\ &\dfrac{\partial\mathcal{L}}{\partial\boldsymbol{\mathrm{\lambda}}}=\boldsymbol{\mathrm{0}}\\ \end{split}\right.\label{eq3}\]

The great advantage of this method is that it allows the optimization to be solved without explicit parameterization in terms of the constraints. As a result, the method of Lagrange multipliers is widely used to solve challenging constrained optimization problems. Further, the method of Lagrange multipliers is generalized by the Karush–Kuhn–Tucker conditions, which can also take into account inequality constraints of the form $\boldsymbol{\mathrm{h}}(\boldsymbol{\mathrm{x}})\le\boldsymbol{\mathrm{c}}$ for a given constant $\boldsymbol{\mathrm{c}}$.

Single constraint (with two variables)

Firstly, let’s take the case of only including one equality constraint and two variables:

\[\begin{split} \max_{x,y}\ &f(x,y)\\ \text{s.t.}\ &g(x,y)=0\\ \end{split}\label{eq1}\]

whose Lagrange function can be expressed as:

\[\mathcal{L}(x,y,\lambda)=f(x,y)+\lambda\cdot g(x,y)\notag\]

In this two-dimensional case, the objective function $f(x,y)$ represents a surface, and the constraint $g(x,y)=0$ represents a curve (in another words, the feasible region of this case is a curve).

Then, if a point is a maximum (or a minimum, or using “optimality” to incorporate these two cases) of optimization $\eqref{eq1}$, it will have either characteristic of following two.

Case 1: We can view the process of finding the maximum of $\eqref{eq1}$ as such: the point $(x,y)$ move along the constraint curve $g(x,y)=0$ (along the tangent line for each point on the curve), and for each point $(x,y)$ on the curve we can calculate the corresponding function value $f(x,y)$, and finally find the maximum $f(x,y)$ and hence whose variable values $(x,y)$.

On the other hand, for the objective function $f(x,y)$, on its each contour, denoted as $f(x,y)=d_i$, all possible $(x,y)$ have the same function value $f(x,y)$ by the definition of the contour. Therefore, we can conclude that when the tangent line of $g(x,y)=0$ and the tangent line of $f(x,y)=d_i$ are parallel at the same point $(x_0,y_0)$ (i.e. $g(x,y)=0$ and $f(x,y)=d_i$ are tangent at $(x_0,y_0)$), the point $(x_0,y_0)$ is probably a maximum. It is because that, at this point, the above mentioned movement of $(x,y)$ walking along the curve $g(x,y)=0$ can be viewed as a movement walking along the contour $f(x,y)=d_i$ in an infinitely small segment, and walking along the contour doesn’t change the function value, so $(x_0,y_0)$ might be a maximum.

Like in this example¹, point A might be a maximum, but B is definitely not.

Next, due to the fact that the gradient of a function is perpendicular to the tangent line of the contour (at each point on the each contour)², aforementioned “the tangent line of $g(x,y)=0$ and the tangent line of $f(x,y)=d_i$ are parallel at the same point $(x_0,y_0)$” can be equivalently expressed as “the gradient of $f(x,y)$ and the gradient of $g(x,y)$ (note, here is not $g(x,y)=0$!) are parallel at the point $(x_0,y_0)$”, which is at the point $(x_0,y_0)$ we have:

\[\begin{bmatrix} \dfrac{\partial f(x,y)}{\partial x}\Big\vert_{(x_0,y_0)}\\ \dfrac{\partial f(x,y)}{\partial y}\Big\vert_{(x_0,y_0)}\\ \end{bmatrix} =\lambda \begin{bmatrix} \dfrac{\partial g(x,y)}{\partial x}\Big\vert_{(x_0,y_0)}\\ \dfrac{\partial g(x,y)}{\partial y}\Big\vert_{(x_0,y_0)}\\ \end{bmatrix}\notag\]

i.e.,

\[\nabla_{x,y}f(x,y)\Big\vert_{(x_0,y_0)} = \lambda\nabla_{x,y}g(x,y)\Big\vert_{(x_0,y_0)}\label{eq2}\]

The constant $\lambda$ is needed because the magnitudes of these two gradient vectors are not generally equal.

