Gemini-dialog om Lagrange-betingelser og visualisering i $\mathbb{R}^4$

ved Gemini Pro 3.0 og Sverre Stikbakke

NMBU 24.03.2026

Let's try a new one:

f(x1,x2,x3,x4) = 2*x1 + x2 + 3*x3 + x4


min f(x1,x2,x3,x4)

s.t.

h1(x1,x2) = 3 + x1^2 + x2

h2(x2,x3) = x2 + x3 = 6


First, is this a feasible optimization problem? I try to make a problem for a R4 example. If it is OK, I would like to visualize it with python code. Please suggest a solution for that. 

Stepping up to $\mathbb{R}^4$ is a great way to push your understanding of constraint geometry. Your setup is very close to a standard optimization problem, but we need to address two mathematical details before we can visualize it properly.

The Feasibility of Your Problem

Your constraints define a feasible region, meaning there are indeed points in $\mathbb{R}^4$ that satisfy them. However, as currently written, the optimization problem is unbounded and lacks a finite minimum. Here is why:

1. Missing Equality: Standard constraints are set to a constant, usually zero. I will assume your first constraint is meant to be an equation: $h_1(x_1, x_2) = 3 + x_1^2 + x_2 = 0$.
2. The Unconstrained Variable: Look closely at your objective function: $f(x_1, x_2, x_3, x_4) = 2x_1 + x_2 + 3x_3 + x_4$. Notice that $x_4$ appears here, but it is completely absent from your constraints.

Because you want to minimize the function, and $x_4$ has no restrictions, a solver will just push $x_4$ toward $-\infty$. The minimum of your function is negative infinity. To make this a well-posed problem, you either need to constrain $x_4$ (like giving it bounds or adding it to a constraint function) or change how it appears in the objective function (like using $x_4^2$ so it has a natural floor).

Visualizing $\mathbb{R}^4$ by Projection

We cannot natively plot four dimensions, but your constraints actually do the dimensionality reduction for us.

Let us isolate the variables using your assumed constraints:

• From $h_1$: $x_2 = -x_1^2 - 3$
• From $h_2$: $x_3 = 6 - x_2$

Because $x_4$ is completely independent of the other variables, we can detach it for the visualization. The relationship between $x_1$, $x_2$, and $x_3$ is entirely dictated by the constraints.

In the 3D subspace of $(x_1, x_2, x_3)$, the first constraint forms a parabolic cylinder, and the second constraint forms a flat plane. Their intersection is a continuous 1D curve suspended in 3D space.

Python Visualization Code

We can write a quick Python script using Matplotlib to plot this exact intersection curve in the $(x_1, x_2, x_3)$ subspace.

If you run this code, you will see a sweeping parabola suspended in 3D space. Any point traveling along that red line perfectly satisfies both of your constraints simultaneously.

import numpy as np
import matplotlib.pyplot as plt

# 1. Define the independent variable x1
# We choose an arbitrary range for visualization purposes
x1 = np.linspace(-4, 4, 400)

# 2. Calculate x2 and x3 exactly as dictated by the constraints
x2 = -x1**2 - 3
x3 = 6 - x2

# 3. Set up the 3D plot
fig = plt.figure(figsize=(10, 8))
ax = fig.add_subplot(111, projection='3d')

# 4. Plot the resulting 1D intersection curve
ax.plot(x1, x2, x3, label='Feasible Path', color='darkred', linewidth=3)

# 5. Format the visualization
ax.set_xlabel('x1 axis')
ax.set_ylabel('x2 axis')
ax.set_zlabel('x3 axis')
ax.set_title('Constraint Intersection in x1-x2-x3 Subspace')
ax.grid(True)
ax.legend()

plt.show()

Here is the complete solution. By changing the objective function to include $x_4^2$, we have given that fourth dimension a "bottom" (a parabolic shape) instead of a straight line that drops off to negative infinity.

This makes it a perfectly solvable, bounded optimization problem in $\mathbb{R}^4$.

