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 Lagrange multiplier method.

Short answer

Answer: The maximum value of the function is \(\frac{1}{4}\).

Step-by-step solution

Solve for one variable in terms of the other using the constraint equation

Since the constraint is \(x + y = 1\), you can solve for \(y\) in terms of \(x\) (or vice versa): \(y = 1-x\).

Substitute the expression for \(y\) into the function \(f(x, y)\)

Now substitute the expression for \(y\) from the constraint equation into the function \(f(x, y) = xy\). This gives us the new function \(f(x) = x(1-x)\).

Find the derivative of the new function

To find the critical points, we need to take the derivative of the function \(f(x)\): \(f'(x) = (1-x)-x\).

Set the derivative to zero and solve for x

Now set \(f'(x)\) to zero and solve for \(x\): \((1-x)-x = 0 \Rightarrow x = \frac{1}{2}\).

Substitute the x-value back into the constraint equation to find the corresponding y-value

Using the value of \(x = \frac{1}{2}\) we found and the constraint equation \(y = 1-x\), we can find the corresponding \(y\) value: \(y = 1-\frac{1}{2} = \frac{1}{2}\).

Find the maximum value of the function

Plug the values of \(x\) and \(y\) we found into the original function \(f(x, y) = xy\) to find the maximum value: \(f(\frac{1}{2}, \frac{1}{2}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\). Method 2: Lagrange Multiplier Method

Set up the Lagrangian function

The Lagrangian function is defined as \(L(x, y, \lambda) = f(x, y) - \lambda(g(x, y) - c)\), where \(\lambda\) is the Lagrange multiplier. In this case, \(f(x, y) = xy\), \(g(x, y) = x + y\), and \(c = 1\). So \(L(x, y, \lambda) = xy - \lambda(x + y - 1)\).

Find the partial derivatives of the Lagrangian function

Calculate the partial derivatives of the Lagrangian function with respect to \(x\), \(y\), and \(\lambda\): \(\frac{\partial L}{\partial x} = y - \lambda\) \(\frac{\partial L}{\partial y} = x - \lambda\) \(\frac{\partial L}{\partial \lambda} = x + y - 1\)

Set the partial derivatives equal to zero and solve the equations

Set the partial derivatives equal to zero: \(\left\{\begin{array}{l} y - \lambda=0\\ x - \lambda=0 \\ x + y - 1=0 \end{array}\right.\) Solve this system of equations to find the values of \(x\), \(y\), and \(\lambda\). Since \(y-\lambda = 0\) and \(x-\lambda = 0\), we know both \(x\) and \(y\) are equal to \(\lambda\). Substitute \(x\) and \(y\) for \(\lambda\) in the third equation to get: \(x+x-1=0 \Rightarrow x = \frac{1}{2}\). Since \(x=y=\lambda\), we also have \(y = \frac{1}{2}\).

Find the maximum value of the function

Plug the values of \(x\) and \(y\) we found into the original function \(f(x, y) = xy\) to find the maximum value: \(f(\frac{1}{2}, \frac{1}{2}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\). In both methods, the maximum value of \(f(x, y) = xy\) with the constraint \(x + y = 1\) is \(\frac{1}{4}\).

Key concepts

Optimization with Constraints

When we attempt to optimize a function subject to a constraint, we're often dealing with a scenario where we want to find the maximum or minimum values of a function within a restricted domain. This typically involves a secondary equation that limits the possible values of the variables we're looking to optimize. A common technique to tackle such problems is the method of substitution, where we use the constraint equation to express one variable in terms of the others and subsequently reduce the number of variables in our function.
In the exercise provided, we're tasked with optimizing the function \( f(x, y) = xy \) under the constraint that the sum of \( x \) and \( y \) equals 1. The constraint \( x + y = 1 \) is utilized to express \( y \) in terms of \( x \), leading to a single-variable function which can then be optimized using standard techniques. By taking this approach, we translate a multivariable problem into a simpler one-dimensional problem, which is sometimes more intuitive to solve.

Partial Derivatives

Partial derivatives are fundamental in the study of multivariable calculus, as they measure how a function changes as one of its variables changes, holding all other variables constant. They play a significant role especially when dealing with functions of multiple variables, such as in the optimization problem where we need to understand the behavior of the function with respect to each variable independently.
To find the maximum value of our function \( f(x, y) = xy \), when applying the Lagrange multiplier method, we calculate the partial derivatives of the Lagrange function \( L(x, y, \lambda) = xy - \lambda(x + y - 1) \). These derivatives represent the rate of change of \( L \) with respect to \( x \), \( y \), and \( \lambda \) – where \( \lambda \) is introduced as a new variable, the Lagrange multiplier. Setting these partial derivatives equal to zero gives us a system of equations whose solutions are critical for finding the maximum value of our original function subject to the constraint.

Critical Points

In the context of calculus, critical points are where the derivative (or gradient) of a function is zero or undefined. These points are essential because they are potential locations where a function may attain maximum or minimum values.
By using the results from taking the partial derivatives in the Lagrange multiplier method and setting them to zero, we are essentially finding the critical points of the extended function that includes the constraint. In our exercise, after finding that both \( x \) and \( y \) equal \( \lambda \), we determine that \( x = y = \frac{1}{2} \) by solving the equations derived from the partial derivatives. By finding these critical points, we are able to locate the specific values for \( x \) and \( y \) that will maximize the function \( f(x, y) = xy \) under the given constraint. The process reveals not only the coordinates of the maximum point but also reaffirms the powerful utility of critical points in solving optimization problems.