91Ó°ÊÓ

Constrained optimization is a method used in mathematical optimization to find the best solution while satisfying certain restrictions. In real-world scenarios, many problems involve optimizing a function subject to constraints. These constraints can be equations or inequalities that the solution must satisfy.

In our exercise, the goal is to minimize the function \(3x^2 - 2xy + x - 3y + 1\) while ensuring the constraint \(x - 3y = 1\) is met. This is a classic example of a constrained optimization problem.

To solve this, we use the method of Lagrange multipliers, which is a strategy for finding the local maxima and minima of a function subject to equality constraints. By introducing a new variable called a Lagrange multiplier (\(\theta\)), we transform the constrained problem into a form that is simpler to handle. The transformed function is called the Lagrangian. In this case, the Lagrangian is:

\[ \text{L}(x, y, \theta) = 3x^2 - 2xy + x - 3y + 1 + \theta (x - 3y - 1) \]The next steps involve finding the partial derivatives and solving the resulting system of equations.

Partial derivatives are a fundamental concept in calculus, representing how a function changes as its variables change. For functions with multiple variables, partial derivatives are taken with respect to one variable at a time, treating other variables as constants.

In the context of our exercise, we compute the partial derivatives of the Lagrangian with respect to \(x\), \(y\), and \(\theta\). These partial derivatives are then set to zero to find the critical points where the function could have minima or maxima:

• \( \frac{\partial \text{L}}{\partial x} = 6x - 2y + 1 + \theta = 0 \)
• \( \frac{\partial \text{L}}{\partial y} = -2x - 3 + \theta(-3) = 0 \)
• \( \frac{\partial \text{L}}{\partial \theta} = x - 3y - 1 = 0 \)

These equations represent a system of simultaneous equations that capture the behavior of the Lagrangian at critical points. By solving them, we can find the values of \(x\), \(y\), and \(\theta\) that minimize or maximize the original function under the given constraint.

A system of equations is a collection of two or more equations with a same set of unknowns. Solving a system involves finding the values of the unknowns that satisfy all the equations simultaneously.

In our exercise, the partial derivatives give us the following system of equations:
\[ 6x - 2y + 1 + \theta = 0 \]\[ -2x - 3 -3\theta = 0 \]\[ x - 3y - 1 = 0 \]Solving this system involves several steps:

First, isolate one variable. From \(-2x -3 -3\theta = 0\), we find \(\theta = \frac{2x + 3}{3}\).

Next, substitute \(\theta\) back into the first equation to get:

\[ 6x - 2y + 1 = -\frac{2x + 3}{3} \]Clearing the fractions and solving, we get:

\[ 18x - 6y + 3 = -2x - 3 \]Adding across, we find:
\[ 20x - 6y = -6 \]\[ 10x - 3y = -3 \]
Finally, using the constraint \(x - 3y = 1\), we solve these two equations simultaneously. Using elimination method:
\[ x - 3y = 1 \]\[ 10x - 3y = -3 \]Subtracting the first equation from the second, we solve for \(x\):

\[ 9x = -4 \]\[ x = -\frac{4}{9} \]Substitute \(x\) back into the constraint to get \(y\):

\[ -\frac{4}{9} - 3y = 1 \]This simplifies to:
\[ y = -\frac{13}{27} \]Thus, the values \( x = -\frac{4}{9} \) and \( y = -\frac{13}{27} \) minimize the original function subject to the constraint.