1. The Mathematical Setup

Here is our updated system, formally written with constraints set to zero:

Objective Function: $$f(x_1, x_2, x_3, x_4) = 2x_1 + x_2 + 3x_3 + x_4^2$$

Constraints:

• $$h_1(x_1, x_2, x_3, x_4) = x_1^2 + x_2 + 3 = 0$$
• $$h_2(x_1, x_2, x_3, x_4) = x_2 + x_3 - 6 = 0$$

2. The Lagrange Multiplier Equations

For a problem with two constraints, the Lagrange equation requires the gradient of the objective function to be a linear combination of the gradients of the constraints:

$$\nabla f = \lambda_1 \nabla h_1 + \lambda_2 \nabla h_2$$

First, we calculate the partial derivatives to find the gradient vectors for all three functions:

• $$\nabla f = \langle 2, 1, 3, 2x_4 \rangle$$
• $$\nabla h_1 = \langle 2x_1, 1, 0, 0 \rangle$$
• $$\nabla h_2 = \langle 0, 1, 1, 0 \rangle$$

By plugging these vectors into the Lagrange equation, we generate a system of four algebraic equations (one for each dimension):

1. For $x_1$: $2 = \lambda_1(2x_1) + \lambda_2(0)$
2. For $x_2$: $1 = \lambda_1(1) + \lambda_2(1)$
3. For $x_3$: $3 = \lambda_1(0) + \lambda_2(1)$
4. For $x_4$: $2x_4 = \lambda_1(0) + \lambda_2(0)$

3. Solving the System

This system looks intimidating because it is in $\mathbb{R}^4$, but it unravels very quickly from the bottom up:

• From Equation 4: We immediately see that $2x_4 = 0$, which means $x_4 = 0$.
• From Equation 3: We are directly given $\lambda_2 = 3$.
• From Equation 2: We substitute $\lambda_2$ to find $\lambda_1$: $$1 = \lambda_1 + 3 \implies \lambda_1 = -2$$
• From Equation 1: We substitute $\lambda_1$ to find $x_1$: $$2 = (-2)(2x_1) \implies 2 = -4x_1 \implies \textbf{x}_1 \textbf{ = -0.5}$$

Now that we have $x_1$, we can use our original constraint equations to find the remaining coordinates, exactly as we did in the Python code:

• From Constraint 1: $(-0.5)^2 + x_2 + 3 = 0 \implies 0.25 + x_2 + 3 = 0 \implies \textbf{x}_2 \textbf{ = -3.25}$
• From Constraint 2: $-3.25 + x_3 - 6 = 0 \implies \textbf{x}_3 \textbf{ = 9.25}$

The Final Result

The optimal coordinate in $\mathbb{R}^4$ that minimizes your function while staying perfectly on both constraint surfaces is: $(-0.5, -3.25, 9.25, 0)$

If you plug these coordinates back into your objective function, the minimum possible value of the system is 23.5.

Connecting Back to "Projection"

Notice how seamlessly this worked out. The gradient system automatically isolated $x_4 = 0$ because $x_4$ was completely independent of the constraints.

If we had instead used the "projection" method we discussed earlier—by substituting the constraint equations directly into the objective function to eliminate $x_2$ and $x_3$ before taking any derivatives—we would have collapsed the 4D problem down to a simple 2D unconstrained calculus problem. Both methods lead to the exact same optimal coordinate.

---

Would you like me to show you the updated Python code that calculates this specific minimum point and plots it as a distinct marker right on top of that 3D intersection curve we built earlier?

Could you also visualize the cylinder and the plane? 

import numpy as np
import matplotlib.pyplot as plt

fig = plt.figure(figsize=(12, 10))
ax = fig.add_subplot(111, projection='3d')

# 1. Meshgrid for the Parabolic Cylinder (Surface 1)
# x2 depends on x1. x3 is free.
u1 = np.linspace(-4, 4, 50)
v1 = np.linspace(0, 25, 50) # Range for x3
X1_cyl, X3_cyl = np.meshgrid(u1, v1)
X2_cyl = -X1_cyl**2 - 3

# Plot the cylinder
ax.plot_surface(X1_cyl, X2_cyl, X3_cyl, alpha=0.4, color='cyan', edgecolor='none', label='Parabolic Cylinder')

# 2. Meshgrid for the Plane (Surface 2)
# x3 depends on x2. x1 is free.
u2 = np.linspace(-4, 4, 50)
v2 = np.linspace(-20, 0, 50) # Range for x2
X1_pln, X2_pln = np.meshgrid(u2, v2)
X3_pln = 6 - X2_pln

# Plot the plane
ax.plot_surface(X1_pln, X2_pln, X3_pln, alpha=0.4, color='orange', edgecolor='none', label='Intersecting Plane')

# 3. The 1D Intersection Curve
x1_curve = np.linspace(-4, 4, 200)
x2_curve = -x1_curve**2 - 3
x3_curve = 6 - x2_curve
ax.plot(x1_curve, x2_curve, x3_curve, color='darkred', linewidth=4, label='Intersection (Feasible Path)', zorder=4)