Case 2: For a $f(x,y)$, if in a small area around $(x_0,y_0)$, $f$ is level, then these points also might be maxima. And at this point, we have:

\[\nabla_{x,y}f(x,y)\Big\vert_{(x_0,y_0)}=0\notag\]

For example, consider the optimization:

\[\begin{split} \max_{x,y}\ &1\\ \text{s.t.}\ & x^2+y^2=1\\ \end{split}\notag\]

We can know that, all points, denoted as $(x_0,y_0)$, on $x^2+y^2=1$ are optimal and they have the same maximum value, $1$. And we also have:

\[\left\{ \begin{split} &\nabla_{x,y}f(x,y)\Big\vert_{(x_0,y_0)} = \boldsymbol{\mathrm{0}}\\ &\nabla_{x,y}g(x,y)\Big\vert_{(x_0,y_0)} = \begin{bmatrix} 2x_0\\ 2y_0\\ \end{bmatrix}\\ \end{split}\right.\notag\]

Therefore, as can be seen, this kind of case, i.e., $f(x,y)$ is level, is also can be depicted using Eq. $\eqref{eq2}$ by setting $\lambda=0$.

Finally, we can get that, to find maxima of $f(x,y)$ constrained by $g(x,y)=0$, we could look for those points on $g(x,y)=0$ that satisfy $\nabla_{x,y}f(x,y) = \lambda\nabla_{x,y}g(x,y)$, and they may be maxima (the reason why we use “may” here is that the method of Lagrange multipliers only yields a necessary, not sufficient, condition for maxima. See below.) So, we can construct a auxiliary function:

\[\mathcal{L}(x,y,\lambda)\equiv f(x,y)+\lambda\cdot g(x,y)\notag\]

and make each partial derivative of $\mathcal{L}$ with respect to $x$, $y$, and $\lambda$, equal to zero, respectively:

\[\left\{ \begin{split} &\dfrac{\partial\mathcal{L}}{\partial x}=0\\ &\dfrac{\partial\mathcal{L}}{\partial y}=0\\ &\dfrac{\partial\mathcal{L}}{\partial \lambda}=0\\ \end{split}\right. \quad\Rightarrow\quad\left\{ \begin{split} &\dfrac{\partial f(x,y)}{\partial x}+\lambda\dfrac{\partial f(x,y)}{\partial x}=0\\ &\dfrac{\partial f(x,y)}{\partial y}+\lambda\dfrac{\partial f(x,y)}{\partial y}=0\\ &g(x,y)=0 \end{split}\right.\label{eq4}\]

By solving this equation system, we can find those points! Actually, $\eqref{eq4}$ is the two-dimensional case of $\eqref{eq3}$.

Besides, it should highlight that, the method of Lagrange multipliers only yields a necessary condition for optimality in constrained problems, that is, other than maxima, there are also other points that satisfy $\eqref{eq4}$, such as minima (see below Example 1, Example 2, and Example 3), or ones neither maxima nor minima (see Example 3).

Multiple constraints

When there are multiple constraints, the method of Lagrange multipliers is also available. We can write Lagrange function like:

\[\mathcal{L}(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{\lambda}})\equiv f(\boldsymbol{\mathrm{x}})+\boldsymbol{\mathrm{\lambda}}\cdot\boldsymbol{\mathrm{g}}(\boldsymbol{\mathrm{x}})\notag\]

and based on $\eqref{eq3}$, we’ll get:

\[\left\{ \begin{split} &\nabla f(\boldsymbol{\mathrm{x}})+\sum_{k=1}^M\lambda_k\nabla g_k(\boldsymbol{\mathrm{x}})=0\\ &g_1(\boldsymbol{\mathrm{x}})=\cdots=g_M(\boldsymbol{\mathrm{x}})=0 \end{split}\right.\notag\]

where note that the first equation denotes such a geometric meaning¹:

… In the case of multiple constraints, that will be what we seek in general: The method of Lagrange seeks points not at which the gradient of $f$ is a multiple of any single constraint’s gradient necessarily, but in which it is a linear combination of all the constraints’ gradients.

