Problem 36 - Calculus for Biology and Medicine

Question

Use Lagrange multipliers to find the maxima and minima of the functions under the given constraints. $$ f(x, y)=2 x-y ; x^{2}+y^{2}=5 $$

Step-by-Step Solution

Answer

Maximum: 5 at (2, -1); Minimum: -5 at (-2, 1).

Step 1: Define the constraint with Lagrange Function

We begin by setting up the Lagrange function, which combines the given function and the constraint using a Lagrange multiplier, \( \lambda \). The Lagrange function is given by: \[ \mathcal{L}(x, y, \lambda) = 2x - y + \lambda (5 - x^2 - y^2) \].

Step 2: Compute the Partial Derivatives

Next, find the partial derivatives of \( \mathcal{L} \) with respect to \( x \), \( y \), and \( \lambda \). Set each derivative to zero to find critical points.- \( \frac{\partial \mathcal{L}}{\partial x} = 2 - 2\lambda x = 0 \)- \( \frac{\partial \mathcal{L}}{\partial y} = -1 - 2\lambda y = 0 \)- \( \frac{\partial \mathcal{L}}{\partial \lambda} = 5 - x^2 - y^2 = 0 \)

Step 3: Solve the System of Equations

Solve the system of equations obtained from the partial derivatives:1. \( 2 - 2\lambda x = 0 \Rightarrow \lambda x = 1 \)2. \(-1 - 2\lambda y = 0 \Rightarrow \lambda y = -\frac{1}{2} \)3. \( x^2 + y^2 = 5 \)From the first two equations, we have \( x = \frac{1}{\lambda} \) and \( y = -\frac{1}{2\lambda} \).

Step 4: Substitute and Solve for \( x \) and \( y \)

Substitute \( x \) and \( y \) from Step 3 into the constraint \( x^2 + y^2 = 5 \):\[ \left(\frac{1}{\lambda}\right)^2 + \left(-\frac{1}{2\lambda}\right)^2 = 5 \]\[ \frac{1}{\lambda^2} + \frac{1}{4\lambda^2} = 5 \]\[ \frac{5}{4\lambda^2} = 5 \]\[ \lambda^2 = \frac{1}{4} \Rightarrow \lambda = \pm \frac{1}{2} \].

Step 5: Find Critical Points

Using \( \lambda = \frac{1}{2} \), we have \( x = 2 \) and \( y = -1 \).For \( \lambda = -\frac{1}{2} \), we have \( x = -2 \) and \( y = 1 \).Therefore, the critical points are \( (2, -1) \) and \( (-2, 1) \).

Step 6: Evaluate the Function at Critical Points

Evaluate \( f(x, y) = 2x - y \) at the critical points:- At \((2, -1)\), \( f(2, -1) = 2(2) - (-1) = 5 \).- At \((-2, 1)\), \( f(-2, 1) = 2(-2) - 1 = -5 \).

Step 7: Determine Maximum and Minimum

The maximum value of \( f(x, y) \) under the constraint is 5, occurring at \((2, -1)\). The minimum value is -5, occurring at \((-2, 1)\).

Key Concepts

Maxima and Minima

Finding the maxima and minima of a function involves identifying the highest and lowest values that the function can attain. These values are critical because they can provide insights into the behavior of the function under certain conditions.
In mathematical terms, maxima refer to the topmost points in the range of the function, where the function reaches its greatest value locally. Conversely, minima represent the lowest points in the function's range, where it attains its smallest value.
For functions involving multiple variables, such as in this exercise, finding maxima and minima requires considering not just the function itself, but also any constraints imposed on it. These constraints, such as the circle equation given in the problem, limit the scope within which we search for these extreme values.

• Identify all potential critical points by setting the derivative or partial derivatives to zero.
• Apply constraints where necessary to determine feasible solutions.
• Evaluate the function at these critical points to find the maximum and minimum values.

In the given problem, using Lagrange multipliers helps efficiently find these values by incorporating the constraint into the solution process.

Constraint Optimization

Constraint optimization is a critical concept when dealing with real-world problems where limitations exist. It refers to the process of optimizing a function under a set of restrictions or constraints.
In this exercise, we are given a function, which we want to maximize or minimize, subjected to a constraint. In mathematical terms, our task is to optimize the function \( f(x, y) = 2x - y \) under the constraint \( x^{2}+y^{2}=5 \).

• This problem setup is a classic example of using Lagrange multipliers, a method specifically designed for solving problems of constrained optimization.
• The constraints act as boundaries within which the solution must lie, shaping the optimization path we take.
• This is particularly useful in real-world problems like economics, where resources might be limited, or in engineering where physical limits may exist.

By understanding constraint optimization, we can tackle complex scenarios where there might be limitations affecting outcomes, ensuring that we find solutions that not only are optimal but also practical and feasible.

Multivariable Calculus

Multivariable calculus extends the basic concepts of calculus to functions of multiple variables. Instead of dealing with a single variable, it handles two or more, such as in the function \( f(x, y) = 2x - y \).
The main tools in multivariable calculus include partial derivatives, which allow us to explore how changes in one variable affect the function, treating other variables as constants.

• Partial derivatives are crucial when using the Lagrange function, which involves multiple variables and a constraint.
• In the Lagrange method, we create a function (here, \( \mathcal{L}(x, y, \lambda) \)) that incorporates an extra variable, \( \lambda \), to account for constraints.
• The solution involves setting partial derivatives with respect to each variable, including \( \lambda \), to zero. This yields a set of equations to solve, uncovering potential maxima or minima.

Understanding multivariable calculus opens up many avenues in mathematical analysis, helping in exploring complex systems impacted by multiple factors, be they mathematical, scientific, or real-world applications.