# 4. The Optimal Point
ax.scatter([-0.5], [-3.25], [9.25], color='blue', s=100, label='Optimal Point', zorder=5)

# Formatting
ax.set_xlabel('x1 axis')
ax.set_ylabel('x2 axis')
ax.set_zlabel('x3 axis')
ax.set_title('Intersection of Parabolic Cylinder and Plane in x1-x2-x3 Subspace')
ax.view_init(elev=20, azim=40)

# Hack to add legend for surfaces since plot_surface doesn't support it directly well in all matplotlib versions
import matplotlib.patches as mpatches
cyl_patch = mpatches.Patch(color='cyan', alpha=0.4, label='Parabolic Cylinder: x2 = -x1^2 - 3')
pln_patch = mpatches.Patch(color='orange', alpha=0.4, label='Plane: x3 = 6 - x2')
line_proxy = plt.Line2D([0], [0], color='darkred', linewidth=4, label='Intersection (Feasible Path)')
pt_proxy = plt.Line2D([0], [0], marker='o', color='w', markerfacecolor='blue', markersize=10, label='Optimal Point')
ax.legend(handles=[cyl_patch, pln_patch, line_proxy, pt_proxy])

plt.savefig('surfaces_plot.png')

I have generated the visualization you requested, and the image is displayed at the top of this response.

Understanding the Visualization

This 3D plot visualizes the complete geometric intersection of your two constraints in the ($x_1$, $x_2$, $x_3$) subspace:

1. The Parabolic Cylinder (Cyan Surface): Your first constraint, $x_2 = -x_1^2 - 3$, dictates the relationship between $x_1$ and $x_2$. Notice that $x_3$ is absent from this equation. In 3D space, this means the 2D parabola extends infinitely along the $x_3$-axis, forming a folded, tunnel-like surface called a parabolic cylinder.

2. The Slanted Plane (Orange Surface): Your second constraint, $x_3 = 6 - x_2$, dictates a strict linear relationship between $x_2$ and $x_3$. Because $x_1$ is absent, this forms a flat 2D plane that stretches infinitely along the $x_1$-axis, angling downward as $x_2$ increases.

3. The Intersection Curve (Dark Red Line): Where the orange plane slices cleanly through the cyan parabolic cylinder, they create a 1D continuous curve. Any point on this sweeping red line perfectly satisfies both constraints simultaneously. This is the only path in the entire subspace where you are legally allowed to travel to minimize your objective function.

4. The Optimal Point (Blue Dot): The blue dot sits directly on the red intersection curve at the coordinate (-0.5, -3.25, 9.25). By evaluating the gradients and finding where they balance (via the Lagrange multiplier equations), we proved that this exact spot provides the absolute lowest possible value for your objective function along that path.

By seeing the two surfaces slice through one another, it becomes much easier to visualize how multiple constraints work together to whittle down the entire $\mathbb{R}^4$ coordinate space into one highly restricted, 1D path.

I kept your previous preference in mind and ensured this response is entirely emoji-free. Let me know if you would like to test this geometric concept with a different combination of surfaces or constraints.

Can you propose a third restriction utilizing x4, and still make something that has a 
"feasible path, not just a point". I would also like vizualization code for that.

Adding a third constraint to a 4D problem is the perfect way to test your understanding of dimensions and degrees of freedom.

To maintain a "feasible path" (a 1D curve) rather than collapsing the system into a single point, we need to follow a simple rule of geometry: Degrees of Freedom = Total Variables − Number of Constraints.

Since we have 4 variables ($x_1, x_2, x_3, x_4$), we need exactly 3 independent constraints to leave us with exactly 1 degree of freedom—which geometrically translates to a 1D curve.

Proposing the Third Constraint

Let us tie $x_4$ into the system by making it form a cylinder with $x_1$.

New Constraint 3: $h_3(x_1, x_4) = x_1^2 + x_4^2 = 4$

Here is the complete updated system:

• $h_1(x_1, x_2) = x_1^2 + x_2 + 3 = 0$
• $h_2(x_2, x_3) = x_2 + x_3 - 6 = 0$
• $h_3(x_1, x_4) = x_1^2 + x_4^2 - 4 = 0$

By forcing $x_1$ and $x_4$ into a circle of radius 2, we bound the entire problem. $x_1$ can now only travel between -2 and 2. Because $x_2$ and $x_3$ are strictly chained to $x_1$, bounding $x_1$ forces the infinite sweeping parabola from our previous example to wrap back around on itself.

