Question

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

Short answer

The maximum of the function \(f(x,y)=x^2 - y^2\), with the constraint \(2y - x^2 = 0\), using Lagrange multipliers is located at (2,-2) with the value 8.

Step-by-step solution

Setup the system of equations

By the Lagrange technique, we set up a system of equations from \(\nabla f = \lambda \nabla g\), where \(f\) is our function, \(g\) is our constraint, and \(\lambda\) is our lagrange multiplier. We find \(\nabla f = (2x, -2y)\) and \(\nabla g = (-2x, 2)\). Equating these gives us: \(2x = -\lambda 2x\) and \(-2y = \lambda 2\). We also need the original constraint \(2y - x^2 = 0\).

Solve the system

We solve the system by equating \(\lambda\) from the two equations. So \(-x = y\), substituting this into our constraint equation \(2(-x) - x^2 = 0\), we get \(2x = x^2\). Solving for \(x\) gives us \(x = 0 , 2\). Then from \(-x = y\), \(y = 0, -2\).

Evaluate solutions

Now we have 2 solutions, (0,0) and (2,-2). We need to determine the value of the function at these points to determine whether they are max, min or saddle points. From \(f(x, y) = x^2 - y^2\), \(f(0, 0) = 0\) and \(f(2, -2) = 8\).

Final answer

With \(f(0, 0) = 0\) and \(f(2, -2) = 8\), we identify (2,-2) as the maximizer of our function, in the positive domain of \(x\).

Key concepts

Extrema Optimization

In the world of mathematics, particularly when delving into calculus, extrema optimization is a vital concept. This topic addresses the practice of finding the maximum or minimum values (collectively known as extrema) of a function within a given set of constraints. When dealing with functions of two variables, like our example function f(x, y) = x² - y², finding its extremum is not as straightforward as it is for single-variable functions. We often have additional equations representing constraints that our solution must satisfy, which complicate the process.

Lagrange multipliers are a powerful tool designed specifically for this type of problem - they allow us to introduce a new variable (the multiplier) and transform the constraint into a form that can be much easier to handle. When we talk about finding the maximum value of our function with the constraint 2y - x² = 0, what we're looking for is the highest value that f(x, y) can reach, considering that x and y must lie on the curve defined by the constraint.

Multivariable Calculus

The study of multivariable calculus extends the concepts of single-variable calculus to functions of multiple variables. The beauty of this branch of calculus is its ability to tackle real-world problems where outcomes depend on various factors, each represented as a different variable. When you're optimizing functions like f(x, y), you're engaging in multivariable calculus. An important tool in this subject is the gradient vector, which points in the direction of the steepest ascent of a function. This vector is central to understanding how the function behaves locally and is a key component when employing the method of Lagrange multipliers.

The magic of the gradient vectors lies in their capacity to summarize the rate of change of a function in all of its input directions. With functions of two variables, gradients add complexity but also allow us to navigate towards the extremum while respecting constraints that occur naturally in many scenarios, such as physics, economics, and engineering problems.

System of Equations

System of equations are collections of two or more equations that share a set of unknowns, and they crop up constantly in mathematics, particularly when dealing with multivariable problems such as optimization. When we use Lagrange multipliers, we derive a system of equations that include both gradient equations and the original constraint.

The goal when solving a system of equations is to find the values of the variables that satisfy all the equations simultaneously. In the context of our Lagrange multipliers problem, after setting up the equations, our next step is to solve for our unknowns, including the Lagrange multiplier itself. One insightful strategy, often used, is substitution or elimination. In our example, after finding relationships between our variables and the multiplier, we can substitute one variable for another, as seen where we equated -x to y, eventually leading us to find the values of x and y that satisfy the constraint.

Gradient Vectors

The term gradient vectors frequently buzzes in the world of multivariable calculus and optimization. A gradient vector of a function is a vector whose components are the partial derivatives of that function with respect to each variable. In simpler terms, it tells us how much the function increases if we take a tiny step in either direction of any of the variables.

For a function f(x, y), its gradient is denoted as ∇f and calculated as the vector (∂f/∂x, ∂f/∂y). In optimization problems, especially when using Lagrange multipliers, we look at the gradient of the function and the gradient of the constraint. The solutions to our optimization problem lie where these gradients are scalar multiples of each other. That is the core principle behind Lagrange multipliers, and it's why in our example, we equate the gradient of f and the gradient of g with the multiplier λ factored in.