Optimization Problems (Maxima/Minima) in Multivariable Functions

Nerd Cafe

Goal

To find maximum or minimum values of a multivariable function, often subject to certain conditions (e.g., within a region or constraint).

Step 1: Understand the Problem

For a function f(x,y), you want to find points where the function reaches:

• Maximum value (peak)

• Minimum value (valley)

• Saddle point (neither max nor min, like a horse saddle)

These occur where the gradient vector is zero:

$$\frac{\partial f}{\partial x}=0 \;\;and\;\; \frac{\partial f}{\partial y}=0$$

Step 2: Mathematical Steps for Finding Critical Points

Let’s say we have:

$$f(x,y)=x^{2}+y^{2}-4x-6y+13$$

Step A: Compute First-Order Partial Derivatives

$$f_{x}=\frac{\partial f}{\partial x}=2x-4$$

and

$$f_{y}=\frac{\partial f}{\partial y}=2y-6$$

Step B: Set the Partial Derivatives to Zero

$$\begin{matrix} 2x-4=0\Rightarrow x=2 \\ \\ 2y-6=0\Rightarrow y=3 \end{matrix}$$

So the critical point is (2,3).

Step C: Use Second Derivative Test

Let’s define:

$$\begin{matrix} f_{xx}=\frac{\partial^{2}f }{\partial x^{2}}=2 \\ \\ f_{yy}=\frac{\partial^{2}f }{\partial y^{2}}=2 \\ \\ f_{xy}=\frac{\partial^{2}f }{\partial x \partial y}=0 \end{matrix}$$

Now compute the discriminant:

$$D=f_{xx}f_{yy}-\left( fxy \right)^{2}=(2\times 2)-0=4$$

Interpretation:

• If D>0 and f_xx>0 → local minimum

• If D>0 and f_xx<0 → local maximum

• If D<0 → saddle point

In our case:

Since D=4>0 and f_xx=2>0, it's a local minimum at (2,3).

Python Code Using SymPy

import sympy as sp

# Define variables
x, y = sp.symbols('x y')

# Define the function
f = x**2 + y**2 - 4*x - 6*y + 13

# First-order partial derivatives
fx = sp.diff(f, x)
fy = sp.diff(f, y)

# Solve for critical points
critical_points = sp.solve([fx, fy], (x, y))
print("Critical Points:", critical_points)

# Second-order partial derivatives
fxx = sp.diff(fx, x)
fyy = sp.diff(fy, y)
fxy = sp.diff(fx, y)

# Discriminant
D = fxx * fyy - fxy**2

# Evaluate at the first critical point
point = critical_points[0]  # it's a dict like {x: 2, y: 3}
D_val = D.subs(point)
fxx_val = fxx.subs(point)

print("D =", D_val)
print("f_xx =", fxx_val)

# Nature of critical point
if D_val > 0:
    if fxx_val > 0:
        print("Local minimum at", point)
    else:
        print("Local maximum at", point)
elif D_val < 0:
    print("Saddle point at", point)
else:
    print("Test inconclusive")

Outline

Critical Points: {x: 2, y: 3}

Visualize the Function in 3D

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

X = Y = np.linspace(-1, 5, 100)
X, Y = np.meshgrid(X, Y)
Z = X**2 + Y**2 - 4*X - 6*Y + 13

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(X, Y, Z, cmap='viridis')
ax.set_title("f(x, y) = x² + y² - 4x - 6y + 13")
ax.set_xlabel("x")
ax.set_ylabel("y")
ax.set_zlabel("f(x, y)")
ax.scatter(2, 3, f.subs({x:2, y:3}), color='r', s=50)  # Critical point
plt.show()

Outline

Summary Table

Step	Action
1	Compute𝑓𝑥,𝑓𝑦​
2	Solve𝑓𝑥=0,𝑓𝑦=0
3	Compute second-order partials: 𝑓 𝑥𝑥 , 𝑓 𝑦𝑦 , 𝑓 𝑥𝑦
4	Find discriminant 𝐷 = 𝑓 𝑥𝑥.𝑓 𝑦𝑦 − ( 𝑓 𝑥𝑦 )2
5	Determine max/min/saddle based on D and 𝑓 𝑥𝑥 ​

Keywords

optimization, multivariable calculus, critical points, maxima, minima, saddle point, gradient, partial derivatives, second derivative test, discriminant, local minimum, local maximum, unconstrained optimization, sympy, Python math, calculus with Python, function analysis, 3D plotting, mathematical optimization, surface plot, nerd cafe