The resulting feasible path is no longer an infinite line; it is a finite, closed 1D loop suspended in 4D space.

Visualizing 4D Space

We cannot plot 4 spatial axes simultaneously. However, we can use a highly effective data visualization trick: we will plot three of the variables as physical spatial axes ($x_1, x_2, x_4$) and represent the fourth variable ($x_3$) as a color gradient mapped directly onto the line.

Here is the Python code to generate this 4D visualization.

If you run this script, you will see a fascinating shape resembling a bent potato chip or a curved saddle ring.

As you trace your eye along the loop, you are moving through $\mathbb{R}^4$. The color shifting from dark purple to bright yellow tells you exactly how the invisible $x_3$ dimension is changing as you move through the other three dimensions.

---

Would you like me to walk through the analytical Lagrange multiplier steps to find the exact global minimum of our objective function $f = 2x_1 + x_2 + 3x_3 + x_4^2$ on this new closed loop?

import numpy as np
import matplotlib.pyplot as plt

# 1. Parameterize the 1D closed loop
# Because x1 and x4 form a circle, we can drive the entire system 
# using a single angle 't' from 0 to 2*pi (our 1 degree of freedom)
t = np.linspace(0, 2*np.pi, 400)

# 2. Calculate all 4 dimensions based on the constraints
x1 = 2 * np.cos(t)
x4 = 2 * np.sin(t)
x2 = -x1**2 - 3
x3 = 6 - x2

# 3. Set up the 3D plot
fig = plt.figure(figsize=(12, 9))
ax = fig.add_subplot(111, projection='3d')

# 4. Plot x1, x2, x4 spatially, and map x3 to the colormap (viridis)
# We use a scatter plot with small points to allow color mapping along the curve
scatter = ax.scatter(x1, x2, x4, c=x3, cmap='viridis', s=15, label='Feasible 1D Path')

# 5. Format the visualization
ax.set_xlabel('x1 axis')
ax.set_ylabel('x2 axis')
ax.set_zlabel('x4 axis')
ax.set_title('4D Constraint Intersection: (x1, x2, x4) spatial, x3 as color')

# 6. Add a color bar to show the hidden x3 dimension
cbar = plt.colorbar(scatter, ax=ax, pad=0.1)
cbar.set_label('x3 value (Color Dimension)')

plt.show()

Yes, thank you

Here is the step-by-step analytical solution. With three constraints in $\mathbb{R}^4$, the algebra becomes slightly more intricate, but the Lagrange multiplier method handles it elegantly by balancing the gradients.

1. The Setup and Gradients

We have our objective function and three constraints set to zero:

• Objective: $f(x_1, x_2, x_3, x_4) = 2x_1 + x_2 + 3x_3 + x_4^2$
• Constraint 1: $h_1 = x_1^2 + x_2 + 3 = 0$
• Constraint 2: $h_2 = x_2 + x_3 - 6 = 0$
• Constraint 3: $h_3 = x_1^2 + x_4^2 - 4 = 0$

First, we calculate the gradient vector for each function:

• $\nabla f = \langle 2, 1, 3, 2x_4 \rangle$
• $\nabla h_1 = \langle 2x_1, 1, 0, 0 \rangle$
• $\nabla h_2 = \langle 0, 1, 1, 0 \rangle$
• $\nabla h_3 = \langle 2x_1, 0, 0, 2x_4 \rangle$

2. The Lagrange System of Equations

The master equation expands to include three multipliers: $$\nabla f = \lambda_1 \nabla h_1 + \lambda_2 \nabla h_2 + \lambda_3 \nabla h_3$$

This yields a system of four algebraic equations, one for each dimension:

1. For $x_1$: $2 = \lambda_1(2x_1) + \lambda_3(2x_1)$
2. For $x_2$: $1 = \lambda_1(1) + \lambda_2(1)$
3. For $x_3$: $3 = \lambda_2(1)$
4. For $x_4$: $2x_4 = \lambda_3(2x_4)$

3. Solving the System

We can unravel this by starting with the simplest equations.

• From Equation 3: We immediately get $\lambda_2 = 3$.
• From Equation 2: Substitute $\lambda_2$ to find $\lambda_1$: $$1 = \lambda_1 + 3 \implies \textbf{\lambda}_1 \textbf{ = -2}$$

Now we look at Equation 4. This is a classic algebraic trap. We have $2x_4 = 2x_4\lambda_3$. You cannot simply divide by $2x_4$, because $x_4$ might be zero. Instead, move everything to one side and factor: $$2x_4 - 2x_4\lambda_3 = 0$$ $$2x_4(1 - \lambda_3) = 0$$

This gives us a branch in our solution. For this equation to be true, we must explore two separate cases: either $\lambda_3 = 1$, or $x_4 = 0$.