Examples

The Wikipedia article¹ also provides some specific examples, and in the following text, I’d like to reproduce some of them.

Example 1

Consider the maximization:

\[\begin{split} \max_{x,y}\ &f(x,y)=x+y\\ \mathrm{s.t.}\ &x^2+y^2=1 \end{split}\notag\]

we can write its Lagrange function is:

\[\mathcal{L}(x,y,\lambda)=(x+y)+\lambda(x^2+y^2-1)\notag\]

then we have:

we can get two solutions:

\[\left\{ \begin{split} &\lambda_1=\dfrac{\sqrt2}{2}\\ &x_1=-\dfrac{\sqrt2}{2}\\ &y_1=-\dfrac{\sqrt2}{2}\\ \end{split}\right.,\quad \left\{\begin{split} &\lambda_2=-\dfrac{\sqrt2}{2}\\ &x_2=\dfrac{\sqrt2}{2}\\ &y_2=\dfrac{\sqrt2}{2}\\ \end{split}\right.\notag\]

and their function values are:

\[f(x_1,y_1)=-\sqrt2,\ f(x_2,y_2)=\sqrt2\notag\]

Finally, we can see that, $f(x_2,y_2)$ is the maximum and $f(x_1,y_1)$ is the minimum.

fig1

clc, clear, close all

fig = figure("Color", "w", "Position", [303,521.66,1900,564.66]);
tiledlayout(1,3)

nexttile
view([32.89,11.58])
sc1 = helperPlot;

nexttile
view([90,0])
helperPlot;

nexttile
view([0,90])
sc2 = helperPlot;
sc2(1).FaceAlpha = 0;

exportgraphics(fig, "fig1.jpg", "Resolution", 600)

function sc = helperPlot
x = -1.5:0.05:1.5;
y = -1.5:0.05:1.5;
[X, Y] = meshgrid(x, y);
fnc = @(x,y) (x+y);

theta = 0:0.01:2*pi;
xx = cos(theta);
yy = sin(theta);
f = fnc(X, Y);
g = xx+yy;

hold(gca, "on")
box(gca, "on")
grid(gca, "on")
sc = surfc(X, Y, f, "EdgeColor", "none");
sc(1).FaceAlpha = 0.7;
sc(2).LineWidth = 1.5;
sc(2).LevelList = min(f,[],"all"):0.2:max(f,[],"all");
plot3(xx, yy, g, "LineWidth", 1.5, "Color", "k")
level = min(min(f))-3;
plot3(xx, yy, level*ones(size(xx)), "LineWidth", 1.5, "Color", "k")

x1 = -sqrt(2)/2;
y1 = -sqrt(2)/2;
fnc1 = fnc(x1, y1);

x2 = sqrt(2)/2;
y2 = sqrt(2)/2;
fnc2 = fnc(x2,y2);

scatter3([x1, x2], [y1, y2], [fnc1, fnc2], 20, "filled", ...
    "MarkerEdgeColor", "k", "MarkerFaceColor", "k")
scatter3([x1, x2], [y1, y2], [level, level], 20, "filled", "MarkerEdgeColor", "k", "MarkerFaceColor", "k")
plot3([x1, x1], [y1, y1], [level, fnc1], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
plot3([x2, x2], [y2, y2], [level, fnc2], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
text(x1, y1, fnc1+0.5, "$(-\sqrt{2}/2,-\sqrt{2}/2,-\sqrt{2})$", "Interpreter", "latex", "FontSize", 12)
text(x2, y2, fnc2+0.5, "$(\sqrt{2}/2,\sqrt{2}/2,\sqrt{2})$", "Interpreter", "latex", "FontSize", 12)

xlabel("$x$", "Interpreter", "latex", "FontSize", 20)
ylabel("$y$", "Interpreter", "latex", "FontSize", 20)
zlabel("$f$", "Interpreter", "latex", "FontSize", 20)
end

