Question

Suppose that \(f(x, y)=x y .\) Find the maximum value for \(f\) if \(x\) and \(y\) are constrained to sum to \(1 .\) Solve this problem in two ways: by substitution and by using the Langrangian multiplier method.

Short answer

- The maximum value of the function is \(\frac{1}{4}\) found using both the substitution and Lagrangian multiplier methods.

Step-by-step solution

Substitution Method

Eliminate one variable using the constraint \(x + y = 1\). We can express \(y\) in terms of \(x\): \(y = 1 - x\). Now substitute this into the function \(f(x, y)\): \(f(x, 1 - x) = x(1 - x)\).

Find the critical points

To find the maximum of the function within the constraint, we need to find the critical points, meaning the points where the derivative of the function is zero or undefined. Find the derivative of the function \(f\) with respect to \(x\): \(f'(x) = \frac{d}{dx} (x(1 - x)) = 1 - 2x\).

Find the maximum value

Now, we need to find the maximum value of \(f(x)\). To do this, set the derivative equal to 0 and solve for x: \(1 - 2x = 0 \Rightarrow x = \frac{1}{2}\). Now, recall that we have the constraint \(x + y = 1\). Substituting \(x = \frac{1}{2}\), we obtain \(y = \frac{1}{2}\). Now, we can find the maximum value of the function: \(f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\).

Lagrangian Multiplier Method

Now, we will use the Lagrangian multiplier method to find the maximum value of the function \(f(x, y) = xy\) with the constraint \(x + y = 1\). First, we must define our Lagrangian function. The Lagrangian function, denoted by \(L\), is defined as follows: \(L(x, y, \lambda) = f(x, y) - \lambda(x + y - 1)\), where \(\lambda\) is the Lagrangian multiplier.

Computing the gradient

To find the critical points, we need to find the gradient of the Lagrangian function and set each component equal to zero: $\nabla L (x, y, \lambda) = \begin{cases} \frac{\partial L}{\partial x} = y - \lambda = 0\\ \frac{\partial L}{\partial y} = x - \lambda = 0 \\ \frac{\partial L}{\partial \lambda} = x + y - 1 = 0 \end{cases}$.

Solving the system of equations

Now, we need to solve the system of equations: \begin{cases} y - \lambda = 0\\ x - \lambda = 0 \\ x + y - 1 = 0 \end{cases}. Statement 1 implies \(\lambda = y\), and statement 2 implies \(\lambda = x\). Therefore, \(x = y\). From statement 3, we obtain \(2x - 1 = 0\) or \(x = \frac{1}{2}\). Thus, \(y = \frac{1}{2}\) as well.

Find the maximum value

With the solutions \(x = \frac{1}{2}\) and \(y = \frac{1}{2}\), we can find the maximum value of the function \(f(x, y)\): \(f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\). Both methods, substitution and Lagrangian multiplier, found the same maximum value for the function, \(\frac{1}{4}\).

Key concepts

Constrained Optimization

Constrained optimization refers to the process of finding an optimal solution for a function under given restrictions or limitations, called constraints. In our example, the function to optimize is denoted as \(f(x, y) = xy\), and the constraint is that the sum of the variables \(x\) and \(y\) must always equal 1, which can be written as \(x + y = 1\). The objective in constrained optimization is to either maximize or minimize the function while satisfying the constraint.

Approaches to solving constrained optimization problems include the Substitution Method and the Lagrangian Multiplier Method. The substitution method directly incorporates the constraint by expressing one variable in terms of another, which simplifies the problem. The Lagrangian multiplier method introduces an auxiliary variable known as the Lagrange multiplier, denoted \(\lambda\), to account for the constraint, transforming the problem into one of finding stationary points of a new function called the Lagrangian.

In practice, constrained optimization is widely applicable across fields like economics, engineering, and operations research, where resources and variables must satisfy certain constraints.

Critical Points

Critical points are values of the variables that can potentially be maximum, minimum, or saddle points of a function. To find critical points, we derive the function with respect to each variable and set those derivatives to zero. This identifies points where the function changes direction, which might be a peak (maximum value), a trough (minimum value), or a saddle.

For our function \(f(x, 1-x) = x(1-x)\), we took the derivative with respect to \(x\), resulting in \(f'(x) = 1 - 2x\). By setting this derivative equal to zero, we solve for \(x = \frac{1}{2}\). With the constraint \(x + y = 1\), this implies \(y = \frac{1}{2}\) as well. Thus, the critical point in the context of our problem is at \((x, y) = (\frac{1}{2}, \frac{1}{2})\).

Understanding the concept of critical points is essential because these are the candidates for any optimization process, particularly in constrained scenarios, where certain limitations bound the variables involved.

Substitution Method

The substitution method is one of the simplest techniques for handling constrained optimization problems. It involves expressing one of the constrained variables in terms of the other(s) to reduce the number of variables in the function. In the given problem, the constraint \(x + y = 1\) allows us to substitute \(y\) with \(1 - x\) in the function \(f(x, y) = xy\).

This substitution simplifies the function to a single variable, resulting in \(f(x, 1-x) = x(1-x)\). We then differentiate this function, \(f'(x) = 1 - 2x\), and set the derivative to zero to find critical points. Solving \(1 - 2x = 0\) yields \(x = \frac{1}{2}\), leading to \(y = \frac{1}{2}\) using the constraint \(x + y = 1\).

The substitution method helps in solving problems with equality constraints more easily by converting them into an unconstrained format. This makes it a handy tool in cases where rewriting the function with fewer variables makes the problem solvable through basic calculus techniques.