Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the minimum of \(f(x, y, z)=4 x-2 y+3 z\) subject to the constraint \(2 x^{2}+y^{2}-3 z=0\)

Step-by-Step Solution

Answer

The minimum value is -3.

Step 1: Understand the Problem

We need to find the critical points of the function \( f(x, y, z) = 4x - 2y + 3z \) subject to the constraint \( g(x, y, z) = 2x^2 + y^2 - 3z = 0 \). We will use the method of Lagrange multipliers.

Step 2: Set Up Lagrange Function

Formulate the Lagrange function \( \mathcal{L}(x, y, z, \lambda) = 4x - 2y + 3z + \lambda (2x^2 + y^2 - 3z) \), where \( \lambda \) is the Lagrange multiplier.

Step 3: Calculate Partial Derivatives

Take the partial derivatives of \( \mathcal{L} \) with respect to \( x, y, z, \) and \( \lambda \):- \( \frac{\partial \mathcal{L}}{\partial x} = 4 + 4\lambda x = 0 \)- \( \frac{\partial \mathcal{L}}{\partial y} = -2 + 2\lambda y = 0 \)- \( \frac{\partial \mathcal{L}}{\partial z} = 3 - 3\lambda = 0 \)- \( \frac{\partial \mathcal{L}}{\partial \lambda} = 2x^2 + y^2 - 3z = 0 \)

Step 4: Solve the Equations

Solve the system of equations derived from the partial derivatives:1. \( 4 + 4\lambda x = 0 \) implies \( \lambda x = -1 \).2. \( -2 + 2\lambda y = 0 \) implies \( \lambda y = 1 \).3. \( 3 - 3\lambda = 0 \) gives \( \lambda = 1 \).4. Substitute \( \lambda = 1 \) into \( \lambda x = -1 \) and \( \lambda y = 1 \) to find \( x = -1 \) and \( y = 1 \).5. Use the constraint \( 2x^2 + y^2 - 3z = 0 \) with \( x = -1 \) and \( y = 1 \) to find \( z = 1 \).

Step 5: Find the Minimum Value

Substitute \( x = -1 \), \( y = 1 \), and \( z = 1 \) back into the original function \( f(x, y, z) = 4x - 2y + 3z \) to find the minimum: \( f(-1, 1, 1) = 4(-1) - 2(1) + 3(1) = -4 - 2 + 3 = -3 \).

Key Concepts

calculus optimization

In calculus, optimization refers to the process of finding the maximum or minimum values of a function. Optimization plays a crucial role in various fields, such as economics, engineering, and the physical sciences, where finding optimal solutions to problems is necessary. When dealing with functions of a single variable, basic derivatives help determine the local maxima and minima through critical points obtained by setting the derivative equal to zero.
However, when dealing with functions of multiple variables, the problem becomes more complex, requiring more advanced techniques such as gradient vectors and Hessian matrices. The critical points in multivariable functions are also found where the partial derivatives equal zero.
In situations where constraints are present, methods like Lagrange multipliers are employed. This transforms a constrained problem into an unconstrained one by introducing an auxiliary function, allowing the constraint to be taken into account in optimization.

constrained optimization

Constrained optimization is an extension of the optimization problem, where the solution to be found must also satisfy some constraints. The constraints are usually expressed as equations or inequalities that limit the possible solutions of the problem. This is common in real-world scenarios where, for instance, resources are limited, or other restrictions are imposed.
The Lagrange multiplier method is a popular technique for constrained optimization, effectively finding local maxima and minima of a function subject to equality constraints. This method involves creating a Lagrangian, which incorporates the constraint into the objective function with the help of a new variable called the Lagrange multiplier.

• It requires setting up a Lagrangian of the form: \( \mathcal{L}(x, y, z, \lambda) = f(x, y, z) + \lambda g(x, y, z) \) where \( f(x, y, z) \) is the function to be optimized and \( g(x, y, z) \) represents the constraint.
• Then, you take partial derivatives with respect to each variable, including the Lagrange multiplier \( \lambda \).
• Solving this system of equations yields the critical points that satisfy both the objective function and the constraint.

This method is powerful as it allows simultaneous optimization of the function while respecting the constraints provided.

multivariable calculus

Multivariable calculus extends the principles of calculus to functions with multiple variables. It consists of studying behaviors, such as rates of change, for multi-dimensional settings. Key components include:

• Partial derivatives, which measure the rate of change of a function with respect to one of its variables, keeping the others constant.
• The gradient, a vector of all the partial derivatives, which points in the direction of the steepest ascent of a function. This is vital in optimization to find the direction where the function increases or decreases most rapidly.
• Double and triple integrals, used to calculate volumes under surfaces, similar to how single-variable integrals calculate areas under curves.

In optimization, multivariable calculus is essential because it provides the necessary tools for analyzing functions with more than one variable. The process typically involves finding and studying the critical points of the function, which are places where the gradient is zero or undefined. As seen in this exercise, these methods are especially important when applying techniques like Lagrange multipliers for constrained optimization, where we must consider both the objective function and additional constraints.