Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \text { Minimize } f(x, y)=x^{2}-y^{2} \quad x-2 y+6=0 $$

Short answer

The minimum of the function \(f(x, y)=x^{2}-y^{2}\) subject to the constraint \(x-2y+6=0\) occurs at \(x=2\), \(y=2\) with a minimum value of 0.

Step-by-step solution

Set Up the Lagrange Function

The Lagrange function is \(\mathcal{L}(x, y, \lambda) = f(x,y) - \lambda g(x,y)\), where \(f(x,y)\) is the original function we want to minimize and \(g(x,y) = x-2y+6\) is the constraint. So, \(\mathcal{L}(x, y, \lambda) = x^{2}-y^{2} - \lambda (x - 2y + 6)\)

Compute the derivatives

Compute the partial derivatives of \(\mathcal{L}\) and set each equal to zero:\n\n\[\frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda = 0 \tag{1}\]\n\n\[\frac{\partial \mathcal{L}}{\partial y} = -2y + 2\lambda = 0 \tag{2}\]\n\n\[\frac{\partial \mathcal{L}}{\partial \lambda} = x - 2y + 6 = 0 \tag{3}\]

Solve the equations

Solve equations (1), (2) and (3) simultaneously to find the values of \(x\), \(y\), and \(\lambda\). Using equation (1), we get \(\lambda = 2x\). Substituting this into equation (2) gives \(y=x\). Substituting \(y=x\) into equation (3) then gives \(x = 2\) and \(\lambda = 4\). Thus, the minimum of the function subject to the constraint occurs at \(x = 2\), \(y = 2\).

Verify the result

Substitute the values of \(x\) and \(y\) into \(f(x,y)\) to verify that this is indeed the minimum value. We find \(f(2,2) = 2^2 - 2^2 = 0\), confirming that this is the minimum of the function subject to the constraint.

Key concepts

Extremum

In mathematics, an extremum is a point at which a function takes either a maximum or a minimum value. When we talk about finding an extremum, we're usually interested in identifying these critical points for a given function. Specifically, in the context of optimization problems, we look for minima or maxima subject to some constraints. For unconstrained functions, this means finding points where the derivative is zero or does not exist.
In constrained problems, like the one in this example, extremums are considered within the allowed region defined by constraints. The Lagrange multiplier technique can be used to solve these kinds of problems effectively.

Partial Derivatives

Partial derivatives are used when we need to compute derivatives of functions with more than one variable. They represent how the function changes as one variable changes while keeping all other variables constant.
In the context of the problem, we have the Lagrange function \[ \mathcal{L}(x, y, \lambda) = x^{2}-y^{2} - \lambda (x - 2y + 6) \] and we find its partial derivatives:

• \( \frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda \)
• \( \frac{\partial \mathcal{L}}{\partial y} = -2y + 2\lambda \)
• \( \frac{\partial \mathcal{L}}{\partial \lambda} = x - 2y + 6 \)

Setting these partial derivatives to zero helps us identify potential locations for the extrema while considering the constraint.

Simultaneous Equations

Simultaneous equations are a set of equations containing multiple variables that you solve together to find a common solution. In this exercise, three simultaneous equations arise from setting the partial derivatives of the Lagrange function to zero.
This leads to the following system of equations:

• \(2x - \lambda = 0\)
• \(-2y + 2\lambda = 0\)
• \(x - 2y + 6 = 0\)

Solving these equations together helps us find the values of \(x\), \(y\), and \(\lambda\) that satisfy all conditions, guiding us to the extremum of the problem. This method is essential for handling more complex constraints in optimization tasks.

Constrained Optimization

Constrained optimization is the process of optimizing a function given specific limitations or conditions. This involves finding the maximum or minimum of a function while ensuring that all constraints are satisfied. In mathematical terms, it means optimizing a primary function subject to equality or inequality constraints.
The Lagrange multipliers approach is a powerful method for addressing these constrained problems. It does so by introducing a new variable, the multiplier \( \lambda \), to account for the constraint.

• Formulate the Lagrange function that combines both the primary function and the constraint.
• Find partial derivatives and solve simultaneously to locate extremum points.

Understanding and applying constrained optimization is valuable in various fields such as economics, engineering, and operations research.