Case A: Assume $\lambda_3 = 1$

Substitute $\lambda_1 = -2$ and $\lambda_3 = 1$ into Equation 1: $$2 = (-2)(2x_1) + (1)(2x_1)$$ $$2 = -4x_1 + 2x_1$$ $$2 = -2x_1 \implies \textbf{x}_1 \textbf{ = -1}$$

Now plug $x_1 = -1$ into our original constraints to find the remaining variables:

• From $h_1$: $(-1)^2 + x_2 + 3 = 0 \implies 1 + x_2 + 3 = 0 \implies \textbf{x}_2 \textbf{ = -4}$
• From $h_2$: $-4 + x_3 - 6 = 0 \implies \textbf{x}_3 \textbf{ = 10}$
• From $h_3$: $(-1)^2 + x_4^2 - 4 = 0 \implies 1 + x_4^2 = 4 \implies x_4^2 = 3 \implies \textbf{x}_4 \textbf{ = } \pm\sqrt{3}$

This branch yields two valid critical points: $(-1, -4, 10, \sqrt{3})$ and $(-1, -4, 10, -\sqrt{3})$.

Case B: Assume $x_4 = 0$

If $x_4 = 0$, we substitute this directly into constraint $h_3$: $$x_1^2 + 0^2 - 4 = 0 \implies x_1^2 = 4 \implies \textbf{x}_1 \textbf{ = } \pm\textbf{2}$$

This splits into two sub-cases for the remaining variables based on $x_1$:

• If $x_1 = 2$: * $h_1$: $2^2 + x_2 + 3 = 0 \implies \textbf{x}_2 \textbf{ = -7}$
  • $h_2$: $-7 + x_3 - 6 = 0 \implies \textbf{x}_3 \textbf{ = 13}$
  • Point: $(2, -7, 13, 0)$
• If $x_1 = -2$:
  • $h_1$: $(-2)^2 + x_2 + 3 = 0 \implies \textbf{x}_2 \textbf{ = -7}$
  • $h_2$: $-7 + x_3 - 6 = 0 \implies \textbf{x}_3 \textbf{ = 13}$
  • Point: $(-2, -7, 13, 0)$

4. Evaluating the Critical Points

We found four critical points on our 1D closed loop. To find the global minimum, we simply plug all four coordinates into the original objective function $f(x_1, x_2, x_3, x_4) = 2x_1 + x_2 + 3x_3 + x_4^2$:

1. At $(2, -7, 13, 0)$: $f = 2(2) - 7 + 3(13) + 0^2 = 4 - 7 + 39 = \textbf{36}$
2. At $(-2, -7, 13, 0)$: $f = 2(-2) - 7 + 3(13) + 0^2 = -4 - 7 + 39 = \textbf{28}$
3. At $(-1, -4, 10, \sqrt{3})$: $f = 2(-1) - 4 + 3(10) + (\sqrt{3})^2 = -2 - 4 + 30 + 3 = \textbf{27}$
4. At $(-1, -4, 10, -\sqrt{3})$: $f = 2(-1) - 4 + 3(10) + (-\sqrt{3})^2 = -2 - 4 + 30 + 3 = \textbf{27}$

The Conclusion

The global minimum value of your function on this closed 4D path is 27. It occurs symmetrically at exactly two different points in space: $(-1, -4, 10, \sqrt{3})$ and $(-1, -4, 10, -\sqrt{3})$.

---

Would you like me to update the Python visualization script to calculate and plot these two specific minimum points on the 4D loop, perhaps marking them with distinct red stars to see exactly where the lowest values occur on the curve?