Example 2

Consider the maximization:

\[\begin{split} \max_{x,y}\ &f(x,y)=(x+y)^2\\ \mathrm{s.t.}\ &x^2+y^2=1 \end{split}\notag\]

whose Lagrange function is:

\[\mathcal{L}(x,y,\lambda)=(x+y)^2+\lambda(x^2+y^2-1)\notag\]

then:

\[\left\{ \begin{split} &\dfrac{\partial\mathcal{L}}{\partial x}=0\\ &\dfrac{\partial\mathcal{L}}{\partial y}=0\\ &\dfrac{\partial\mathcal{L}}{\partial \lambda}=0\\ \end{split}\right. \quad\Rightarrow\quad\left\{ \begin{split} &(1+\lambda)x+y=0\\ &x+(1+\lambda)y=0\\ &x^2+y^2-1=0\\ \end{split}\right.\notag\]

and:

\[\left\{ \begin{split} &\lambda_1=-2\\ &x_1=\dfrac{\sqrt2}{2}\\ &y_1=\dfrac{\sqrt2}{2}\\ \end{split}\right.,\quad \left\{\begin{split} &\lambda_2=-2\\ &x_2=-\dfrac{\sqrt2}{2}\\ &y_2=-\dfrac{\sqrt2}{2}\\ \end{split}\right.,\quad \left\{ \begin{split} &\lambda_3=0\\ &x_3=-\dfrac{\sqrt2}{2}\\ &y_3=\dfrac{\sqrt2}{2}\\ \end{split}\right.,\quad \left\{\begin{split} &\lambda_4=0\\ &x_4=\dfrac{\sqrt2}{2}\\ &y_4=-\dfrac{\sqrt2}{2}\\ \end{split}\right.\notag\]

i.e.

\[\begin{split} &f(\lambda_1,x_1,y_1)=2\\ &f(\lambda_2,x_2,y_2)=2\\ &f(\lambda_3,x_3,y_3)=0\\ &f(\lambda_4,x_4,y_4)=0\\ \end{split}\notag\]

where $f(x_1,y_1)$ and $f(x_2,y_2)$ are maxima, whereas $f(x_3,y_3)$ and $f(x_4,y_4)$ are minima.

fig2

clc, clear, close all

fig = figure("Color", "w", "Position", [303,521.66,1900,564.66]);
tiledlayout(1,3)

nexttile
view([32.89,11.58])
sc1 = helperPlot;

nexttile
view([90,0])
helperPlot;

nexttile
view([0,90])
sc2 = helperPlot;
sc2(1).FaceAlpha = 0;

exportgraphics(fig, "fig2.jpg", "Resolution", 600)

function sc = helperPlot
x = -1.5:0.05:1.5;
y = -1.5:0.05:1.5;
[X, Y] = meshgrid(x, y);
fnc = @(x,y) (x+y).^2;
f = fnc(X,Y);

theta = 0:0.01:2*pi;
xx = cos(theta);
yy = sin(theta);
g = fnc(xx,yy);

hold(gca, "on")
box(gca, "on")
grid(gca, "on")
sc = surfc(X, Y, f, "EdgeColor", "none");
sc(1).FaceAlpha = 0.7;
sc(2).LineWidth = 1.5;
sc(2).LevelList = min(f,[],"all"):0.2:max(f,[],"all");
plot3(xx, yy, g, "LineWidth", 1.5, "Color", "k")
level = min(min(f))-3;
plot3(xx, yy, level*ones(size(xx)), "LineWidth", 1.5, "Color", "k")

x1 = sqrt(2)/2;
y1 = sqrt(2)/2;
fnc1 = fnc(x1,y1);

x2 = -sqrt(2)/2;
y2 = -sqrt(2)/2;
fnc2 = fnc(x2,y2);

x3 = -sqrt(2)/2;
y3 = sqrt(2)/2;
fnc3 = fnc(x3,y3);

x4 = sqrt(2)/2;
y4 = -sqrt(2)/2;
fnc4 = fnc(x4,y4);

