Maximize F(x): Finding The Optimal Point X*

October 14, 2025 by TextBrain Team 44 views

Hey guys! Let's dive into a fun optimization problem where we need to find the maximum point of a function. This type of problem pops up in all sorts of fields, from engineering to economics, so understanding how to tackle it is a super valuable skill. We'll break down the problem step-by-step, making it easy to follow even if you're not a math whiz. So, buckle up and let's get started!

Problem Setup: Our Optimization Challenge

Our mission, should we choose to accept it (and we do!), is to find the values of three variables – x₁, x₂, and x₃ – that maximize the following function:

f(x) = -3/2 * x₁² - 2 * x₂² - 3/2 * x₃² - x₁x₃ - 2x₂x₃ + 3x₁ + x₃

Think of this function as a landscape with hills and valleys. Our goal is to find the very highest peak (the maximum point) in this landscape. The variables x₁, x₂, and x₃ represent our coordinates in this landscape. To make things a bit more compact, we can represent these variables as a vector:

x = [x₁ x₂ x₃]ᵀ

The superscript ᵀ indicates the transpose of the vector, which just means we're writing it as a column instead of a row. This notation will help us keep things organized as we move through the solution.

Why is This Important?

You might be wondering, "Why should I care about maximizing this function?" Well, optimization problems like this are everywhere! Imagine you're designing a bridge and need to minimize the amount of material used while ensuring it can still handle the load. Or maybe you're an economist trying to figure out the optimal way to allocate resources. Or perhaps you're a machine learning engineer trying to train a model to be as accurate as possible. All of these scenarios involve finding the maximum or minimum of some function.

Cracking the Code: Finding the Maximum Point

So, how do we actually find the maximum point, x*? We'll use a classic technique from calculus: finding the critical points of the function. A critical point is a point where the gradient of the function is zero or undefined. In simpler terms, it's where the slope of the landscape is flat in all directions. These critical points are potential maxima, minima, or saddle points.

Step 1: Calculate the Gradient

The gradient of a function tells us the direction of the steepest ascent. It's a vector containing the partial derivatives of the function with respect to each variable. In our case, we need to calculate the partial derivatives of f(x) with respect to x₁, x₂, and x₃.

Let's break it down:

Partial derivative with respect to x₁ (∂f/∂x₁):
```
∂f/∂x₁ = -3x₁ - x₃ + 3
```
This tells us how the function changes as we change x₁, keeping x₂ and x₃ constant.
Partial derivative with respect to x₂ (∂f/∂x₂):
```
∂f/∂x₂ = -4x₂ - 2x₃
```
This tells us how the function changes as we change x₂, keeping x₁ and x₃ constant.
Partial derivative with respect to x₃ (∂f/∂x₃):
```
∂f/∂x₃ = -3x₃ - x₁ - 2x₂ + 1
```
This tells us how the function changes as we change x₃, keeping x₁ and x₂ constant.

Now we can write the gradient as a vector:

∇f(x) = [∂f/∂x₁ ∂f/∂x₂ ∂f/∂x₃]ᵀ = [-3x₁ - x₃ + 3  -4x₂ - 2x₃  -3x₃ - x₁ - 2x₂ + 1]ᵀ

Step 2: Set the Gradient to Zero

To find the critical points, we need to find the values of x₁, x₂, and x₃ that make the gradient equal to the zero vector:

∇f(x) = [0 0 0]ᵀ

This gives us a system of three equations:

-3x₁ - x₃ + 3 = 0
-4x₂ - 2x₃ = 0
-3x₃ - x₁ - 2x₂ + 1 = 0

Step 3: Solve the System of Equations

Now we have a system of three linear equations with three unknowns. There are several ways to solve this, such as substitution, elimination, or using matrices. Let's use substitution to make it easier to understand. From equation (2), we can express x₂ in terms of x₃:

x₂ = -1/2 * x₃

Now, let’s substitute this expression for x₂ into equation (3):

-3x₃ - x₁ - 2(-1/2 * x₃) + 1 = 0
-3x₃ - x₁ + x₃ + 1 = 0
-2x₃ - x₁ + 1 = 0

We can rewrite this as:

x₁ = 1 - 2x₃

Now we have x₁ and x₂ expressed in terms of x₃. Let’s substitute x₁ into equation (1):

-3(1 - 2x₃) - x₃ + 3 = 0
-3 + 6x₃ - x₃ + 3 = 0
5x₃ = 0

So, x₃ = 0. Now we can find x₁ and x₂:

x₁ = 1 - 2(0) = 1
x₂ = -1/2 * (0) = 0

Therefore, the critical point is:

x* = [1 0 0]ᵀ

Step 4: Check for Maximum

We've found a critical point, but how do we know if it's a maximum, minimum, or saddle point? We can use the second derivative test. This involves calculating the Hessian matrix, which is a matrix of second partial derivatives.

The Hessian matrix for our function is:

H(x) = | ∂²f/∂x₁²   ∂²f/∂x₁∂x₂   ∂²f/∂x₁∂x₃ |
       | ∂²f/∂x₂∂x₁   ∂²f/∂x₂²   ∂²f/∂x₂∂x₃ |
       | ∂²f/∂x₃∂x₁   ∂²f/∂x₃∂x₂   ∂²f/∂x₃² |

Let's calculate the second partial derivatives:

∂²f/∂x₁² = -3
∂²f/∂x₂² = -4
∂²f/∂x₃² = -3
∂²f/∂x₁∂x₂ = ∂²f/∂x₂∂x₁ = 0
∂²f/∂x₁∂x₃ = ∂²f/∂x₃∂x₁ = -1
∂²f/∂x₂∂x₃ = ∂²f/∂x₃∂x₂ = -2

So the Hessian matrix is:

H(x) = | -3   0   -1 |
       |  0  -4   -2 |
       | -1  -2   -3 |

To determine if the critical point is a maximum, we need to check the eigenvalues of the Hessian matrix. If all eigenvalues are negative, then the critical point is a maximum. You can calculate the eigenvalues using various methods, but for this matrix, the eigenvalues are all negative. This indicates that we have indeed found a maximum point.

The Grand Finale: The Maximum Point

After all that math, we've arrived at the solution! The maximum point of the function f(x) is:

x* = [1 0 0]ᵀ

This means that the function f(x) reaches its maximum value when x₁ = 1, x₂ = 0, and x₃ = 0.

Plugging it In

If we substitute these values back into the original function, we can find the maximum value of f(x):

f(x*) = -3/2 * (1)² - 2 * (0)² - 3/2 * (0)² - (1)(0) - 2(0)(0) + 3(1) + (0)
f(x*) = -3/2 + 3
f(x*) = 3/2

So the maximum value of the function is 3/2.

Key Takeaways: What We Learned

We tackled an optimization problem by finding the critical points of a function.
We calculated the gradient of the function and set it to zero to find these critical points.
We solved a system of linear equations to find the values of the variables at the critical point.
We used the second derivative test (Hessian matrix) to confirm that we found a maximum point.

Wrapping Up: You're an Optimization Pro!

Congratulations! You've successfully navigated a tricky optimization problem. You've learned how to find the maximum of a function using calculus techniques. Keep practicing, and you'll become an optimization pro in no time! Remember, these skills are valuable in many different fields, so keep exploring and applying what you've learned. You got this!