Problem 38 - Calculus for Biology and Medicine

Question

Use Lagrange multipliers to find the maxima and minima of the functions under the given constraints. $$f(x, y)=x y ; x+y=4$$

Step-by-Step Solution

Answer

The maximum value of the function is 4 at the point (2, 2). There is no minimum due to symmetry and linearity of the constraint.

Step 1: Understand the Problem

We are given a function to maximize or minimize, \(f(x, y) = x y\), subject to a constraint \(x + y = 4\). We will use the method of Lagrange multipliers, which introduces a new variable, \(\lambda\), to solve this problem.

Step 2: Formulate the Lagrangian

The Lagrangian \(L(x, y, \lambda)\) is given by combining the function and the constraint with the Lagrange multiplier: \[L(x, y, \lambda) = x y + \lambda(4 - x - y)\] Here, \(\lambda\) is our Lagrange multiplier.

Step 3: Find Partial Derivatives

Compute the partial derivatives of the Lagrangian with respect to \(x\), \(y\), and \(\lambda\):1. \( \frac{\partial L}{\partial x} = y - \lambda \)2. \( \frac{\partial L}{\partial y} = x - \lambda \)3. \( \frac{\partial L}{\partial \lambda} = 4 - x - y\)

Step 4: Set Partial Derivatives to Zero

Set each of the partial derivatives equal to zero to get the system of equations:1. \( y - \lambda = 0 \)2. \( x - \lambda = 0 \)3. \( 4 - x - y = 0\)

Step 5: Solve the System of Equations

From equations 1 and 2, we have \( y = \lambda \) and \( x = \lambda \), so \( x = y \). Substitute \( x = y \) into equation 3:\[ 4 - x - x = 0 \Rightarrow 4 - 2x = 0 \Rightarrow x = 2 \]Since \( x = y \), \( y = 2 \). Thus, the point is \((x, y) = (2, 2)\).

Step 6: Evaluate Function at Critical Point

Substitute the critical point \((2, 2)\) into the original function:\[ f(2, 2) = 2 \times 2 = 4 \]

Step 7: Determine if Maximum or Minimum

Since the function is quadratic and the constraint linear, this solution represents both the maximum and minimum possible values for \(f(x, y)\) under the constraint \(x + y = 4\).

Key Concepts

Constrained Optimization

In mathematics, constrained optimization is about finding maximum or minimum values of functions under specific restrictions or conditions. These restrictions are known as constraints and could shape the feasible region of your search. The method we often use to tackle such problems is the method of Lagrange multipliers.

• It allows us to convert a constrained problem into an unconstrained one by introducing a new variable, the Lagrange multiplier (\(\lambda\)).
• This method helps in finding the optimal points which otherwise could be tough to determine directly due to the constraints.
• In this exercise, our main focus is optimizing the function \(f(x, y) = xy\) subjected to a linear constraint \(x + y = 4\).

By introducing the Lagrange multiplier, the constraint is incorporated into the objective function, creating what is known as the Lagrangian. This effectively transforms the problem, allowing us to explore solutions using calculus. The critical aspect is understanding how these constraints can narrow down potential solutions.

Partial Derivatives

Partial derivatives are fundamental in the study of functions of multiple variables. They help us understand how a function changes as only one of its variables changes, while holding others constant.

• In this optimization method, partial derivatives are calculated for each variable, including the Lagrange multiplier (\(\lambda\)).
• For the function \(L(x, y, \lambda) = xy + \lambda(4 - x - y)\), we calculate the derivatives with respect to \(x\), \(y\), and \(\lambda\).
• The partial derivatives give us: \(\frac{\partial L}{\partial x} = y - \lambda\), \(\frac{\partial L}{\partial y} = x - \lambda\), and \(\frac{\partial L}{\partial \lambda} = 4 - x - y\).

By setting these derivatives to zero, as you'll find in typical optimization problems, we generate a system of equations. This system provides critical relationships between the variables and the Lagrange multiplier that will later help us locate optimal points.

Critical Points

In calculus, critical points are where the derivative is zero or undefined, indicating potential maxima, minima, or saddle points for a function. When dealing with constrained optimization problems, critical points—solutions to the system of equations derived from our partial derivatives—are crucial.

• These points reveal where the trade-off between our objective function and constraints reaches its peak efficiency, balancing maximization or minimization with restriction satisfaction.
• For \(f(x, y) = xy\) with constraint \(x + y = 4\), the derivative equations give \(y = \lambda\) and \(x = \lambda\), leading us to \(x = y\).
• Substituting back, we find that \((x, y) = (2, 2)\) is a critical point, where both our function value constraints are satisfied.

Evaluating the function at this critical point, we obtain \(f(2, 2) = 4\), representing the combined maximum and minimum possible under the given constraint. Critical points are thus essential in understanding constrained problems and assessing their potential optimal solutions.