scatter3([x1, x2, x3, x4], [y1, y2, y3, y4], [fnc1, fnc2, fnc3, fnc4], 20, "filled", ...
    "MarkerEdgeColor", "k", "MarkerFaceColor", "k")
scatter3([x1, x2, x3, x4], [y1, y2, y3, y4], [level, level, level, level], 20, "filled", ...
    "MarkerEdgeColor", "k", "MarkerFaceColor", "k")
plot3([x1, x1], [y1, y1], [level, fnc1], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
plot3([x2, x2], [y2, y2], [level, fnc2], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
plot3([x3, x3], [y3, y3], [level, fnc3], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
plot3([x4, x4], [y4, y4], [level, fnc4], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
text(x1, y1, fnc1+0.5, "$(\sqrt{2}/2,\sqrt{2}/2,2)$", "Interpreter", "latex", "FontSize", 12)
text(x2, y2, fnc2+0.5, "$(-\sqrt{2}/2,-\sqrt{2}/2,2)$", "Interpreter", "latex", "FontSize", 12)
text(x3, y3, fnc1+0.5, "$(-\sqrt{2}/2,\sqrt{2}/2,0)$", "Interpreter", "latex", "FontSize", 12)
text(x4, y4, fnc1+0.5, "$(\sqrt{2}/2,-\sqrt{2}/2,0)$", "Interpreter", "latex", "FontSize", 12)

xlabel("$x$", "Interpreter", "latex", "FontSize", 20)
ylabel("$y$", "Interpreter", "latex", "FontSize", 20)
zlabel("$f$", "Interpreter", "latex", "FontSize", 20)
end

Example 3

Consider the maximization:

\[\begin{split} \max_{x,y}\ &f(x,y)=xy^2\\ \mathrm{s.t.}\ &x^2+y^2=3 \end{split}\notag\]

whose Lagrange function is:

\[\mathcal{L}(x,y,\lambda)=xy^2+\lambda(x^2+y^2-3)\]

then:

\[\left\{ \begin{split} &\dfrac{\partial\mathcal{L}}{\partial x}=0\\ &\dfrac{\partial\mathcal{L}}{\partial y}=0\\ &\dfrac{\partial\mathcal{L}}{\partial \lambda}=0\\ \end{split}\right. \quad\Rightarrow\quad\left\{ \begin{split} &y^2+2\lambda x=0\\ &xy+\lambda y=0\\ &x^2+y^2-3=0\\ \end{split}\right.\notag\]

then we’ll get six critical points:

\[\left\{ \begin{split} &\lambda_1=0\\ &x_1=\sqrt3\\ &y_1=0\\ \end{split}\right.,\quad \left\{\begin{split} &\lambda_2=0\\ &x_2=-\sqrt3\\ &y_2=0\\ \end{split}\right.\\ \left\{ \begin{split} &\lambda_3=-1\\ &x_3=1\\ &y_3=\sqrt2\\ \end{split}\right.,\quad \left\{\begin{split} &\lambda_4=-1\\ &x_4=1\\ &y_4=-\sqrt2\\ \end{split}\right.\\ \left\{\begin{split} &\lambda_5=1\\ &x_5=-1\\ &y_5=\sqrt2\\ \end{split}\right.,\quad \left\{\begin{split} &\lambda_6=1\\ &x_6=-1\\ &y_6=-\sqrt2\\ \end{split}\right.\notag\]

and their function values are:

\[\begin{split} &f(\lambda_1,x_1,y_1)=0\\ &f(\lambda_2,x_2,y_2)=0\\ &f(\lambda_3,x_3,y_3)=2\\ &f(\lambda_4,x_4,y_4)=2\\ &f(\lambda_5,x_5,y_5)=-2\\ &f(\lambda_6,x_6,y_6)=-2\\ \end{split}\notag\]

As can be seen in this case, $f(x_5,y_5)$ and $f(x_6,y_6)$ are minima, $f(x_3,y_3)$ and $f(x_4,y_4)$ are maxima, and $f(x_1,y_1)$ and $f(x_2,y_2)$ are neither minima nor maxima.