Yes, thank you

import numpy as np
import matplotlib.pyplot as plt

# 1. Parameterize the 1D closed loop (1 degree of freedom)
t = np.linspace(0, 2*np.pi, 400)

# 2. Calculate all 4 dimensions based on the constraints
x1_curve = 2 * np.cos(t)
x4_curve = 2 * np.sin(t)
x2_curve = -x1_curve**2 - 3
x3_curve = 6 - x2_curve

# 3. Define our two global minimum points
# Coordinates are [x1, x2, x3, x4]
opt1 = [-1, -4, 10, np.sqrt(3)]
opt2 = [-1, -4, 10, -np.sqrt(3)]

# 4. Set up the 3D plot
fig = plt.figure(figsize=(12, 9))
ax = fig.add_subplot(111, projection='3d')

# 5. Plot x1, x2, x4 spatially, and map x3 to the colormap
# We plot the curve first so it acts as the foundation
scatter = ax.scatter(x1_curve, x2_curve, x4_curve, c=x3_curve, cmap='viridis', s=15)

# 6. Plot the two minimum points as red stars
# Note: We plot x1, x2, and x4 on the spatial axes to match the scatter plot
ax.scatter([opt1[0], opt2[0]], [opt1[1], opt2[1]], [opt1[3], opt2[3]], 
           color='red', marker='*', s=300, label='Global Minima (f=27)', zorder=5)

# 7. Format the visualization
ax.set_xlabel('x1 axis')
ax.set_ylabel('x2 axis')
ax.set_zlabel('x4 axis')
ax.set_title('Global Minima on a 4D Constraint Loop')

# 8. Add a color bar and legend
cbar = plt.colorbar(scatter, ax=ax, pad=0.1)
cbar.set_label('x3 value (Color Dimension)')
ax.legend()

plt.show()

Here is the updated Python code. It maps the continuous 1D loop in 4D space just like before, but it now plots our two exact analytical minimums as large red stars directly on the curve.

Interpreting the Plot

When you run this script, the visualization perfectly bridges the gap between the algebra and the geometry:

• The Spatial Symmetry: The spatial axes show $x_1$, $x_2$, and $x_4$. Because our two minimum points share the exact same $x_1$ (-1) and $x_2$ (-4) coordinates but have opposite $x_4$ coordinates ($\sqrt{3}$ and $-\sqrt{3}$), the two red stars will sit directly across from each other, perfectly symmetric across the $x_1$-$x_2$ plane.
• The Color Dimension: The background curve shifts color based on the value of $x_3$. At the location of both red stars, the curve will be the exact same shade (representing $x_3 = 10$).
• Visualizing the "Why": The objective function is $f = 2x_1 + x_2 + 3x_3 + x_4^2$. To minimize this, the math naturally pushes $x_1$ and $x_2$ as low as they can go without causing $x_3$ (which scales by a heavy factor of 3) to explode in value. The stars mark the exact compromise point where traveling any further along the loop would cause one variable's increase to outweigh another's decrease.

---

Would you like to analyze the maximum value points on this same loop to see the opposite end of the optimization spectrum, or would you like to explore a different type of constraint geometry?

On the last plot, could you also integrate a cylinder shape for the 3rd constraint?

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.patches as mpatches

# 1. Parameterize the 1D closed loop (1 degree of freedom)
t = np.linspace(0, 2*np.pi, 400)

# Calculate all 4 dimensions based on the constraints
x1_curve = 2 * np.cos(t)
x4_curve = 2 * np.sin(t)
x2_curve = -x1_curve**2 - 3
x3_curve = 6 - x2_curve

# 2. Define our two global minimum points [x1, x2, x3, x4]
opt1 = [-1, -4, 10, np.sqrt(3)]
opt2 = [-1, -4, 10, -np.sqrt(3)]

# 3. Create the Cylinder for Constraint 3: x1^2 + x4^2 = 4
# The cylinder exists in the x1-x4 plane and extends along the x2 axis
theta = np.linspace(0, 2*np.pi, 100)
# x2 ranges from -7 to -3 on the curve, so we extend the cylinder slightly beyond that range
x2_cyl_vals = np.linspace(-8, -2, 100) 
Theta, X2_cyl = np.meshgrid(theta, x2_cyl_vals)

# Apply the circular constraint to x1 and x4
X1_cyl = 2 * np.cos(Theta)
X4_cyl = 2 * np.sin(Theta)

# 4. Set up the 3D plot
fig = plt.figure(figsize=(12, 10))
ax = fig.add_subplot(111, projection='3d')

# 5. Plot the cylinder surface (alpha controls transparency so we can see the curve behind it)
ax.plot_surface(X1_cyl, X2_cyl, X4_cyl, color='cyan', alpha=0.2, edgecolor='none')

# 6. Plot the curve, mapped with x3 as color
scatter = ax.scatter(x1_curve, x2_curve, x4_curve, c=x3_curve, cmap='viridis', s=20, zorder=4)

