Question

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

Short answer

The maximum value of the function \(f(x, y) = xy\) under the constraint \(2x + y = 4\) is 2 and it occurs at the point \(x = 1, y = 2\).

Step-by-step solution

Defining the Lagrange Function

Given \(f(x, y) = xy\) and the constraint \(g(x, y) = 2x + y - 4 = 0\), we can form the Lagrange function \(L(x, y, \lambda) = xy - \lambda(2x+y-4)\). The subtractions of the lambda term establishes the constraint.

Differentiating the Lagrange Function

Now we take the partial derivatives of the Lagrange function \(L(x, y, \lambda)\) with respect to \(x\), \(y\), and \(\lambda\) and set them to zero to find extremum which therefore forms a system of equations: \[\begin{cases} \frac{\partial L}{\partial x} = y - 2\lambda = 0 \\ \frac{\partial L}{\partial y} = x - \lambda = 0 \\ \frac{\partial L}{\partial \lambda} = 2x + y - 4 = 0. \\ \end{cases}\] This is because at the points where the function \(f(x,y)\) has a maximum or minimum under the constraints, the change in \(L(x, y, \lambda)\) should be zero.

Solving the Equations

Solving the equations from step 2: \[\begin{cases} y = 2\lambda \\ x = \lambda \\ 2x + y = 4 \\ \end{cases}\] We substitute \(x\) and \(y\) into the third equation, then we find that \(\lambda = 1\). Substituting \(\lambda = 1\) into the first two equations, we get \(x = 1\) and \(y = 2\).

Finding the Maximum Value

Finally, substitute \(x = 1\) and \(y = 2\) back into the original function \(f(x, y) = xy\), we get \(f(1,2) = 1*2 = 2\).

Key concepts

Extrema of Functions

In the study of calculus and mathematical optimization, finding the extrema (either maximum or minimum values) of a function is a fundamental problem. An extremum of a function can be either a local maximum or minimum, where the function's value is greater or smaller than all other values in the immediate vicinity, or it can be a global extremum, where the function's value is the highest or the lowest over its entire domain. To locate these valuable points, one usually calculates the derivative of the function and looks for where this derivative is zero—that is, points where the function doesn't increase or decrease, indicating a potential extremum.

In our example, we seek to maximize the function \(f(x, y) = xy\) while maintaining the constraint \(2x + y = 4\). The extrema of this function considering the constraint, are what the use of the Lagrange multipliers method is aimed at finding. By finding the points where the function has its extreme values, we will have achieved the goal of the exercise.

Partial Derivatives

Partial derivatives come into play when dealing with functions of several variables—in our case, the function \(f(x, y) = xy\). A partial derivative represents the rate at which the function changes as one of the variables changes, while the others are held constant. Just as regular derivatives are crucial for optimization in single-variable calculus, partial derivatives are essential in multivariable calculus.

The method of Lagrange multipliers requires us to take partial derivatives of a specially constructed function, called the Lagrange function, with respect to each variable. For the given optimization problem, we define the Lagrange function \(L(x, y, \lambda) = xy - \lambda(2x+y-4)\), including the constraint scaled by a new variable \(\lambda\). Taking the partial derivatives of this Lagrange function with respect to \(x\), \(y\), and \(\lambda\) and setting them to zero allows us to find where the function's rate of change respects the constraint and is stationary—indicative of an extremum.

Optimization with Constraints

Optimization problems often come with constraints—conditions that the solution must satisfy. The Lagrange multipliers method is a strategy used in calculus for finding the local maxima and minima of a function subject to equality constraints. The beauty of this technique is that it converts an optimization problem with restrictions into a problem without constraints. By introducing auxiliary variables known as Lagrange multipliers (denoted as \(\lambda\)), we adjust the optimization process to take the constraints into account.

In our exercise, we are trying to maximize the function \(f(x, y) = xy\), while the constraint is \(2x + y = 4\). Using Lagrange multipliers, we incorporate this constraint into the function we wish to optimize, creating an extended function—the Lagrange function—where the constraint is weighted by the multiplier \(\lambda\). Solving the system of equations yielded by taking the partial derivatives of the Lagrange function gives us the values that maximize (or minimize) the original function within the constraint's boundaries. This approach transforms a potentially complex optimization situation into a more manageable form that can be tackled with the standard tools of calculus.