fig3

clc, clear, close all

fig = figure("Color", "w", "Position", [303,521.66,1900,564.66]);
tiledlayout(1,3)

nexttile
view([32.89,11.58])
sc1 = helperPlot;

nexttile
view([90,0])
helperPlot;

nexttile
view([0,90])
sc2 = helperPlot;
sc2(1).FaceAlpha = 0;

exportgraphics(fig, "fig3.jpg", "Resolution", 600)

function sc = helperPlot
x = -3:0.05:3;
y = -3:0.05:3;
[X, Y] = meshgrid(x, y);
fnc = @(x,y) x.*y.^2;
f = fnc(X,Y);
theta = 0:0.01:2*pi;
xx = sqrt(3)*cos(theta);
yy = sqrt(3)*sin(theta);
g = fnc(xx,yy);

hold(gca, "on")
box(gca, "on")
grid(gca, "on")
colormap("jet")
sc = surfc(X, Y, f, "EdgeColor", "none");
sc(1).FaceAlpha = 0.7;
sc(2).LevelList = min(f,[],"all"):0.4:max(f,[],"all");
sc(2).LineWidth = 1.5;
plot3(xx, yy, g, "LineWidth", 1.5, "Color", "k")
level = min(min(f))-3;
plot3(xx, yy, level*ones(size(xx)), "LineWidth", 1.5, "Color", "k")

x1 = sqrt(3);
y1 = 0;
fnc1 = fnc(x1,y1);

x2 = -sqrt(3);
y2 = 0;
fnc2 = fnc(x2,y2);

x3 = 1;
y3 = sqrt(2);
fnc3 = fnc(x3,y3);

x4 = 1;
y4 = -sqrt(2);
fnc4 = fnc(x4,y4);

x5 = -1;
y5 = sqrt(2);
fnc5 = fnc(x5,y5);

x6 = -1;
y6 = -sqrt(2);
fnc6 = fnc(x6,y6);

scatter3([x1, x2, x3, x4, x5, x6], [y1, y2, y3, y4, y5, y6], [fnc1, fnc2, fnc3, fnc4, fnc5, fnc6], 20, "filled", ...
    "MarkerEdgeColor", "k", "MarkerFaceColor", "k")
scatter3([x1, x2, x3, x4, x5, x6], [y1, y2, y3, y4, y5, y6], [level, level, level, level, level, level], 20, "filled", ...
    "MarkerEdgeColor", "k", "MarkerFaceColor", "k")
plot3([x1, x1], [y1, y1], [level, fnc1], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
plot3([x2, x2], [y2, y2], [level, fnc2], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
plot3([x3, x3], [y3, y3], [level, fnc3], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
plot3([x4, x4], [y4, y4], [level, fnc4], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
plot3([x5, x5], [y5, y5], [level, fnc5], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
plot3([x6, x6], [y6, y6], [level, fnc6], "LineWidth", 1.5, "LineStyle", "--", "Color", "k")
text(x1, y1, fnc1+0.5, "$(\sqrt{3},0,0)$", "Interpreter", "latex", "FontSize", 12)
text(x2, y2, fnc2+0.5, "$(-\sqrt{3},0,0)$", "Interpreter", "latex", "FontSize", 12)
text(x3, y3, fnc3+0.5, "$(1,\sqrt2,2)$", "Interpreter", "latex", "FontSize", 12)
text(x4, y4, fnc4+0.5, "$(1,-\sqrt2,2)$", "Interpreter", "latex", "FontSize", 12)
text(x5, y5, fnc5+0.5, "$(-1,\sqrt2,-2)$", "Interpreter", "latex", "FontSize", 12)
text(x6, y6, fnc6+0.5, "$(-1,-\sqrt2,-2)$", "Interpreter", "latex", "FontSize", 12)

xlabel("$x$", "Interpreter", "latex", "FontSize", 20)
ylabel("$y$", "Interpreter", "latex", "FontSize", 20)
zlabel("$f$", "Interpreter", "latex", "FontSize", 20)
end

References