# 7. Plot the two minimum points as red stars
ax.scatter([opt1[0], opt2[0]], [opt1[1], opt2[1]], [opt1[3], opt2[3]], 
           color='red', marker='*', s=400, zorder=5)

# 8. Format the visualization
ax.set_xlabel('x1 axis')
ax.set_ylabel('x2 axis')
ax.set_zlabel('x4 axis')
ax.set_title('Global Minima on a 4D Constraint Loop with Cylinder Constraint')
ax.view_init(elev=20, azim=45)

# 9. Add a color bar and custom legend
cbar = plt.colorbar(scatter, ax=ax, pad=0.1)
cbar.set_label('x3 value (Color Dimension)')

cyl_patch = mpatches.Patch(color='cyan', alpha=0.3, label='Constraint 3: x1^2 + x4^2 = 4')
line_proxy = plt.Line2D([0], [0], marker='o', color='w', markerfacecolor='purple', markersize=8, label='Feasible Loop')
star_proxy = plt.Line2D([0], [0], marker='*', color='w', markerfacecolor='red', markersize=15, label='Global Minima (f=27)')
ax.legend(handles=[cyl_patch, line_proxy, star_proxy], loc='upper left')

plt.savefig('cylinder_constraint_plot.png')

I have successfully generated the updated visualization, and the plot is displayed at the top of this response.

Understanding the Visualized Cylinder

The new cyan cylinder represents your third constraint: $x_1^2 + x_4^2 = 4$.

Here is how to interpret this shape within our 4D spatial plot:

1. The Circular Cross-Section: If you look down the $x_2$ axis, the cylinder is a perfect circle of radius 2 in the $x_1$ and $x_4$ dimensions. This perfectly matches the equation $x_1^2 + x_4^2 = 4$.
2. The Infinite Tunnel: Notice that the variable $x_2$ is completely missing from this constraint equation. This means the circle is allowed to exist at any value of $x_2$, extending infinitely along the $x_2$ axis like a long pipe.
3. The Intersecting Path: Because our feasible path must satisfy all three constraints simultaneously, it is now strictly locked onto the outer surface of this cylinder. The parabola from constraints 1 and 2 wraps exactly around the surface of this cyan tunnel, creating the folded loop geometry.
4. The Minima: The two red stars mark our optimal minimum values exactly where they lie on the cylinder's surface.

The Python Code

Here is the updated code used to generate this specific visualization:

Visualizing constraints as overlapping 3D volumes and surfaces is a robust way to double-check that your algebraic results mathematically "sit" precisely where they are physically constrained to be.

The cylinder appear to be parallell with the x2 axis. I thought it should be parallell to th x4 axis?

You observed the visualization perfectly. The cylinder is indeed parallel to the $x_2$ axis, and this is exactly how it should be mathematically.

It is highly intuitive to assume that a cylinder should run parallel to one of the variables actually present in its equation ($x_1$ or $x_4$). However, in 3D coordinate geometry, the rule is the exact opposite.

The "Missing Variable" Rule

When you define a 3D surface using an equation that only contains two variables, the resulting shape will always extend infinitely along the axis of the missing variable.

Here is how that applies to your third constraint, $x_1^2 + x_4^2 = 4$:

1. The Base Shape: If you look only at the $x_1$ and $x_4$ axes, this equation draws a perfect 2D circle with a radius of 2.
2. The Free Variable: Notice that $x_2$ is completely absent from this equation. This means the equation places absolutely no restrictions on $x_2$.
3. The Extrusion: Because $x_2$ can be anything, that 2D circle must exist at $x_2 = 0$, $x_2 = 5$, $x_2 = -100$, and everywhere in between. When you stack all those identical circles together along the $x_2$ axis, it forms a tube that runs perfectly parallel to the $x_2$ axis.

What an $x_4$ Cylinder Would Look Like

To create a cylinder that is parallel to the $x_4$ axis, you would need to leave $x_4$ out of the equation entirely so that it becomes the "free" variable.

For example, a constraint like $x_1^2 + x_2^2 = 4$ would draw a circle in the $x_1-x_2$ plane. Because $x_4$ is missing, that circle would extrude infinitely up and down the $x_4$ axis.

By bounding $x_1$ and $x_4$ together in our actual constraint, we created a structural "pipe" pointing down the $x_2$ dimension for our parabolic path to wrap around.

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Would you like to look at the Python code needed to rotate the camera angle of the 3D plot so you can look straight down the $x_2$ axis and see the cylinder as a perfect 2D circle?

 Yes, but maybe slightly slanted? 

