Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \text { Minimize } f(x, y)=x^{2}+y^{2} \quad-2 x-4 y+5=0 $$

Short answer

The minimum of the function \( f(x, y) = x^{2}+y^{2} \) subject to the constraint \( -2x -4y + 5 = 0 \), is found at the point (0.5, 1).

Step-by-step solution

Set up the Lagrange function

Firstly, form a new function \( L(x, y, \lambda) = f(x,y) - \lambda g(x,y) \), where \( f(x, y) = x^{2}+y^{2} \) is the original function to be minimized and \( g(x, y) = -2x -4y +5 \) is the constraint. This results in \( L(x, y, \lambda) = x^{2}+y^{2} + \lambda( 2x +4y - 5) \).

Compute the partial derivatives and set them equal to zero

Compute the partial derivatives of \( L \) with respect to each variable (\(x\), \(y\), \(\lambda\)) and set them equal to zero. The equations are as follows: \(\frac{\partial L}{\partial x} = 2x + 2\lambda = 0\), \(\frac{\partial L}{\partial y} = 2y + 4\lambda = 0\), \(\frac{\partial L}{\partial \lambda} = 2x + 4y - 5 = 0\).

Solve the system of equations

Solve the system of equations obtained in the previous step. From the first two equations, we get \(x = -\lambda \) and \(y = -2\lambda \). Substituting these into the third equation gives us:\(2(-\lambda) + 4(-2\lambda) - 5 = 0\), which simplifies to \( \lambda = -\frac{5}{10} = -0.5\).Substitute \(\lambda = -0.5\) back into \(x = -\lambda\) and \(y = -2\lambda\), to find \(x = 0.5\) and \(y = 1\).

Key concepts

Understanding Extremum Problems

Extremum problems involve finding the minimum or maximum values of a function subject to certain conditions or constraints. In this context, the extremum we're interested in is the minimum value of the function. When dealing with a function of several variables, the extremum is where the function increases or decreases most rapidly, and this often occurs under some constraint, such as a specific curve or line. In this exercise, we are tasked with minimizing the function \(f(x, y) = x^2 + y^2\), subject to the linear constraint \(-2x - 4y + 5 = 0\). To effectively tackle extremum problems like this, Lagrange multipliers provide a powerful strategy. This method helps us locate points of extremum when the normal calculus approach of setting derivatives to zero is complicated by the presence of a constraint.

The Role of Partial Derivatives

Partial derivatives help us understand how a function responds to changes in one variable while holding others constant. They are essential tools in multivariable calculus and play a crucial role in methods like Lagrange multipliers.For a function \(L(x, y, \lambda)\), where \(L\) is called the Lagrange function, partial derivatives with respect to each variable \(x\), \(y\), and \(\lambda\) are computed. This allows us to assess the sensitivity of our function to changes in each input.

• To find extrema, set each partial derivative to zero. This signifies that the function's rate of change is balanced out across its variables, indicating a potential extremum point.
• In the problem, we find the partial derivatives by computing \(\frac{\partial L}{\partial x}\), \(\frac{\partial L}{\partial y}\), and \(\frac{\partial L}{\partial \lambda}\), which yield a system of equations guiding us toward the extremum point.

Exploring Constraint Optimization

Constraint optimization is a method used to find the best solution to a problem subject to limitations. When a problem involves constraints, such as the equation \(-2x - 4y + 5 = 0\) in this exercise, not just any function value on the entire plane can be considered.Instead, the solution must lie on the curve defined by the constraint.

• The Lagrange multiplier technique helps seamlessly incorporate the constraint into the problem-solving process by converting it into a system of equations derived from the function and the constraint.
• By solving this system, we ensure that our solution meets both the goal of extremizing the function and adhering to the constraint.

This methodology is essential in many fields, including economics, engineering, and operations research, whenever an optimal solution must be found within specific boundaries.