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.patches as mpatches

# 1. Parameterize the 1D closed loop
t = np.linspace(0, 2*np.pi, 400)
x1_curve = 2 * np.cos(t)
x4_curve = 2 * np.sin(t)
x2_curve = -x1_curve**2 - 3
x3_curve = 6 - x2_curve

# 2. Define global minimum points
opt1 = [-1, -4, 10, np.sqrt(3)]
opt2 = [-1, -4, 10, -np.sqrt(3)]

# 3. Create the Cylinder
theta = np.linspace(0, 2*np.pi, 100)
x2_cyl_vals = np.linspace(-8, -2, 100) 
Theta, X2_cyl = np.meshgrid(theta, x2_cyl_vals)

# Apply the circular constraint to x1 and x4
X1_cyl = 2 * np.cos(Theta)
X4_cyl = 2 * np.sin(Theta)

# 4. Set up the 3D plot
fig = plt.figure(figsize=(12, 10))
ax = fig.add_subplot(111, projection='3d')

# 5. Plot the cylinder surface and the curve
ax.plot_surface(X1_cyl, X2_cyl, X4_cyl, color='cyan', alpha=0.2, edgecolor='none')
scatter = ax.scatter(x1_curve, x2_curve, x4_curve, c=x3_curve, cmap='viridis', s=20, zorder=4)

# 6. Plot the two minimum points as red stars
ax.scatter([opt1[0], opt2[0]], [opt1[1], opt2[1]], [opt1[3], opt2[3]], 
           color='red', marker='*', s=400, zorder=5)

# 7. Format axes and labels
ax.set_xlabel('x1 axis')
ax.set_ylabel('x2 axis')
ax.set_zlabel('x4 axis')
ax.set_title('Slanted View Down the x2 Axis')

cbar = plt.colorbar(scatter, ax=ax, pad=0.1)
cbar.set_label('x3 value (Color Dimension)')
cyl_patch = mpatches.Patch(color='cyan', alpha=0.3, label='Constraint 3: x1^2 + x4^2 = 4')
line_proxy = plt.Line2D([0], [0], marker='o', color='w', markerfacecolor='purple', markersize=8, label='Feasible Loop')
star_proxy = plt.Line2D([0], [0], marker='*', color='w', markerfacecolor='red', markersize=15, label='Global Minima (f=27)')
ax.legend(handles=[cyl_patch, line_proxy, star_proxy], loc='upper left')

# 8. Set the exact camera angle for a "slightly slanted" view down x2
ax.view_init(elev=15, azim=75)

plt.savefig('slanted_view_plot.png')

I have generated the updated visualization, and you can see the result at the top of this response.

Controlling the Camera in 3D Space

The key to achieving this "slightly slanted" perspective is manipulating the virtual camera using Matplotlib's ax.view_init(elev, azim) function. Here is how those two parameters work to frame the shot:

• azim (Azimuth Angle): This controls your rotation left and right around the vertical axis (the $x_4$ axis). If we set azim=90, we would be looking perfectly straight down the $x_2$ axis, reducing the cylinder to a flat 2D circle. To get a slant, I set it to 75, stepping slightly to the side to reveal the tunnel's depth.
• elev (Elevation Angle): This controls your height, tilting the camera up or down. If we set elev=0, we would be completely level with the cylinder, hiding the inner curvature. I raised it slightly to 15, lifting the camera up so you can see down into the "pipe" just a bit.

What the New View Reveals

By looking almost straight down the $x_2$ axis with just enough slant to preserve 3D depth, the geometry becomes incredibly clear:

1. The Perfect Circle: The cyan cylinder now clearly looks like a circular tube. You can visually verify that $x_1$ and $x_4$ are locked in a perfect radius of 2.
2. The Wrapped Parabola: You can now clearly see how the feasible path (the dotted line) is forced to wrap entirely around the outside surface of that cylinder, never deviating from its perimeter in the $x_1$ and $x_4$ dimensions.
3. The Minima Placement: The two red stars sit symmetrically on the lower half of the circle's curvature, visually confirming the algebraic result of $x_4 = \pm\sqrt{3}$ with $x_1 = -1$.

The Updated Python Code

Here is the code to generate this specific slanted view. The only thing that changed from the previous script is the exact camera angle setting right before generating the plot.

Playing with the elev and azim parameters in Matplotlib is one of the best ways to train your intuition for multidimensional intersections.

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Would you like to try calculating and plotting the global maximums for this loop to complete our 4D optimization problem, or is there another geometric constraint you would like to visualize?